Chapter 5: Prime Time

Step 1  – Recall the rules of the “idli-vada” game
Choose two whole numbers. While counting 1, 2, 3, …

• say idli for every multiple of the first number;
• say vada for every multiple of the second number;
• say idli-vada for a number that is a multiple of both.

Here the first number is $$4$$ and the second number is yet to be chosen.

Step 2  – Interpret the given clue
“Nobody said just ‘vada’” means there was no counting number that was a multiple of the second number alone.

So, whenever a number was a multiple of the second number, it was also a multiple of $$4$$. In symbols:

if a number $$n$$ is a multiple of the second number, then $$n$$ is a multiple of $$4$$.

This can happen only when every multiple of the second number is contained in the set of multiples of $$4$$. Therefore

second number must be a multiple of $$4$$.

Step 3  – Examine the given options

Step 4  – Conclusion
Only $$8$$ is a multiple of $$4$$, so choosing $$8$$ as the second number guarantees that every “vada” will automatically be an “idli-vada”. Hence nobody will ever say just “vada”.

Step 1  Understand the rule of the game
Whenever a number is:

• a multiple of 3  →  players say Idli
• a multiple of 5  →  players say Vada
• a multiple of both 3 and 5  →  players say Idli-Vada

Step 2  Find a quick way to get common multiples of 3 and 5
The common multiples of two numbers are the multiples of their LCM (Least Common Multiple).
Because 3 and 5 have no common factor other than 1, their LCM is simply their product:
$$\text{LCM}(3,5)=3\times5=15$$ Therefore every multiple of 15 gives an Idli-Vada.

Step 3  List the first ten multiples of 15

Step 4  State the required number
The 10^th time Idli-Vada is spoken when the number $$150$$ is reached.

Step 1 | The rules of the game stay the same; only the upper limit changes.
In the ‘idli-vada’ game the players say
• idli for every multiple of $$3$$,
• vada for every multiple of $$5$$, and
• idli-vada for every number that is a multiple of both $$3$$ and $$5$$, that is, every multiple of $$\operatorname{lcm}(3,5)=15$$.

Earlier we played from $$1$$ to $$90$$; this time we play from $$1$$ to $$900$$.

Step 2 | Number of times “idli” is said (multiples of 3)
The multiples of $$3$$ between $$1$$ and $$900$$ are
$$3,\;6,\;9,\;\dots ,\;900.$$
How many such numbers are there? Divide $$900$$ by $$3$$:
\[\dfrac{900}{3}=300.\]
So “idli” is said $$300$$ times (this already includes the occasions when the children say “idli-vada”).

Step 3 | Number of times “vada” is said (multiples of 5)
The multiples of $$5$$ between $$1$$ and $$900$$ are
$$5,\;10,\;15,\;\dots ,\;900.$$
Their count is
\[\dfrac{900}{5}=180.\]
So “vada” is said $$180$$ times (again, “idli-vada”s are included in this count).

Step 4 | Number of times “idli-vada” is said (multiples of 15)
The multiples of $$15$$ between $$1$$ and $$900$$ are
$$15,\;30,\;45,\;\dots ,\;900.$$
Their count is
\[\dfrac{900}{15}=60.\]
So “idli-vada” is said $$60$$ times.

Step 5 | How the answers of question 2 change
The new range $$1$$–$$900$$ is exactly $$10$$ times the old range $$1$$–$$90$$. Because counting multiples of a fixed number depends only on the size of the range, every count is simply multiplied by $$10$$:

Conclusion
If the game is played from $$1$$ to $$900$$, then “idli” is said $$300$$ times, “vada” is said $$180$$ times and “idli-vada” is said $$60$$ times. Each of the earlier answers (for $$1$$–$$90$$) gets multiplied by $$10$$, but the method of finding them does not change.

Step 1 – Recall the rules of the “idli-vada” game
• When you clap out the numbers in order, you say
  • “idli” instead of every multiple of 3;
  • “vada” instead of every multiple of 5;
  • “idli-vada” instead of every number that is a multiple of both 3 and 5.

Step 2 – Relate the given Venn diagram to the game
• The left circle is the list of numbers that would be called “idli”.
• The right circle is the list of numbers that would be called “vada”.
• The overlapping part shows the numbers for which we say “idli-vada”.
Thus the figure is exactly the record of an “idli-vada” game played up to 30.

Step 3 – Play the game further, up to 60
(a) Multiples of 3 up to 60:
$$3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60$$
(b) Multiples of 5 up to 60:
$$5,10,15,20,25,30,35,40,45,50,55,60$$
(c) Common multiples (multiples of 15) up to 60:
$$15,30,45,60$$

Step 4 – Sort them for the Venn diagram
Only “idli” (multiples of 3, not of 5)
$$3,6,9,12,18,21,24,27,33,36,39,42,48,51,54,57$$
Only “vada” (multiples of 5, not of 3)
$$5,10,20,25,35,40,50,55$$
“Idli-vada” (common multiples)
$$15,30,45,60$$

Step 5 – Describe the new figure
Draw two equal, overlapping circles.
• Title of the left circle: “Multiples of 3 (idli)”. Write inside it:
  3, 6, 9, 12, 18, 21, 24, 27, 33, 36, 39, 42, 48, 51, 54, 57.
• Title of the right circle: “Multiples of 5 (vada)”. Write inside it:
  5, 10, 20, 25, 35, 40, 50, 55.
• In the overlapping strip between the two circles write:
  15, 30, 45, 60.
Label that strip “Common multiples of 3 and 5 (idli-vada)”.
This is the required diagram for an “idli-vada” game played up to 60.

Step 1 | Understand “jump size”
If we start at 0 on a number line and keep adding the same number each time, we make equal jumps. A jump size $$k$$ reaches a point like $$15$$ exactly when $$15$$ is a multiple of $$k$$, that is, when $$k$$ is a factor of $$15$$. To reach both $$15$$ and $$30$$ the same size $$k$$ must be a factor of both numbers.

Step 2 | List all factors of each number

• Factors of $$15$$: check divisors one-by-one.
    $$15\div1=15$$ (no remainder) → $$1$$ is a factor.
    $$15\div2=7.5$$ (remainder) → $$2$$ is not a factor.
    $$15\div3=5$$ (no remainder) → $$3$$ is a factor.
    $$15\div4$$ leaves a remainder → $$4$$ not a factor.
    $$15\div5=3$$ (no remainder) → $$5$$ is a factor.
    Continuing up to $$15$$ we also get $$15$$ itself.
  Hence the factor list of $$15$$ is
  $$1,\;3,\;5,\;15$$.
• Factors of $$30$$: repeat the test.
  $$30\div1=30\;(\checkmark),\;30\div2=15\;(\checkmark),\;30\div3=10\;(\checkmark),\;30\div4$$ has remainder, $$30\div5=6\;(\checkmark),\;30\div6=5\;(\checkmark)$$ and finally $$10,\;15,\;30$$ itself.
  So the factor list of $$30$$ is
  $$1,\;2,\;3,\;5,\;6,\;10,\;15,\;30$$.

Step 3 | Common factors
Compare the two lists and keep only the numbers that appear in both:

$$1,\;3,\;5,\;15$$

Step 4 | Interpretation
Each common factor is a possible positive jump size. For example, jumps of size $$5$$ land on $$0,5,10,15,20,25,30,\dots$$ so we indeed reach both $$15$$ and $$30$$.

Hence all possible jump sizes that reach both 15 and 30 are

\[1,\;3,\;5,\;15\]

Step 1 – Express a general multiple of 40
Any multiple of 40 can be written as
$$40\times n$$
where $$n$$ is a whole (natural) number.

Step 2 – Translate the condition “between 310 and 410”
We need those multiples that satisfy
$$310 < 40\times n < 410$$

Step 3 – Isolate $$n$$ by dividing every part by 40
$$\frac{310}{40} < n < \frac{410}{40}$$
$$7.75 < n < 10.25$$

Step 4 – List the integer values of $$n$$ that fit
Because $$n$$ must be a whole number, the only possibilities in the range $$7.75<n<10.25$$ are
$$n = 8,\;9,\;10$$

Step 5 – Write the required multiples

• For $$n=8$$: $$40\times8 = 320$$
• For $$n=9$$: $$40\times9 = 360$$
• For $$n=10$$: $$40\times10 = 400$$

The multiples of 40 that lie between 310 and 410 are therefore

\[320,\;360,\;400\]

Step 1 : Recall the definition

A number is called a perfect number if

$$\text{sum of all its factors}=2\times(\text{the number}).$$

So we must look for a number between $$1$$ and $$10$$ whose factor–sum equals twice the number itself.

Step 2 : List every number between $$1$$ and $$10$$ and write all its factors

Step 3 : Find the sum of the factors of each number

Step 4 : Compare each sum with $$2\times\text{(the number)}$$

• For $$1$$: $$2\times1=2\;(\neq1)$$
• For $$2$$: $$2\times2=4\;(\neq3)$$
• For $$3$$: $$2\times3=6\;(\neq4)$$
• For $$4$$: $$2\times4=8\;(\neq7)$$
• For $$5$$: $$2\times5=10\;(\neq6)$$
• For $$6$$: $$2\times6=12\;(=12)$$ ✔
• For $$7$$: $$2\times7=14\;(\neq8)$$
• For $$8$$: $$2\times8=16\;(\neq15)$$
• For $$9$$: $$2\times9=18\;(\neq13)$$
• For $$10$$: $$2\times10=20\;(\neq18)$$

The only number between $$1$$ and $$10$$ whose factor–sum equals twice the number is $$6$$.

Step 5 : Write the conclusion

Hence, the perfect number between $$1$$ and $$10$$ is $$6$$.

Step 1 — Recall the definition of a multiple.
A number is a multiple of $$25$$ if it can be written in the form $$25 \times n$$ where $$n$$ is a whole number (1, 2, 3, …).

Step 2 — Write down a few multiples of $$25$$.

$$25 \times 1 = 25$$,
$$25 \times 2 = 50$$,
$$25 \times 3 = 75$$,
$$25 \times 4 = 100$$,
$$25 \times 5 = 125$$,
$$25 \times 6 = 150$$, …

Step 3 — Understand the second condition.
A number is a multiple of $$50$$ if it can be written as $$50 \times m$$ for some whole number $$m$$. Notice that $$50 = 25 \times 2$$, so any multiple of $$50$$ is actually $$25 \times (2m)$$, meaning the factor of $$25$$ is being multiplied by an even number.

Step 4 — Pick the multiples of $$25$$ that are not multiples of $$50$$.
From the list in Step 2, remove those that are divisible by $$50$$ (that is, whose factor $$n$$ is even).

• $$25 = 25 \times 1$$  (factor 1 is odd; not a multiple of $$50$$)
• $$50 = 25 \times 2$$  (factor 2 is even; is a multiple of $$50$$) » reject
• $$75 = 25 \times 3$$  (factor 3 is odd; not a multiple of $$50$$)
• $$100 = 25 \times 4$$  (factor 4 is even; is a multiple of $$50$$) » reject
• $$125 = 25 \times 5$$  (factor 5 is odd; not a multiple of $$50$$)

Step 5 — State any three acceptable numbers.
Three numbers that satisfy both conditions are:

$$25,\; 75,\; 125$$

Step 1 – Understand the condition
The shout “idli-vada” is made exactly at the common multiples of the two chosen numbers.
The first time it can occur is therefore the least common multiple (LCM) of the two numbers.
The question tells us that this first shout is after the number 50, so we need

$$\text{LCM of the two numbers} > 50$$

Both numbers are whole numbers smaller than 10, so each of them can only be one of
$$2,\;3,\;4,\;5,\;6,\;7,\;8,\;9.$$

Step 2 – Note a quick fact
For any number $$n$$ not exceeding 5 and for every number $$m<10$$ we always have $$\text{LCM}(n,m)\le 45,$$ because the largest product that can appear is $$5\times9=45$$ (and many LCMs are even smaller).
Since we need an LCM bigger than 50, at least one of the two numbers must be $$7,\;8,\text{ or }9$$.

Step 3 – Examine all possible pairs

Step 4 – Conclusion
Exactly three pairs of numbers below 10 have an LCM greater than 50, hence any one of these pairs can be the numbers assigned “idli” and “vada”.

They are:

• $$7 \text{ and } 8\;(\text{LCM}=56),$$
• $$7 \text{ and } 9\;(\text{LCM}=63),$$
• $$8 \text{ and } 9\;(\text{LCM}=72).$$

With any of these choices the first shout of “idli-vada” comes only after the number 50.

To reach a number by equal jumps, the jump size must be a factor of that number.
Therefore, a jump size that lands on both 28 and 70 must be a factor of 28 and a factor of 70 — in other words, a common factor.

1. List all factors of each number.

   Number	Factors (pair-checking)	Complete list
   28	$$1\times28=28$$, $$2\times14=28$$, $$4\times7 =28$$	1, 2, 4, 7, 14, 28
   70	$$1\times70=70$$, $$2\times35=70$$, $$5\times14=70$$, $$7\times10=70$$	1, 2, 5, 7, 10, 14, 35, 70

2. Pick the factors that appear in both lists:

   Common factors = $$1,\;2,\;7,\;14$$

3. Interpretation for the game:

   Any of these four numbers can be chosen as the jump size, because starting from 0 they will land exactly on both 28 and 70:

   • Jump of $$1$$: 0, 1, 2, …, 28, …, 70
   • Jump of $$2$$: 0, 2, 4, …, 28, …, 70
   • Jump of $$7$$: 0, 7, 14, 21, 28, …, 70
   • Jump of $$14$$: 0, 14, 28, 42, 56, 70

   Hence these four jump sizes meet the requirement.

Step 1  Which two tables is the diagram talking about?
The three numbers written in the overlapping part are $$24,\;48\text{ and }72$$.
If they lie in the overlap, each of them must be a multiple of both of the hidden numbers.
So every hidden number must divide 24, 48 and 72.

• The common factors of 24, 48 and 72 are
$$1,\;2,\;3,\;4,\;6,\;8,\;12\text{ and }24.$$

• We now need two different factors whose first common multiple is 24 (because 24 is the smallest number lying in the overlap).
The neatest such pair is $$6$$ and $$8$$ because

• $$6$$ and $$8$$ are both in the factor-list above,
• their least common multiple is
  \[\operatorname{LCM}(6,8)=24\]
• and hence the first three common multiples of 6 and 8 are
  $$24,\;48,\;72,$$ exactly the numbers already written in the overlap.

Therefore the two headings must be “Multiples of 6” and “Multiples of 8”.

Step 2  List the multiples and fill the diagram (taking numbers below 75, so that the given 24, 48, 72 are the only common ones that appear).

Step 3  Write the numbers in their proper regions.

• Left-only part (multiples of 6 but not 8):
  $$6,\;12,\;18,\;30,\;36,\;42,\;54,\;60,\;66$$
• Overlap (already given):
  $$24,\;48,\;72$$
• Right-only part (multiples of 8 but not 6):
  $$8,\;16,\;32,\;40,\;56,\;64$$

After filling, the Venn diagram will show:

• Left circle caption → Multiples of 6
• Right circle caption → Multiples of 8
• Numbers placed exactly as listed above.

Step 1 — Recall the definitions

• A prime number has exactly two distinct factors: $$1$$ and the number itself.
• A composite number has more than two factors.
• The number $$1$$ is neither prime nor composite, but it does not appear in the present range, so we need not worry about it here.

Step 2 — Write all whole numbers from $$21$$ to $$30$$

$$21,\;22,\;23,\;24,\;25,\;26,\;27,\;28,\;29,\;30$$

Step 3 — Test each number for primality

For every number we only have to divide by primes that do not exceed its square root.

Step 4 — Count them

• Prime numbers: $$23,\;29$$ → 2 primes
• Composite numbers: all the remaining eight → 8 composites

Answer

Between $$21$$ and $$30$$ (both inclusive) there are 2 prime numbers and 8 composite numbers.

Step 1 – Recall the meanings of “even” and “prime”.

• An even number is any number that has $$2$$ as a factor, i.e. it can be written as $$2 \times k$$ where $$k$$ is a whole number.
• A prime number is a number greater than $$1$$ that has exactly two factors: $$1$$ and the number itself.

Step 2 – Consider an arbitrary even number larger than 2.

Take an even number $$n$$ such that $$n > 2$$. Because it is even,

$$n = 2 \times k,$$

where $$k$$ is a whole number and $$k \ge 2$$.

Step 3 – Identify the factors of $$n$$.

Writing $$n = 2 \times k$$ shows that $$2$$ and $$k$$ are both factors of $$n$$. Thus the factors of $$n$$ include

• $$1$$,
• $$2$$,
• $$n$$ itself (because $$1 \times n = n$$).

So $$n$$ has at least three factors: $$1$$, $$2$$ and $$n$$.

Step 4 – Use the definition of prime.

A prime number must have only two factors. Since our even number $$n$$ has more than two, it cannot be prime.

Step 5 – State the result.

Every even number greater than $$2$$ is composite. Therefore the only even number that is prime is $$2$$ itself.

Hence,

there is no other even prime number besides $$2$$.

Step 1 – Recall what “successive primes” means
When we write the primes in increasing order, each number is called the successor of the one just before it. For example, after $$5$$ the next prime is $$7$$, so $$5$$ and $$7$$ are successive primes.

Step 2 – List every prime number from $$2$$ to $$100$$
The NCERT book already shows this list, but let us write it out once more so we can work with it:

$$2,\;3,\;5,\;7,\;11,\;13,\;17,\;19,\;23,\;29,\;31,\;37,\;41,\;43,\;47,\;53,\;59,\;61,\;67,\;71,\;73,\;79,\;83,\;89,\;97$$

There are $$25$$ primes in all up to $$100$$.

Step 3 – Find the difference between every pair of successive primes

Step 4 – Read the smallest and largest differences
Going through the “Difference” column:

• The smallest number we see is $$1$$ (between $$2$$ and $$3$$).
• The largest number we see is $$8$$ (between $$89$$ and $$97$$).

Conclusion
Hence, among all successive primes not exceeding $$100$$, the smallest gap is $$1$$ and the largest gap is $$8$$.

First recall that a prime number has only two factors, $$1$$ and the number itself. We scan every number from $$1$$ to $$100$$, decide whether it is prime or not and then count how many primes appear in each horizontal row (each “decade”) of the 10 × 10 hundred-chart.

Observations

• The counts are $$4,4,2,2,3,2,2,3,2,1$$, so the numbers of primes are not equal in every row.
• The decade with the least primes is $$91$$–$$100$$ (only $$97$$), i.e. just $$1$$ prime.
• The decades with the most primes are $$1$$–$$10$$ and $$11$$–$$20$$, each containing $$4$$ primes.

Definition of a prime number
A number greater than $$1$$ that has exactly two factors – $$1$$ and the number itself – is called prime.

To test any number we only have to try dividing it by all prime numbers that are not larger than the square-root of the number. If none of those primes divides it, the number is prime.

Below we examine each of the given numbers.

Therefore, among the numbers given, the primes are $$23$$ and $$37$$.

Step 1   List all prime numbers smaller than 20.
They are $$2,\;3,\;5,\;7,\;11,\;13,\;17,\;19$$.

Step 2   Recall when a sum is a multiple of 5.
A number is a multiple of 5 when it leaves no remainder on dividing by 5; that is, when it is congruent to $$0$$ (mod 5).

Step 3   Find the “remainder 5” (modulus 5) of every prime found.

Step 4   Work out which pairs of remainders add up to $$0$$ (mod 5).
To make a sum divisible by 5, the two remainders must be one of the following combinations:

• $$2 + 3 = 5 \;(\equiv 0\;\text{mod }5)$$
• $$1 + 4 = 5 \;(\equiv 0\;\text{mod }5)$$
• $$0 + 0 = 0 \;(\equiv 0\;\text{mod }5)$$

Step 5   Match the actual primes using these combinations.

• From the pair $$2$$ (remainder 2) and $$3$$ (remainder 3):
  $$2 + 3 = 5$$, which is a multiple of 5.
• From the pair $$7$$ (remainder 2) and $$13$$ (remainder 3):
  $$7 + 13 = 20$$, which is a multiple of 5.
• From the pair $$11$$ (remainder 1) and $$19$$ (remainder 4):
  $$11 + 19 = 30$$, which is a multiple of 5.

Step 6   State any three suitable pairs.

\[ (2,3),\; (7,13),\; (11,19) \] All six numbers are prime, each is below 20, and every pair’s sum is a multiple of 5.

Step 1 : List all prime numbers up to 100

A prime number has exactly two factors, $$1$$ and the number itself. The primes up to $$100$$ are

$$2,\;3,\;5,\;7,\;11,\;13,\;17,\;19,\;23,\;29,\;31,\;37,\;41,\;43,\;47,\;53,\;59,\;61,\;67,\;71,\;73,\;79,\;83,\;89,\;97.$$

Step 2 : Keep only the two-digit primes

The one-digit primes ($$2,3,5,7$$) cannot form a pair like $$13\;\text{and}\;31$$ because they have only one digit. The two-digit primes are

$$11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97.$$

Step 3 : Reverse the digits of each two-digit prime

Step 4 : Collect the required pairs

Whenever both the original number and its reverse are prime (and the two numbers are different), we get a valid pair:

• $$(13,31)$$
• $$(17,71)$$
• $$(37,73)$$
• $$(79,97)$$

Each pair contains the same two digits in opposite order, and every number in the list is less than $$100$$.

Step 1 — Recall the meaning of “composite”.

A composite number is a whole number that has more than two factors; in other words, it can be divided exactly by some whole number other than $$1$$ and itself.

Step 2 — Plan how to find seven such numbers in a row.

• The numbers must lie between $$1$$ and $$100$$.
• The seven numbers must be consecutive (one after another in counting order).
• Every one of the seven must be composite (none may be prime or $$1$$).

Instead of checking every block of seven numbers, notice that primes become rarer among larger numbers. So it is sensible to begin our search near $$100$$ and move backwards.

Step 3 — Test the block from $$90$$ to $$96$$.

Every number in the list $$90,91,92,93,94,95,96$$ has a factor other than $$1$$ and itself. Therefore each is composite.

Step 4 — State the required block.

We have found seven consecutive composite numbers:

\[90,\;91,\;92,\;93,\;94,\;95,\;96\]

All of them lie between $$1$$ and $$100$$, and there is no prime within the block. Hence the task is complete.

Step 1 : Recall the definition

A pair of prime numbers is called twin primes if the difference between the two numbers is $$2$$.

We have to find all such pairs that lie between $$1$$ and $$100$$ (both numbers of the pair must themselves be within this range).

Step 2 : List all prime numbers up to $$100$$

First, write every prime number from $$2$$ to $$100$$.

$$2,\;3,\;5,\;7,\;11,\;13,\;17,\;19,\;23,\;29,\;31,\;37,\;41,\;43,\;47,\;53,\;59,\;61,\;67,\;71,\;73,\;79,\;83,\;89,\;97$$

Step 3 : Check successive primes for a difference of $$2$$

Start from the first prime and move to the next, subtracting the two each time.

• $$3-2 = 1$$ (not $$2$$)  →  no twin pair here.
• $$5-3 = 2$$  →  $$(3,5)$$ is a twin-prime pair. (Already given in the question.)
• $$7-5 = 2$$  →  $$(5,7)$$ is a twin-prime pair.
• $$11-7 = 4$$  →  not twin.
• $$13-11 = 2$$  →  $$(11,13)$$ is a twin-prime pair.
• $$17-13 = 4$$  →  not twin.
• $$19-17 = 2$$  →  $$(17,19)$$ is a twin-prime pair. (Given in the question.)
• $$23-19 = 4$$  →  not twin.
• $$29-23 = 6$$  →  not twin.
• $$31-29 = 2$$  →  $$(29,31)$$ is a twin-prime pair.
• $$37-31 = 6$$  →  not twin.
• $$41-37 = 4$$  →  not twin.
• $$43-41 = 2$$  →  $$(41,43)$$ is a twin-prime pair.
• $$47-43 = 4$$  →  not twin.
• $$53-47 = 6$$  →  not twin.
• $$59-53 = 6$$  →  not twin.
• $$61-59 = 2$$  →  $$(59,61)$$ is a twin-prime pair.
• $$67-61 = 6$$  →  not twin.
• $$71-67 = 4$$  →  not twin.
• $$73-71 = 2$$  →  $$(71,73)$$ is a twin-prime pair.
• All later differences between successive primes up to $$100$$ are larger than $$2$$.

Step 4 : Collect only the pairs not already mentioned in the question

The question already lists $$(3,5)$$ and $$(17,19)$$. Therefore, the other twin-prime pairs between $$1$$ and $$100$$ are:

$$(5,7),\;(11,13),\;(29,31),\;(41,43),\;(59,61),\;(71,73).$$

Result

Hence, excluding the examples given, there are six more twin-prime pairs between $$1$$ and $$100$$.

Step 1 – Understand the requirement
We have to find the number that can be written in the form $$p\times q\times r$$ where $$p,q,r$$ are different prime numbers. Each prime must occur once.

Step 2 – Prime-factorise each candidate

Step 3 – Selection
Only $$105$$ is the product of exactly three distinct prime numbers and each prime appears once: $$105 = 3\times 5\times 7$$.

Answer
The required number is $$105$$.

Step 1 – List all the three-digit numbers we can form.

We must use each of the digits $$2,4,5$$ once. The total number of different arrangements is

$$3! = 3 \times 2 \times 1 = 6$$.

Writing them out gives:

• $$245$$
• $$254$$
• $$425$$
• $$452$$
• $$524$$
• $$542$$

Step 2 – Check each number for primality.

A prime number greater than $$10$$ cannot end in $$0,2,4,5,6$$ or $$8$$ because:

• If it ends in $$0,2,4,6,8$$ it is even, so it is divisible by $$2$$.
• If it ends in $$5$$ it is divisible by $$5$$.

Now observe the last digit of every number in our list:

• $$245$$ ends in $$5 \Rightarrow$$ divisible by $$5$$.
• $$254$$ ends in $$4 \Rightarrow$$ even, divisible by $$2$$.
• $$425$$ ends in $$5 \Rightarrow$$ divisible by $$5$$.
• $$452$$ ends in $$2 \Rightarrow$$ even, divisible by $$2$$.
• $$524$$ ends in $$4 \Rightarrow$$ even, divisible by $$2$$.
• $$542$$ ends in $$2 \Rightarrow$$ even, divisible by $$2$$.

Each of the six numbers has a factor other than $$1$$ and itself, so none of them is prime.

Step 3 – Count the prime numbers.

Hence, we cannot make any three-digit prime number using the digits $$2,4,5$$ exactly once.

Step 1 – Recall the task
We have to look for prime numbers $$p$$ such that after we double the number and add $$1$$, the new number $$2p + 1$$ is also prime.

Step 2 – List the first few primes
The first few prime numbers are
$$2,\;3,\;5,\;7,\;11,\;13,\;17,\;19,\;23,\;29,\;31,\;37,\;41,\;43,\;47,\ldots$$

Step 3 – Test each prime one by one

Step 4 – Collect at least five successful pairs
From the table we see that the following primes work:

• $$p = 2: \;2p+1 = 5$$
• $$p = 3: \;2p+1 = 7$$
• $$p = 5: \;2p+1 = 11$$
• $$p = 11: \;2p+1 = 23$$
• $$p = 23: \;2p+1 = 47$$
• $$p = 29: \;2p+1 = 59$$

That already gives more than the required five examples.

Step 5 – State the result
So, besides $$3$$, there are many other primes with the same property. Six of them are listed above, and you can continue checking bigger primes in exactly the same way if you wish.

Key idea recalled from the chapter
When we list the multiples of two natural numbers, the first (i.e. smallest) common multiple is called their least common multiple, written LCM. The chapter also showed that

$$\text{LCM of }a\text{ and }b\;=\;\dfrac{a\times b}{\text{HCF}(a,b)}$$

so the HCF (highest common factor) decides whether the LCM equals, or is smaller than, the product $$a\times b$$.

1. When the first common multiple equals the product

Take the pair 4 and 9.

• Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, …
• Multiples of 9: 9, 18, 27, 36, …

The first common multiple is $$36$$.

Product of the two numbers: $$4\times 9 = 36$$.

Since the LCM is the same as the product, Anshu’s first observation is illustrated.

Why did this happen? List their factors:

• Factors of 4: 1, 2, 4
• Factors of 9: 1, 3, 9

The only common factor is 1, so the numbers are co-prime. For co-prime numbers the HCF is 1, and the formula above gives

\[\text{LCM}=\dfrac{4\times 9}{1}=36,\] which is exactly the product.

2. When the first common multiple is smaller than the product

Take the pair 6 and 8.

• Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, …
• Multiples of 8: 8, 16, 24, 32, 40, 48, …

The first common multiple is $$24$$.

Product of the two numbers: $$6\times 8 = 48\; (>24)$$.

Why is the LCM smaller here? Look at their common factor:

• Factors of 6: 1, 2, 3, 6
• Factors of 8: 1, 2, 4, 8

The HCF is 2 (greater than 1). Using the formula,

\[\text{LCM}=\dfrac{6\times 8}{2}=24,\] clearly less than the product.

Connection with being co-prime

• If the two numbers are co-prime (HCF = 1), no prime factor is repeated in both, so none can be “cancelled” in the fraction above. Hence $$\text{LCM}=a\times b$$.
• If the numbers share any common factor (>1), that factor divides the product in the formula, making the LCM smaller than the product.

Therefore, Anshu’s observation is exactly explained by whether the number pair is co-prime or not.

Step 1 : Recall the meaning of “co-prime”
Two natural numbers are co-prime if the only common factor between them is $$1$$.
So, to decide whether $$56$$ and $$63$$ are co-prime we must look at all their factors (or at least all their prime factors).

Step 2 : List (or factorise) the numbers correctly

• Prime factorisation of $$56$$:
  $$56 = 2 \times 2 \times 2 \times 7 = 2^{3}\,\times 7$$.
• Prime factorisation of $$63$$:
  $$63 = 3 \times 3 \times 7 = 3^{2}\,\times 7$$.

Step 3 : Find the common factor(s)
Both numbers contain the prime $$7$$, so $$7$$ is a common factor. Hence their greatest common factor (HCF) is $$7$$, not $$1$$, which means the numbers are not co-prime.

Step 4 : What was Anshu’s mistake?
Anshu wrote only one convenient factor pair for each number: $$56 = 14 \times 4$$ and $$63 = 21 \times 3$$. Because neither of these particular pairs shows a common factor, Anshu hastily concluded that no common factor exists.
However, a number usually has several factor pairs. To decide about co-primeness we must look at all the factors (or do a complete prime factorisation). By ignoring other factor pairs Anshu missed the factor $$7$$ that both numbers share.

Therefore Anshu went wrong because he did not list all factors (or do a complete prime-factorisation); he checked only one pair of factors and, as a result, overlooked the common factor $$7$$.

The best way to obtain the prime factorisation of any whole number is to keep dividing it by the smallest possible prime number until we finally reach 1. At each step we write down the prime we have divided by. Repeating this gives every prime that occurs in the factorisation.

1. Number : 64

• $$64 \div 2 = 32$$ (2 is prime)
• $$32 \div 2 = 16$$
• $$16 \div 2 = 8$$
• $$8 \div 2 = 4$$
• $$4 \div 2 = 2$$
• $$2 \div 2 = 1$$

We used the prime 2 exactly six times, so
$$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{6}.$$

2. Number : 104

• $$104 \div 2 = 52$$
• $$52 \div 2 = 26$$
• $$26 \div 2 = 13$$ (13 is prime and cannot be divided further except by itself)
• $$13 \div 13 = 1$$

Hence
$$104 = 2 \times 2 \times 2 \times 13 = 2^{3} \times 13.$$

3. Number : 105

• $$105 \div 3 = 35$$
• $$35 \div 5 = 7$$
• $$7 \div 7 = 1$$

The prime factorisation is
$$105 = 3 \times 5 \times 7.$$

4. Number : 243

• $$243 \div 3 = 81$$
• $$81 \div 3 = 27$$
• $$27 \div 3 = 9$$
• $$9 \div 3 = 3$$
• $$3 \div 3 = 1$$

We have used the prime 3 five times, so
$$243 = 3^{5}.$$

5. Number : 320

• $$320 \div 2 = 160$$
• $$160 \div 2 = 80$$
• $$80 \div 2 = 40$$
• $$40 \div 2 = 20$$
• $$20 \div 2 = 10$$
• $$10 \div 2 = 5$$
• $$5 \div 5 = 1$$

There are six 2’s and one 5, so
$$320 = 2^{6} \times 5.$$

6. Number : 141

• $$141 \div 3 = 47$$ (47 is a prime)
• $$47 \div 47 = 1$$

Thus
$$141 = 3 \times 47.$$

7. Number : 1728

• $$1728 \div 2 = 864$$
• $$864 \div 2 = 432$$
• $$432 \div 2 = 216$$
• $$216 \div 2 = 108$$
• $$108 \div 2 = 54$$
• $$54 \div 2 = 27$$
• $$27 \div 3 = 9$$
• $$9 \div 3 = 3$$
• $$3 \div 3 = 1$$

This gives six 2’s and three 3’s, so
$$1728 = 2^{6} \times 3^{3}.$$

8. Number : 729

• $$729 \div 3 = 243$$
• $$243 \div 3 = 81$$
• $$81 \div 3 = 27$$
• $$27 \div 3 = 9$$
• $$9 \div 3 = 3$$
• $$3 \div 3 = 1$$

We used the prime 3 six times, so
$$729 = 3^{6}.$$

9. Number : 1024

This number is a known power of 2, but we can still show the divisions:

• $$1024 \div 2 = 512$$
• $$512 \div 2 = 256$$
• $$256 \div 2 = 128$$
• $$128 \div 2 = 64$$
• $$64 \div 2 = 32$$
• $$32 \div 2 = 16$$
• $$16 \div 2 = 8$$
• $$8 \div 2 = 4$$
• $$4 \div 2 = 2$$
• $$2 \div 2 = 1$$

Hence
$$1024 = 2^{10}.$$

10. Number : 1331

• $$1331 \div 11 = 121$$
• $$121 \div 11 = 11$$
• $$11 \div 11 = 1$$

Thus
$$1331 = 11^{3}.$$

11. Number : 1000

• $$1000 \div 2 = 500$$
• $$500 \div 2 = 250$$
• $$250 \div 2 = 125$$
• $$125 \div 5 = 25$$
• $$25 \div 5 = 5$$
• $$5 \div 5 = 1$$

Therefore
$$1000 = 2^{3} \times 5^{3}.$$

Final summary

• $$64 = 2^{6}$$
• $$104 = 2^{3} \times 13$$
• $$105 = 3 \times 5 \times 7$$
• $$243 = 3^{5}$$
• $$320 = 2^{6} \times 5$$
• $$141 = 3 \times 47$$
• $$1728 = 2^{6} \times 3^{3}$$
• $$729 = 3^{6}$$
• $$1024 = 2^{10}$$
• $$1331 = 11^{3}$$
• $$1000 = 2^{3} \times 5^{3}$$

We know the required number is obtained by multiplying together all its prime factors.

• one $$2$$,
• two $$3$$s,
• one $$11$$.

Write them side by side (order does not matter when we multiply):

$$2 \times 3 \times 3 \times 11$$

Multiply step by step.

1. First the two equal factors:
   $$3 \times 3 = 9$$
2. Multiply this result by $$2$$:
   $$9 \times 2 = 18$$
3. Finally multiply by $$11$$:
   $$18 \times 11 = 198$$

So the number whose prime factorisation contains one $$2$$, two $$3$$s and one $$11$$ is

We must split the number $$1955$$ into three prime factors. Because each factor has to be less than $$30$$, the easiest way is to carry out an ordinary prime-factorisation and then read off the three primes we get.

Step 1 – Check divisibility by the smallest primes.

• $$1955$$ is odd, so it is not divisible by $$2$$.
• The sum of its digits is $$1+9+5+5 = 20$$, which is not a multiple of $$3$$, so it is not divisible by $$3$$.
• Its last digit is $$5$$, so it is divisible by $$5$$.

Step 2 – Divide by $$5$$.

Perform the division:

$$1955 \div 5 = 391$$

Thus we already have one required prime:

$$1955 = 5 \times 391$$

Step 3 – Factorise $$391$$.

We again test the smaller primes.

• $$391$$ is odd  ⇒  not divisible by $$2$$.
• Digit-sum $$3+9+1 = 13$$  ⇒  not divisible by $$3$$.
• It does not end in $$0$$ or $$5$$  ⇒  not divisible by $$5$$.
• Try $$7$$:  $$391 \div 7 = 55.857\dots$$ (not an integer).
• Try $$11$$:  $$391 \div 11 = 35.545\dots$$ (not an integer).
• Try $$13$$:  $$391 \div 13 = 30.076\dots$$ (not an integer).
• Try $$17$$:  $$391 \div 17 = 23$$ — an exact division!

So

$$391 = 17 \times 23$$

Both $$17$$ and $$23$$ are primes, and both are certainly less than $$30$$ (a quick glance at the prime table for numbers below $$30$$ confirms this).

Step 4 – Write the full prime factorisation.

Combining all the steps we have

\[1955 = 5 \times 17 \times 23\]

Step 5 – Answer the question.

The required three prime numbers, each less than $$30$$, are therefore:

$$5,\;17,\;23$$

Step 1 : Write down the two numbers from their prime-factor form.

First number  $$=2 \times 3 \times 7$$
Second number $$=3 \times 7 \times 11$$

Multiplying we get the ordinary (composite) numbers:

• First number  $$2 \times 3 \times 7 = 42$$
• Second number $$3 \times 7 \times 11 = 231$$

Step 2 : Find the common prime factors.

The prime factors of the first number are $$2,3,7$$.
The prime factors of the second number are $$3,7,11$$.

The common prime factors are $$3$$ and $$7$$. Their product is

$$3 \times 7 = 21$$.

Step 3 : Decide whether the numbers are co-prime.

Two numbers are co-prime if their HCF (highest common factor) is $$1$$. Here the common factor is $$21\,(\neq1)$$, so

The two numbers are not co-prime.

Step 4 : Check whether one number divides the other.

For $$42$$ to divide $$231$$, the second number would need the factor $$2$$ (because $$42=2\times3\times7$$). But $$231$$ has no factor $$2$$, so $$42\nmid231$$.

For $$231$$ to divide $$42$$, the first number would need the factor $$11$$ (because $$231=3\times7\times11$$). The number $$42$$ has no factor $$11$$, so $$231\nmid42$$.

Therefore neither number divides the other.

Step 1  Recall the meanings of the two words.

• A prime number has exactly two distinct factors, namely $$1$$ and the number itself.
• Two whole numbers are called co-prime (or relatively prime) when their only common factor is $$1$$. In other words, their HCF (highest common factor) is $$1$$.

Step 2  Check the truth of Guna’s statement with examples.

(i) Take two different prime numbers, say $$2$$ and $$3$$.

• Factors of $$2: \;1,2$$
• Factors of $$3: \;1,3$$
• The only common factor is $$1$$, so $$2$$ and $$3$$ are co-prime.

This looks as if Guna might be right, but we must test all possibilities, not just one pair.

(ii) Now take two same prime numbers, for example $$2$$ and $$2$$.

• Factors of the first $$2: \;1,2$$
• Factors of the second $$2: \;1,2$$
• Common factors: $$1$$ and $$2$$
• Therefore the HCF is $$2\neq1$$, so the numbers are not co-prime.

This single counter-example is enough to show Guna’s statement is false.

Step 3  State the conclusion clearly.

Only two different prime numbers are always co-prime; if the two numbers are the same prime, they are not co-prime. Hence, saying “any two prime numbers are co-prime” is incorrect.

Step 1 :  Understand the statement
We have to check whether the following is always true or not:
“Numbers that are divisible by $$10$$ are exactly those numbers that end with $$0$$.”

Step 2 :  Recall the meaning of “divisible by $$10$$”
A whole number $$N$$ is divisible by $$10$$ if, when we divide $$N$$ by $$10$$, the remainder is $$0$$. In symbols, we can write
\[ N = 10 \times q + r \quad (0 \le r < 10) \]
where $$q$$ is the quotient and $$r$$ is the remainder. If $$r = 0$$, then $$N$$ is divisible by $$10$$.

Step 3 :  What does the remainder $$r$$ represent?
Because $$10$$ is the base of our place-value system, the remainder $$r$$ is precisely the ones-place digit (the last digit) of the number $$N$$. Therefore

• if the last digit is $$0$$, then $$r = 0$$;
• if the last digit is anything other than $$0$$, then $$r \ne 0$$.

Step 4 :  Prove each direction

• Direction A   (If divisible by $$10$$, then last digit is $$0$$)
  Suppose $$N$$ is divisible by $$10$$. That means $$r = 0$$ in the equation $$N = 10q + r$$. Hence the ones-place digit of $$N$$ is $$0$$. So the number ends in $$0$$.
• Direction B   (If last digit is $$0$$, then divisible by $$10$$)
  Suppose the last digit of $$N$$ is $$0$$. Then the remainder $$r$$, when $$N$$ is divided by $$10$$, is $$0$$. Therefore $$N$$ is divisible by $$10$$.

Step 5 :  State the conclusion
Both directions are true, so the given statement is correct. Every number ending in $$0$$ is divisible by $$10$$, and no other numbers are.

Statement to check: “Numbers that are divisible by $$5$$ are exactly those that end with either a $$0$$ or a $$5$$.”

Idea of the proof
Write every whole number $$N$$ in the “tens + units” form

$$N = 10q + r$$   where  $$q$$ is the number of tens and $$r$$ (the units digit) satisfies $$0 \le r \le 9$$.

The two directions we must justify are

1. If the units digit $$r$$ is $$0$$ or $$5$$, then $$N$$ is divisible by $$5$$.
2. If $$N$$ is divisible by $$5$$, then its units digit $$r$$ can be only $$0$$ or $$5$$.

Direction 1 – From last digit to divisibility

• Suppose $$r = 0$$. Then $$N = 10q + 0 = 5(2q)$$, which is a multiple of $$5$$.
• Suppose $$r = 5$$. Then $$N = 10q + 5 = 5(2q + 1)$$, again a multiple of $$5$$.

In both cases $$5$$ divides $$N$$.

Direction 2 – From divisibility to last digit

If $$N$$ is divisible by $$5$$, we have $$N = 5k$$ for some whole number $$k$$. Rewrite $$N$$ as $$10q + r$$ again. Because $$10q = 5(2q)$$ is already a multiple of $$5$$, the only possible remainder when we divide by $$5$$ comes from $$r$$. For divisibility we need that remainder to be $$0$$, so $$r$$ itself must give remainder $$0$$ on division by $$5$$. Among the one-digit numbers $$0,1,2,3,4,5,6,7,8,9$$, only $$0$$ and $$5$$ do that. Hence the units digit must be $$0$$ or $$5$$.

Conclusion

Both directions are true, so the original statement is correct.

Quick check with examples

• $$45 = 5 \times 9$$ (ends in $$5$$)  ✓
• $$120 = 5 \times 24$$ (ends in $$0$$)  ✓
• $$67$$ ends in $$7$$, not divisible by $$5$$  ✓

Therefore, yes, a number is divisible by $$5$$ exactly when it ends in $$0$$ or $$5$$.

Step 1 · Understanding the statement

A number is divisible by $$2$$ when it gives remainder $$0$$ on division by $$2$$. Such numbers are called even numbers.

When we divide a number by $$10$$, its ones digit (last digit) tells us what remainder the number will give on division by $$10$$. In the same way, to see the remainder on division by $$2$$ we need to look only at the ones digit, because $$10 = 2 \times 5$$ and every full group of $$10$$ already contains a factor $$2$$.

• If the ones digit is $$0,2,4,6$$ or $$8$$, that ones digit itself is divisible by $$2$$, so the whole number is divisible by $$2$$.
• If the ones digit is $$1,3,5,7$$ or $$9$$, that ones digit leaves remainder $$1$$ on division by $$2$$, so the whole number also leaves remainder $$1$$ and is not divisible by $$2$$.

Therefore the statement is correct: numbers whose last digit is $$0,2,4,6$$ or $$8$$ are exactly the numbers divisible by $$2$$.

Step 2 · Listing numbers from 399 to 411

The numbers in this range are

$$399,\;400,\;401,\;402,\;403,\;404,\;405,\;406,\;407,\;408,\;409,\;410,\;411.$$

Step 3 · Selecting the multiples of 2

Using the test explained in Step 1, pick out the numbers whose ones digit is $$0,2,4,6$$ or $$8$$:

• $$400$$ — ends in $$0$$
• $$402$$ — ends in $$2$$
• $$404$$ — ends in $$4$$
• $$406$$ — ends in $$6$$
• $$408$$ — ends in $$8$$
• $$410$$ — ends in $$0$$

None of the other numbers (399, 401, 403, 405, 407, 409, 411) qualify because their last digits are odd.

Step 4 · Final list

Hence the multiples of $$2$$ between $$399$$ and $$411$$ are

Divisibility rule for 4
A whole number is divisible by 4 if and only if the number formed by its last two digits is divisible by 4.

We shall use this rule in every part.

(1) Numbers between $$330$$ and $$340$$

• Look at the last two digits 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40.
• Among these, the numbers divisible by 4 are $$32$$, $$36$$ and $$40$$, because $$32 \div 4 = 8$$, $$36 \div 4 = 9$$, $$40 \div 4 = 10$$.
• Hence the required whole numbers are
  $$332,\;336,\;340.$$

(2) Numbers between $$1730$$ and $$1740$$

• Consider the last two digits again: 30, 31, 32, … , 40.
• The same two–digit multiples of 4 (32, 36, 40) now give the four–digit numbers
  $$1732,\;1736,\;1740.$$

(3) Numbers between $$2030$$ and $$2040$$

• Using the very same last two digits 32, 36, 40 we get
  $$2032,\;2036,\;2040.$$

Observation

• In every case the numbers that work end in the same two–digit endings: 32, 36 and 40.
• This shows clearly that only the last two digits decide divisibility by 4; the other digits (hundreds, thousands, …) make no difference.
• So, between any consecutive tens $$n30$$ and $$n40$$, the numbers divisible by 4 will always be $$n32, n36, n40$$.

To find out whether $$8536$$ is divisible by $$4$$, we can use the divisibility rule for 4.

Rule: A whole number is divisible by $$4$$ iff (if and only if) the number formed by its last two digits is divisible by $$4$$.

Write the number and highlight its last two digits:

$$8536 = 85\,\mathbf{36}$$

1. Consider only the last two digits: $$36$$.

2. Check whether $$36$$ is divisible by $$4$$:

   Divide: $$36 \div 4 = 9$$ exactly, with remainder $$0$$.

3. Because the remainder is $$0$$, $$36$$ is divisible by $$4$$.

4. Therefore, by the rule, the whole number $$8536$$ is also divisible by $$4$$.

We can state the conclusion in one line:

\[ 4 \mid 8536 \quad \text{(read: 4 divides 8536)} \]

Step 1 : Recall the divisibility test for 8
A number is divisible by 8 if the number formed by its last three digits is divisible by 8. (Because $$1000 = 8 \times 125$$, every extra thousand only adds a multiple of 8.)

Step 2 : List the multiples of 8 that lie in each required range

• Multiples of 8 near 120: $$8 \times 15 = 120, \; 8 \times 16 = 128, \; 8 \times 17 = 136, \; 8 \times 18 = 144$$.

The numbers between 120 and 140 are therefore 120, 128 and 136.

• To get the numbers between 1120 and 1140 we simply add 1000 to each of the above, because 1000 is itself divisible by 8:
  1120 = 120 + 1000,  1128 = 128 + 1000,  1136 = 136 + 1000.

Hence the numbers between 1120 and 1140 that are divisible by 8 are 1120, 1128 and 1136.

• Likewise, for the range 3120–3140 add 3000 (which is three thousands, and $$3 \times 1000$$ is also divisible by 8):
  3120 = 120 + 3000,  3128 = 128 + 3000,  3136 = 136 + 3000.

Thus the numbers between 3120 and 3140 that are divisible by 8 are 3120, 3128 and 3136.

Step 3 : Observation
Adding 1000, 2000, 3000, … (any whole number of thousands) to a number does not change its divisibility by 8, because $$1000 = 8 \times 125$$ is itself a multiple of 8. Therefore the pattern of numbers divisible by 8 repeats every 1000.

Step 1 — Recall the test for divisibility by 8
A whole number is a multiple of 8 exactly when the three–digit number formed by its last three digits is a multiple of 8.

Step 2 — Keep thousands and hundreds places fixed
In 8560 the thousands and hundreds digits (8 and 5) will stay as they are. Write the required new number as $$85ab$$, where $$a$$ is the tens digit and $$b$$ is the ones digit we have to choose.

Step 3 — Set up the divisibility condition
The last three digits of $$85ab$$ are $$5ab$$. Numerically, that three–digit number is $$500 + ab$$. We need

$$500 + ab\text{ to be divisible by }8.$$

Step 4 — Find the required remainder
First divide 500 by 8:

$$500 = 8 \times 62 + 4 \quad\Longrightarrow\quad 500 \equiv 4 \pmod 8.$$

Therefore we must have

$$ab \equiv 4 \pmod 8.$$

Step 5 — List all two–digit numbers with remainder 4 on division by 8

$$04,\;12,\;20,\;28,\;36,\;44,\;52,\;60,\;68,\;76,\;84,\;92$$

Step 6 — Write the corresponding four–digit multiples of 8

• 8504
• 8512
• 8520
• 8528
• 8536
• 8544
• 8552
• 8560
• 8568
• 8576
• 8584
• 8592

Any one of these answers the question. For example, replacing the last two digits by 04 gives 8504, and

$$8504 \div 8 = 1063,$$

so 8504 is indeed a multiple of 8.

Step 1 – Form of a 4-digit palindrome
Any 4-digit palindrome looks like $$a b b a$$. Here $$a$$ and $$b$$ are single digits, with $$a\neq 0$$ so that the number really has four digits.

Step 2 – Use the rule for divisibility by 4
A whole number is divisible by 4 when the number made by its last two digits is divisible by 4.
For $$a b b a$$ the last two digits are $$b$$ (tens place) and $$a$$ (ones place).
Their value is $$10b + a$$.
So we must have
\[10b + a \text{ is a multiple of } 4.\]

Step 3 – Find the smallest required number

The smallest odd digit is $$b=1$$. Hence the smallest 4-digit palindrome divisible by 4 is $$2\,1\,1\,2 = 2112$$.

Step 4 – Find the largest required number

The largest even digit is $$b=8$$. Hence the largest 4-digit palindrome divisible by 4 is $$8\,8\,8\,8 = 8888$$.

Step 5 – Quick check

• $$2112 \div 4 = 528$$ → divisible
• $$8888 \div 4 = 2222$$ → divisible

Therefore, the required numbers are correctly found.

The division relation. When a whole number $$N$$ is divided by a non-zero whole number $$d$$, we obtain a quotient $$q$$ and a remainder $$r$$ satisfying

That is, the remainder is at least $$0$$ and strictly less than the divisor $$d$$.

Below we use this relation to find the remainder when each given number is divided by (a) $$10$$, (b) $$5$$ and (c) $$2$$. The following shortcuts will be handy:

• Dividing by $$10$$ — the remainder is the units digit (a value from $$0$$ to $$9$$).
• Dividing by $$5$$ — subtract $$5$$ from the units digit if the units digit is $$5$$ or more; the result (always between $$0$$ and $$4$$) is the remainder.
• Dividing by $$2$$ — the remainder is $$0$$ if the number is even and $$1$$ if it is odd.

1. $$78$$
   (a) $$78 = 10 \times 7 + 8$$  ⇒  remainder $$8$$.
   (b) $$78 = 5 \times 15 + 3$$  ⇒  remainder $$3$$.
   (c) $$78 = 2 \times 39 + 0$$  ⇒  remainder $$0$$.
2. $$99$$
   (a) $$99 = 10 \times 9 + 9$$  ⇒  remainder $$9$$.
   (b) $$99 = 5 \times 19 + 4$$  ⇒  remainder $$4$$.
   (c) $$99 = 2 \times 49 + 1$$  ⇒  remainder $$1$$.
3. $$173$$
   (a) $$173 = 10 \times 17 + 3$$  ⇒  remainder $$3$$.
   (b) $$173 = 5 \times 34 + 3$$  ⇒  remainder $$3$$.
   (c) $$173 = 2 \times 86 + 1$$  ⇒  remainder $$1$$.
4. $$572$$
   (a) $$572 = 10 \times 57 + 2$$  ⇒  remainder $$2$$.
   (b) $$572 = 5 \times 114 + 2$$  ⇒  remainder $$2$$.
   (c) $$572 = 2 \times 286 + 0$$  ⇒  remainder $$0$$.
5. $$980$$
   (a) $$980 = 10 \times 98 + 0$$  ⇒  remainder $$0$$.
   (b) $$980 = 5 \times 196 + 0$$  ⇒  remainder $$0$$.
   (c) $$980 = 2 \times 490 + 0$$  ⇒  remainder $$0$$.
6. $$1111$$
   (a) $$1111 = 10 \times 111 + 1$$  ⇒  remainder $$1$$.
   (b) $$1111 = 5 \times 222 + 1$$  ⇒  remainder $$1$$.
   (c) $$1111 = 2 \times 555 + 1$$  ⇒  remainder $$1$$.
7. $$2345$$
   (a) $$2345 = 10 \times 234 + 5$$  ⇒  remainder $$5$$.
   (b) $$2345 = 5 \times 469 + 0$$  ⇒  remainder $$0$$.
   (c) $$2345 = 2 \times 1172 + 1$$  ⇒  remainder $$1$$.

In every line the remainder $$r$$ satisfies $$0 \le r \lt d$$ — when dividing by $$10$$ the remainders lie in $$\{0,1,\dots,9\}$$; when dividing by $$5$$ they lie in $$\{0,1,2,3,4\}$$; and when dividing by $$2$$ they are $$0$$ or $$1$$.

Summary of all remainders

Step 1 – Recall the divisibility rules

• $$n$$ is divisible by $$2$$  &hspace;{1em}⟺  the last digit of $$n$$ is even.
• $$n$$ is divisible by $$4$$  &hspace;{1em}⟺  the number formed by the last two digits of $$n$$ is divisible by $$4$$.
• $$n$$ is divisible by $$5$$  &hspace;{1em}⟺  the last digit of $$n$$ is $$0$$ or $$5$$.
• $$n$$ is divisible by $$8$$  &hspace;{1em}⟺  the number formed by the last three digits of $$n$$ is divisible by $$8$$.
• $$n$$ is divisible by $$10$$  &hspace;{1em}⟺  the last digit of $$n$$ is $$0$$.

Step 2 – Which two tests are enough?
Guna wants two tests whose success automatically guarantees the other three.

(i) One of the tests must be for $$8$$.
  • If a number is divisible by $$8$$, it is automatically divisible by $$4$$ and by $$2$$ because $$8=2\times4$$.
So, checking $$8$$ at once settles divisibility by $$2$$ and $$4$$ as well.

(ii) To settle the remaining numbers, we still need to make sure the given number is divisible by $$5$$ (so that the condition for $$10$$ is fulfilled).
There are two ways to do this from the given list:

• Check divisibility by $$5$$ itself, or
• Check divisibility by $$10$$ (because a number divisible by $$10$$ is automatically divisible by both $$2$$ and $$5$$; the extra factor $$2$$ is already handled by the test for $$8$$).

Therefore, any one of the two pairs below will do the job:

• $$8$$ and $$5$$   (guarantees $$2,4,8,5,10$$)
• $$8$$ and $$10$$  (guarantees $$2,4,8,5,10$$ as well)

Step 3 – Verify with the actual number $$14560$$

• Last three digits are $$560$$; since $$560\div8 = 70$$, $$14560$$ is divisible by $$8$$.
• Last digit is $$0$$; hence the number is divisible by both $$5$$ and $$10$$.

Because the number passes the two selected tests, Guna could confidently announce that it is divisible by all the given numbers.

Conclusion
The two tests Guna could have used are either $$8$$ and $$5$$, or $$8$$ and $$10$$. (The textbook usually expects the first pair, $$8$$ and $$5$$.)

Step 1 : Recall the divisibility rules we need.

• Divisible by $$2$$ ⇔ last digit is even (i.e. $$0,2,4,6,8$$).
• Divisible by $$5$$ ⇔ last digit is $$0$$ or $$5$$.
• Divisible by $$10$$ ⇔ last digit is $$0$$ (this at once makes the number divisible by $$2$$ and $$5$$).
• Divisible by $$4$$ ⇔ the number formed by the last two digits is a multiple of $$4$$.
• Divisible by $$8$$ ⇔ the number formed by the last three digits is a multiple of $$8$$.

Because a number must be divisible by all of $$2,4,5,8,10$$, it has to pass every one of these tests.

Step 2 : Quick elimination using $$10$$.
Any number divisible by $$10$$ must end in $$0$$. So we can discard at once every number that does not end in $$0$$:

• $$572$$ ends in $$2\;\Rightarrow\;$$ not divisible by $$10$$ ⇒ reject.
• $$2352$$ ends in $$2\;\Rightarrow\;$$ not divisible by $$10$$ ⇒ reject.

The remaining candidates are therefore only:

• $$5600$$
• $$6000$$
• $$77622160$$

(All three end in $$0$$, so they are already divisible by $$2,5$$ and $$10$$.)

Step 3 : Check divisibility by $$4$$ and $$8$$ for the remaining numbers.

All three numbers satisfy the conditions for $$4$$ and $$8$$, so each of them is divisible by every one of $$2,4,5,8,10$$.

Conclusion

The numbers that are divisible by all of $$2,4,5,8$$ and $$10$$ are $$5600,\;6000\text{ and }77622160$$.

Step 1 – Prime-factorise 10000

Write 10000 as a product of tens:

$$10000 = 10 \times 10 \times 10 \times 10$$

Each 10 is $$2 \times 5$$, so

$$10000 = (2 \times 5)^4 = 2^4 \times 5^4.$$

Step 2 – Understand the "units digit 0" condition

A number ends in 0 exactly when it has one factor 2 and one factor 5 (because $$2 \times 5 = 10$$ places a 0 at the end).

Therefore, to stop a number from ending in 0, it must miss at least one of those two factors.

Step 3 – Separate the factors

• Put all the 2’s (and none of the 5’s) in the first number.
  First number  = $$2^4 = 16.$$
• Put all the 5’s (and none of the 2’s) in the second number.
  Second number = $$5^4 = 625.$$

Step 4 – Check

$$16 \times 625 = 2^4 \times 5^4 = 10000.$$

The units digits are 6 and 5, neither of which is 0. Hence the pair works.

So one possible answer is $$\boxed{16 \text{ and } 625}$$ (many other pairs are also possible).

First list the factors of each number (remember: a factor is a whole number that divides exactly with no remainder).

Observation:

• $$9=3\times3=3^2$$, $$16=4\times4=2^4$$ and $$25=5\times5=5^2$$ are all perfect squares (and therefore composite).
• $$43$$ is not a perfect square; it has no factor other than $$1$$ and itself, so it is a prime number.

Hence $$43$$ stands out (is special) because it is the only prime number in the box, whereas the other three numbers are composite perfect squares.

The question asks: “Look at the solved picture and guess the rules of the puzzle.” We therefore have to explain what players are allowed (or required) to do and why the nine numbers written in the solved grid satisfy those rules.

Step 1  Examine the information printed on the frame

The three numbers written on the right of the blank squares are called row-products: they are what you should get after multiplying the three numbers that will finally stand in that row.

The three numbers written under the blank squares are called column-products: they are what you should get after multiplying the three numbers that will finally stand in that column.

Step 2  Prime-factorise every product

• Row products
    $$75 = 3 \times 5 \times 5$$
    $$42 = 2 \times 3 \times 7$$
    $$102 = 2 \times 3 \times 17$$
• Column products
    $$170 = 2 \times 5 \times 17$$
    $$30 = 2 \times 3 \times 5$$
    $$63 = 3 \times 3 \times 7$$

Observation All factors that appear are prime numbers (2, 3, 5, 7, 17). That strongly suggests the puzzle is to place only prime numbers inside the boxes.

Step 3  Count how often each prime occurs

Across the three rows we have:

• $$2: 2\text{ times}$$
• $$3: 3\text{ times}$$
• $$5: 2\text{ times}$$
• $$7: 1\text{ time}$$
• $$17: 1\text{ time}$$

Exactly the same list is obtained from the three columns. That is necessary, because whatever numbers we write must be counted both in a row and in a column.

Step 4  Start with a prime that is unique

The only prime that appears once in the whole list is $$17$$. It must sit in the row whose product is $$102$$ (because that row contains 17) and in the column whose product is $$170$$ (because that column also contains 17). Those two intersect at the square in the 3rd row, 1st column, so we must write

Third row, first column → $$17$$.

Step 5  Update the remaining factors

• Row 3 now still needs $$2 \times 3 = 6$$.
• Column 1 now still needs $$2 \times 5 = 10$$.

Step 6  Look for the next forced move

Column 2 has the product $$30 = 2 \times 3 \times 5$$. Row 1 also contains two copies of 5. A convenient place to continue is Row 1, because two of its three factors are equal.

Try Row 1: we must arrange 5, 5 and 3, in some order.

If we place one of the 5’s in Column 1 we would violate Column 1’s need for the factor 5 later (it already needs 5). So the safest decision is

• Row 1, Column 1 → $$5$$ (now Column 1 need is $$2$$);
• Row 1, Column 2 → $$5$$ (now Column 2 need is $$2 \times 3$$);
• Row 1, Column 3 → $$3$$ (Column 3 need is $$3 \times 7$$).

Step 7  Complete Column 1

Column 1 still requires a factor 2. The only empty position in that column is Row 2, Column 1, so we must write

Row 2, Column 1 → $$2$$.

Column 1 is now $$5 \times 2 \times 17 = 170$$ ✔️

Step 8  Finish Row 2

Row 2 already contains $$2$$ and must end at $$42 = 2 \times 3 \times 7$$. The missing factors are therefore $$3$$ and $$7$$.

• Row 2, Column 2 → $$3$$ (Column 2 now satisfied: $$5 \times 3 \times 2 = 30$$ ✔️)
• Row 2, Column 3 → $$7$$ (Column 3 now also satisfied: $$3 \times 7 \times 3 = 63$$ ✔️)

Step 9  Finish Row 3

Row 3 already has $$17$$; the remaining product required is $$2 \times 3$$, exactly the two empty places:

• Row 3, Column 2 → $$2$$
• Row 3, Column 3 → $$3$$

Row 3 is $$17 \times 2 \times 3 = 102$$ ✔️. Every row and column now matches the given product, so the filled grid is unique and complete.

Final filled grid

\[ \begin{array}{|c|c|c||c|} 5 & 5 & 3 & 75\\ 2 & 3 & 7 & 42\\ 17 & 2 & 3 & 102\\ \hline 170 & 30 & 63 & \; \end{array} \]

The RULES of the puzzle (deduced)

• Only prime numbers may be written in the nine blank squares.
• For every row, the product of the three primes in that row must equal the number printed at the end of that row.
• For every column, the product of the three primes in that column must equal the number printed below that column.
• The same prime may be used more than once if the factorisation demands it.

If a player follows precisely these rules, the unique solution reproduced above is obtained.