Chapter 5: Number Play

1. Which natural numbers are sums of consecutive naturals?

Every odd number $$2n+1$$ equals $$n + (n+1)$$. For example, $$9 = 4 + 5$$, $$21 = 10 + 11$$.

A little checking shows the only natural numbers that cannot be written as a sum of two or more consecutive naturals are the powers of $$2$$: $$1,\ 2,\ 4,\ 8,\ 16,\ 32,\ \ldots$$ Try any of these — no consecutive sum works. Every other natural number can be written as such a sum.

2. Numbers that can be written in more than one way.

Numbers like $$15,\ 21,\ 45$$ have several representations, e.g. $$15 = 7+8 = 4+5+6 = 1+2+3+4+5$$. Numbers with several odd factors tend to have several representations. Prime odd numbers (like $$7$$, $$11$$, $$13$$) have only one representation ($$n + (n+1)$$).

3. Even numbers as sums of consecutive numbers.

Not every even number works. For instance $$4$$ and $$8$$ cannot be written as a sum of consecutive naturals. But $$6 = 1+2+3$$, $$10 = 1+2+3+4$$, $$12 = 3+4+5$$, $$14 = 2+3+4+5$$, $$18 = 3+4+5+6$$, and so on. So even numbers that are not powers of $$2$$ can always be written as a sum of consecutive naturals.

4. Can $$0$$ be a sum of consecutive numbers (allowing negatives)?

Yes. For example $$0 = (-1) + 0 + 1$$, or $$0 = (-2) + (-1) + 0 + 1 + 2$$. Any symmetric block $$-n, -(n-1), \ldots, -1, 0, 1, \ldots, n$$ adds up to $$0$$.

Key idea: a run of $$k$$ consecutive integers starting at $$a$$ sums to $$\dfrac{k(2a + k - 1)}{2}$$. Choosing $$k$$ and $$a$$ cleverly explains all the observations above.

There are $$3$$ gaps between the four numbers $$3,\ 4,\ 5,\ 6$$, and each gap can be filled with either $$+$$ or $$-$$. So the number of possibilities is $$2 \times 2 \times 2 = 8$$.

The eight expressions are:

1. $$3 + 4 + 5 + 6$$
2. $$3 + 4 + 5 - 6$$
3. $$3 + 4 - 5 + 6$$
4. $$3 + 4 - 5 - 6$$
5. $$3 - 4 + 5 + 6$$
6. $$3 - 4 + 5 - 6$$
7. $$3 - 4 - 5 + 6$$
8. $$3 - 4 - 5 - 6$$

Using the eight expressions from the previous question with $$3,4,5,6$$:

Every result is even. That is the interesting observation.

Take $$7,\ 8,\ 9,\ 10$$. Compute every sign choice:

Again every result is even. The parity is the same for all eight expressions.

Take $$10, 11, 12, 13$$:

Every value is again even. This confirms the pattern: no matter which $$4$$ consecutive numbers we take, all $$8$$ signed sums have the same parity — they are all even.

Let the four consecutive numbers be $$n,\ n+1,\ n+2,\ n+3$$. Then the eight signed sums are:

$$n + (n+1) + (n+2) + (n+3) = 4n + 6$$
$$n + (n+1) + (n+2) - (n+3) = 2n$$
$$n + (n+1) - (n+2) + (n+3) = 2n + 2$$
$$n + (n+1) - (n+2) - (n+3) = -4$$
$$n - (n+1) + (n+2) + (n+3) = 2n + 4$$
$$n - (n+1) + (n+2) - (n+3) = -2$$
$$n - (n+1) - (n+2) + (n+3) = 0$$
$$n - (n+1) - (n+2) - (n+3) = -2n - 6$$

Every result is of the form $$2 \times (\text{integer})$$, so every signed sum is even for every integer $$n$$. This confirms the pattern by pure reasoning.

Among any four consecutive numbers, exactly two are even and two are odd. Grouping consecutive pairs, adjacent numbers have opposite parity, so switching the sign of an odd/even pair only changes the answer by an even amount.

More directly: recall the parity rules — $$\text{odd} \pm \text{odd} = \text{even}$$, $$\text{even} \pm \text{even} = \text{even}$$, $$\text{odd} \pm \text{even} = \text{odd}$$. In our four consecutive numbers, the two odd values combine (with whatever signs) to give an even number, and the two even values combine to give an even number. Even + even = even. So the final result is always even, no matter which signs are chosen.

Start with $$E_1 = a + b - c - d$$. Say we flip the sign of $$c$$ (a negative becoming positive) to get $$E_2 = a + b + c - d$$.

The difference is

$$E_2 - E_1 = (a + b + c - d) - (a + b - c - d) = 2c.$$

So changing exactly one sign from $$-$$ to $$+$$ (say the sign of the number $$c$$) shifts the value by $$2c$$ — an even number.

Similarly, if we flip the sign of $$d$$ instead, the change is $$2d$$. In every case, flipping any one sign changes the value by twice one of the four numbers — always an even amount.

Every time we flip one sign in a signed sum, we change the result by twice one of the numbers — an even number. So whatever the value of one signed sum, every other signed sum obtained by flipping signs differs from it by an even amount.

Numbers that differ by an even amount always have the same parity. That is why all $$8$$ expressions must be simultaneously even (or simultaneously odd). Since one of them ($$a+b+c+d$$ for four consecutives) is even, they are all even.

No — the argument in the previous question depended only on the fact that flipping one sign changes the value by twice some term (an even number). This is true for a signed sum of any length.

So if we take $$5, 6, 7, \ldots$$ or any fixed collection of numbers $$a_1, a_2, \ldots, a_n$$ and place $$+$$ or $$-$$ signs between them, all the resulting signed sums must share the same parity — even or odd — determined by the parity of $$a_1 + a_2 + \cdots + a_n$$.

Use parity rules: $$\text{odd}\pm\text{odd}=\text{even}$$, $$\text{even}\pm\text{even}=\text{even}$$, $$\text{odd}\pm\text{even}=\text{odd}$$, and a product is even iff at least one factor is even.

An algebraic expression is always even if we can factor a $$2$$ out of it, whatever the letters stand for. Let's check each one:

• $$2a + 2b = 2(a+b)$$ — always even.
• $$3g + 5h$$: if $$g = 1$$, $$h = 0$$ gives $$3$$ (odd); if $$g = 1, h = 1$$ gives $$8$$ (even). Depends on $$g, h$$ — not always.
• $$4m + 2n = 2(2m + n)$$ — always even.
• $$2u - 4v = 2(u - 2v)$$ — always even.
• $$13k - 5k = 8k = 2 \cdot 4k$$ — always even.
• $$6m - 3n = 3(2m - n)$$: parity depends on $$n$$. E.g. $$m=1, n=1$$ gives $$3$$ (odd). Not always even.
• $$x^{2} + 2$$: parity is the parity of $$x^2$$, which matches $$x$$. $$x=1$$ gives $$3$$ (odd). Not always even.
• $$b^{2} + 1$$: $$b=1$$ gives $$2$$ (even), $$b=2$$ gives $$5$$ (odd). Not always even.
• $$4k \times 3j = 12kj = 2(6kj)$$ — always even.

Considering the four expressions from the previous question that are not always even:

$$3g + 5h$$. Try $$g=2, h=2$$: $$3(2)+5(2)=16$$ (even). Try $$g=1, h=2$$: $$3+10=13$$ (odd). So it is sometimes even, sometimes odd — the parity is exactly the parity of $$g+h$$.

$$6m - 3n$$. Try $$m=2, n=2$$: $$12-6 = 6$$ (even). Try $$m=1, n=1$$: $$6-3=3$$ (odd). Parity depends on $$n$$ ($$6m$$ is always even, so the parity of $$6m - 3n$$ is the parity of $$3n$$, i.e. of $$n$$).

$$x^{2} + 2$$. Try $$x=2$$: $$4+2=6$$ (even). Try $$x=3$$: $$9+2=11$$ (odd). Same parity as $$x^2$$, hence same as $$x$$.

$$b^{2} + 1$$. Try $$b=1$$: $$1+1=2$$ (even). Try $$b=2$$: $$4+1=5$$ (odd). Same parity as $$b+1$$: even when $$b$$ is odd, odd when $$b$$ is even.

Any expression that has $$2$$ as a factor is always even. Examples for integer letter-values:

• $$2a$$
• $$2a + 2b$$
• $$6a + 4b - 10c$$
• $$4a^{2} + 8b$$
• $$a(a+1)$$ — the product of two consecutive integers is always even, since one of them is even.
• $$a^{2} - a = a(a-1)$$ — again a product of two consecutives.
• $$(a + b)(a - b) + (a + b)(a + b) = 2a(a+b)$$

An even number is either a multiple of $$4$$ (like $$4, 8, 12$$) or two more than a multiple of $$4$$ (like $$2, 6, 10, 14$$). Call these two types 4k and 4k+2.

Try some pairs:

Algebraically:

• $$4a + 4b = 4(a+b)$$: multiple of $$4$$.
• $$(4a+2) + (4b+2) = 4(a+b) + 4 = 4(a+b+1)$$: multiple of $$4$$.
• $$4a + (4b+2) = 4(a+b) + 2$$: not a multiple of $$4$$.

So the sum of two even numbers is a multiple of $$4$$ exactly when both are of the same type — either both multiples of $$4$$ or both of the form $$4k+2$$.

From the previous exploration, an even number has two possible 'flavours':

• Type A: multiple of $$4$$, i.e. $$4k$$.
• Type B: two more than a multiple of $$4$$, i.e. $$4k+2$$.

Adding two even numbers:

$$4a + 4b = 4(a+b) \ \ \checkmark$$
$$(4a+2) + (4b+2) = 4(a+b+1) \ \ \checkmark$$
$$4a + (4b+2) = 4(a+b) + 2 \ \ \times$$

Hence the sum is a multiple of $$4$$ iff the two even numbers are of the same type — both are $$4k$$, or both are $$4k+2$$.

Picture a multiple of $$4$$ as a rectangle with $$4$$ dots in every row, and a number of the form $$4q+2$$ as a similar rectangle with two extra dots at the top.

Case 1 — both are multiples of $$4$$. Stacking a $$4p$$ rectangle above a $$4q$$ rectangle gives a $$4(p+q)$$ rectangle — again a multiple of $$4$$.
Example: $$8 + 12 = 20 = 4 \times 5$$.

Case 2 — both are of the form $$4k+2$$. Each has $$2$$ extra dots. When stacked, the two extras combine into one full row of $$4$$, so the total is $$4(p+q) + 4 = 4(p+q+1)$$ — a multiple of $$4$$.
Example: $$6 + 10 = 16 = 4 \times 4$$.

Case 3 — one of each type: $$4p + (4q + 2)$$. The two extras cannot fill a row of $$4$$, so the picture is a $$4(p+q)$$ rectangle with $$2$$ dots left over:

$$4p + (4q + 2) = 4(p + q) + 2.$$

The sum leaves remainder $$2$$ when divided by $$4$$.
Example: $$4 + 10 = 14 = 4 \times 3 + 2$$.

Let the two numbers be $$8a$$ and $$8b$$ (multiples of $$8$$). Their sum is

$$8a + 8b = 8(a+b),$$

which is a multiple of $$8$$.

Example: $$16 + 24 = 40 = 8 \times 5$$.

The number $$16$$ is divisible by $$8$$, and $$16 = 7 + 9$$, but neither $$7$$ nor $$9$$ is divisible by $$8$$.

Yet $$16 = 8 + 8$$ splits into two multiples of $$8$$. So sometimes both summands are multiples of $$8$$ and sometimes they are not — the statement is not always true.

Let the number be $$n = 7k$$. Any multiple of $$n$$ is of the form $$mn$$ for some integer $$m$$. Then

$$mn = m \cdot 7k = 7(mk),$$

which is a multiple of $$7$$.

Example: $$14 = 7\times 2$$ is divisible by $$7$$, and any multiple of $$14$$ — like $$28, 42, 56, 70, \ldots$$ — is also divisible by $$7$$.

Factors of $$12$$ are $$1, 2, 3, 4, 6, 12$$. If $$n = 12k$$, then

$$n = 12k = 2(6k) = 3(4k) = 4(3k) = 6(2k) = 12k,$$

so $$n$$ is divisible by each of $$2, 3, 4, 6, 12$$ (and trivially by $$1$$).

Example: $$60 = 12 \times 5$$ — check $$60$$ is divisible by $$1, 2, 3, 4, 6, 12$$.

Take $$n = 7$$. It is divisible by $$7$$ but not by $$14$$ (a multiple of $$7$$). So the statement fails.

However, if we take $$n = 42$$, it is divisible by $$7$$ and also by $$14, 21, 42$$ — so it works for some cases.

Being divisible by $$7$$ does not force divisibility by larger multiples of $$7$$: for that we would additionally need extra factors.

The numbers $$9 = 3\times 3$$ and $$4 = 2\times 2$$ share no common prime factor (they are coprime). When a number is divisible by two coprime numbers, it is divisible by their product.

Since $$\gcd(9, 4) = 1$$ and $$9 \times 4 = 36$$, any number divisible by both $$9$$ and $$4$$ is also divisible by $$36$$.

Example: $$72$$ is divisible by $$4$$ ($$72 = 4 \times 18$$) and by $$9$$ ($$72 = 9 \times 8$$), and $$72 = 36 \times 2$$ — divisible by $$36$$.

$$6$$ and $$4$$ share the common factor $$2$$, so they are not coprime; the argument used above does not apply.

Take $$n = 12$$: divisible by $$6$$ ($$12 = 6 \times 2$$) and by $$4$$ ($$12 = 4 \times 3$$), but $$12$$ is not divisible by $$24$$.

The correct conclusion is that $$n$$ is divisible by the LCM, $$\text{lcm}(6,4) = 12$$, not by $$24$$.

Odd + even = odd. A multiple of $$6$$ is always even. So the sum can never be a multiple of $$6$$ — with one exception… actually there is no exception, since 'multiple of $$6$$' includes $$0, 6, 12, \ldots$$, all of which are even, and odd + even is always odd.

Even the simplest examples fail: $$1 + 2 = 3$$, $$3 + 4 = 7$$, $$5 + 6 = 11$$ — none are multiples of $$6$$.

A number leaves remainder $$3$$ on division by $$5$$ iff it is $$3$$ more than a multiple of $$5$$. So take any multiple of $$5$$ and add $$3$$:

$$0 + 3 = 3,\quad 5 + 3 = 8,\quad 10 + 3 = 13,\quad 15 + 3 = 18,\quad 20 + 3 = 23,\ldots$$

All such numbers form the list $$3, 8, 13, 18, 23, 28, 33, 38, 43, 48, \ldots$$ — each is $$5$$ more than the previous.

We want numbers of the form 'multiple of $$5$$, plus $$3$$'.

• (i) $$3k + 5$$: for $$k=1,2,3$$ gives $$8, 11, 14$$. But $$11$$ leaves remainder $$1$$ on division by $$5$$. Not correct.
• (ii) $$3k - 5$$: same problem — doesn't produce only 'remainder 3' numbers.
• (iii) $$\dfrac{3k}{5}$$: not even always an integer. Not correct.
• (iv) $$5k + 3$$: for $$k = 0, 1, 2, 3, \ldots$$ gives $$3, 8, 13, 18, \ldots$$ — every number is $$3$$ more than a multiple of $$5$$. Correct.
• (v) $$5k - 2$$: for $$k = 1, 2, 3, \ldots$$ gives $$3, 8, 13, 18, \ldots$$ — same list. Since $$5k - 2 = 5(k-1) + 3$$, this captures the same numbers. Correct.
• (vi) $$5k - 3$$: for $$k = 1, 2, 3, \ldots$$ gives $$2, 7, 12, \ldots$$ — these leave remainder $$2$$, not $$3$$. Not correct.

Substitute $$k = 1, 2, 3, 4, \ldots$$ into $$5k - 2$$:

The values are exactly $$3, 8, 13, 18, 23, \ldots$$ — every value is $$3$$ more than a multiple of $$5$$. Indeed $$5k - 2 = 5(k - 1) + 3$$, so $$5k - 2$$ generates precisely the same numbers as $$5k + 3$$.

Yes, many. Any expression of the form $$5k + (5m + 3)$$ for a fixed integer $$m$$ produces the same list, because $$5k + (5m+3) = 5(k+m) + 3$$.

Examples:

• $$5k + 8$$ — gives $$13, 18, 23, \ldots$$
• $$5k - 7$$ — gives $$-2, 3, 8, 13, \ldots$$
• $$5k + 13$$ — gives $$18, 23, 28, \ldots$$

In fact every expression $$5k + c$$ where $$c$$ itself leaves remainder $$3$$ when divided by $$5$$ works.

Let the four consecutive numbers be $$n,\ n+1,\ n+2,\ n+3$$. Their sum is

$$n + (n+1) + (n+2) + (n+3) = 4n + 6.$$

Set this equal to $$34$$:

$$4n + 6 = 34$$
$$4n = 28$$
$$n = 7.$$

So the four numbers are $$7, 8, 9, 10$$.

Check: $$7 + 8 + 9 + 10 = 34$$. $$\checkmark$$

If $$p$$ is the greatest of five consecutive numbers, the number just below it is $$p - 1$$, the next $$p - 2$$, and so on. So the five consecutive numbers, from smallest to largest, are:

$$p - 4,\ p - 3,\ p - 2,\ p - 1,\ p.$$

The other four numbers (apart from $$p$$) are $$p-1,\ p-2,\ p-3,\ p-4$$.

We want a number $$N$$ such that $$N = 3k + 2$$ and $$N = 4m + 2$$ simultaneously. Rearranging, $$N - 2$$ is a multiple of both $$3$$ and $$4$$, so $$N - 2$$ is a multiple of $$\text{lcm}(3, 4) = 12$$.

Hence $$N = 12t + 2$$ for some non-negative integer $$t$$.

Substituting $$t = 0, 1, 2, 3, \ldots$$ gives the numbers

$$2,\ 14,\ 26,\ 38,\ 50,\ 62, \ldots$$

Check ($$N = 14$$): $$14 = 3\times 4 + 2$$ and $$14 = 4\times 3 + 2$$. $$\checkmark$$

Let $$N$$ be the number of pebbles. The clues say:

• $$N$$ leaves remainder $$1$$ on division by $$3$$.
• $$N$$ leaves remainder $$1$$ on division by $$2$$.
• $$N$$ leaves remainder $$1$$ on division by $$5$$.
• $$N$$ is divisible by $$7$$ (no remainder).
• $$N \le 100$$.

The first three conditions say $$N - 1$$ is a multiple of $$2$$, $$3$$, and $$5$$, i.e. a multiple of $$\text{lcm}(2,3,5) = 30$$.

So $$N = 30k + 1$$ for some integer $$k$$. The candidates below $$100$$ are:

$$1,\ 31,\ 61,\ 91.$$

Now use the last condition — $$N$$ must be divisible by $$7$$:

• $$1 \div 7$$: remainder $$1$$. No.
• $$31 \div 7$$: remainder $$3$$. No.
• $$61 \div 7$$: remainder $$5$$. No.
• $$91 \div 7 = 13$$. Yes.

So $$N = 91$$.

Verify: $$91 = 3\cdot 30 + 1$$, so remainder $$1$$ on $$\div 3, \div 5, \div 2$$; and $$91 = 7\times 13$$ divides evenly. And $$91 < 100$$. $$\checkmark$$

Let three numbers of this form be $$6a + 2,\ 6b + 2,\ 6c + 2$$. Add them:

$$(6a + 2) + (6b + 2) + (6c + 2) = 6a + 6b + 6c + 6 = 6(a + b + c + 1).$$

This is a multiple of $$6$$.

Example: $$2 + 8 + 14 = 24 = 6 \times 4$$.
Another: $$8 + 20 + 26 = 54 = 6 \times 9$$.

So Tathagat is right — the sum of any three such numbers is always a multiple of $$6$$.

Observe a nice pattern: in each condition, the remainder is exactly $$1$$ less than the divisor.

• Remainder $$2$$ on division by $$3$$ means $$N + 1$$ is a multiple of $$3$$.
• Remainder $$3$$ on division by $$4$$ means $$N + 1$$ is a multiple of $$4$$.
• Remainder $$4$$ on division by $$5$$ means $$N + 1$$ is a multiple of $$5$$.

So $$N + 1$$ is a common multiple of $$3, 4, 5$$, i.e. a multiple of $$\text{lcm}(3, 4, 5) = 60$$.

The smallest positive such $$N$$ comes from the smallest positive multiple of $$60$$, namely $$N + 1 = 60$$, giving

$$N = 59.$$

Verify: $$59 = 3\times 19 + 2$$; $$59 = 4\times 14 + 3$$; $$59 = 5\times 11 + 4$$. $$\checkmark$$

Why smallest? Because $$N + 1$$ must be a common multiple of $$3, 4, 5$$, and the smallest positive common multiple is their LCM, $$60$$. Any smaller $$N + 1$$ would miss one of the divisibilities.

Any natural number can be written by separating its last few digits. For example, a $$4$$-digit number $$\overline{abcd}$$ is

$$\overline{abcd} = 1000a + 100b + 10c + d.$$

Divisibility by 2. $$1000a + 100b + 10c$$ is a multiple of $$2$$ (each term has factor $$10$$, hence $$2$$). So $$\overline{abcd}$$ is a multiple of $$2$$ iff $$d$$ is a multiple of $$2$$, i.e. the last digit is even.

Divisibility by 5. Similarly $$1000a + 100b + 10c$$ is a multiple of $$5$$. So $$\overline{abcd}$$ is a multiple of $$5$$ iff $$d \in \{0, 5\}$$.

Divisibility by 4. $$100b, 1000a$$ are multiples of $$4$$ (since $$100 = 4 \times 25$$ and $$1000 = 4 \times 250$$). So $$\overline{abcd} = (\text{multiple of }4) + 10c + d.$$ The number is divisible by $$4$$ iff the number formed by the last two digits, $$10c + d$$, is divisible by $$4$$.

Divisibility by 8. $$1000 = 8 \times 125$$, so $$1000a$$ is a multiple of $$8$$. Hence $$\overline{abcd} = 1000a + (100b + 10c + d)$$ is divisible by $$8$$ iff the number formed by the last three digits, $$100b + 10c + d$$, is divisible by $$8$$.

Use the shortcut: a number is divisible by $$9$$ iff the sum of its digits is divisible by $$9$$.

All five numbers are divisible by $$9$$.

Yes. If a number uses only the digits $$0$$ and $$9$$, the sum of its digits is $$9 \times (\text{number of } 9\text{s})$$, which is always a multiple of $$9$$. By the divisibility rule for $$9$$, the number itself is a multiple of $$9$$.

Examples: $$9009,\ 90000,\ 99900909,\ 909090$$ — each has a digit sum that is $$0, 9, 18, 27, \ldots$$, always a multiple of $$9$$.

$$10 = 9 \times 1 + 1$$, so $$10$$ is not divisible by $$9$$; the remainder is $$1$$.

Since $$100 = 99 + 1 = 9 \times 11 + 1$$, $$100$$ leaves remainder $$1$$ on division by $$9$$.

For any multiple of $$100$$, $$100k = 9(11k) + k$$, so $$100k$$ leaves the same remainder as $$k$$ does on division by $$9$$. In particular:

Pattern: $$100k$$ leaves remainder $$k$$ (mod $$9$$) when divided by $$9$$.

Split $$427 = 400 + 20 + 7$$.

• $$400 = 9 \times 44 + 4$$, remainder $$4$$.
• $$20 = 9 \times 2 + 2$$, remainder $$2$$.
• $$7$$ leaves remainder $$7$$.

Adding remainders: $$4 + 2 + 7 = 13$$, and $$13 = 9 + 4$$, so the total remainder is $$4$$.

Equivalently, the digit sum $$4 + 2 + 7 = 13$$ leaves remainder $$4$$ on division by $$9$$.

Verify: $$427 = 9 \times 47 + 4$$. $$\checkmark$$

Yes. Every power of $$10$$ leaves remainder $$1$$ on division by $$9$$:

$$10 = 9 + 1$$
$$100 = 99 + 1$$
$$1000 = 999 + 1$$
$$10000 = 9999 + 1$$

So for a general number $$\overline{a_n a_{n-1} \ldots a_1 a_0}$$ the place values $$10^k a_k$$ contribute a remainder of $$a_k$$ each. Adding all these remainders gives the digit sum $$a_0 + a_1 + \cdots + a_n$$. The full number and its digit sum have the same remainder when divided by $$9$$.

Example: $$54321$$ has digit sum $$5+4+3+2+1 = 15$$, which leaves remainder $$6$$ (since $$15 = 9 + 6$$). Indeed $$54321 = 9 \times 6035 + 6$$.

From the digit-sum rule we have proved: a number and the sum of its digits leave the same remainder on division by $$9$$. This gives an if-and-only-if:

$$9 \mid N \iff 9 \mid (\text{digit sum of }N).$$

1. Correct. If $$9 \mid N$$ then $$9 \mid$$ digit sum.
2. Correct. If $$9 \mid$$ digit sum then $$9 \mid N$$.
3. Correct. This is the contrapositive of (2): if $$9 \nmid N$$, then $$9 \nmid$$ digit sum.
4. Correct. This is the contrapositive of (1): if $$9 \nmid$$ digit sum, then $$9 \nmid N$$.

All four statements are correct. They are just the four equivalent forms (P⇒Q, Q⇒P, ¬P⇒¬Q, ¬Q⇒¬P) of the same biconditional.

We need a multiple of $$9$$ whose digits are all from $$\{0, 2, 4, 6, 8\}$$.

One-digit multiples of $$9$$: $$9$$ — odd digit. No good.

Two-digit multiples of $$9$$: $$18, 27, 36, 45, 54, 63, 72, 81, 90, 99$$. Each contains at least one odd digit ($$1, 3, 5, 7, 9$$). None work.

Three-digit multiples of $$9$$: we need digits from $$\{0, 2, 4, 6, 8\}$$ with digit sum a multiple of $$9$$. Try to make the smallest such number.

To be smallest, use the smallest possible number of digits and the smallest leading digits. With $$3$$ digits, we want digit sum $$9$$ or $$18$$. But $$9$$ is odd — we cannot get an odd sum from only even digits! So we need the digit sum $$= 18$$.

Now we want the smallest $$3$$-digit number using only even digits summing to $$18$$. To minimize the hundreds digit, try hundreds $$= 2$$: then remaining two digits sum to $$16$$ from $$\{0,2,4,6,8\}$$ — smallest option is $$88$$. Gives $$288$$. Check: $$2+8+8=18$$ ✓, and $$288 = 9\times 32$$. All digits even ✓.

Could hundreds be $$0$$? No — then it isn't a $$3$$-digit number. So $$288$$ is the smallest three-digit candidate. (Any $$2$$-digit candidate is impossible, as shown above.)

Divide $$6000$$ by $$9$$: $$6000 = 9 \times 666 + 6$$, so the nearest multiples of $$9$$ are

$$9 \times 666 = 5994 \quad \text{and} \quad 9 \times 667 = 6003.$$

Distances: $$6000 - 5994 = 6$$ and $$6003 - 6000 = 3$$. Since $$3 < 6$$, the closer one is $$6003$$.

Verify with digit sums: $$5994: 5+9+9+4 = 27$$ ✓; $$6003: 6+0+0+3 = 9$$ ✓.

Find the smallest multiple of $$9$$ that is $$\ge 4300$$. Since $$4300 \div 9 = 477.\overline{7}$$, the smallest multiple $$\ge 4300$$ is $$9 \times 478 = 4302$$.

Find the largest multiple of $$9$$ that is $$\le 4400$$. Since $$4400 \div 9 = 488.\overline{8}$$, the largest multiple $$\le 4400$$ is $$9 \times 488 = 4392$$.

So the multiples of $$9$$ in the range are $$9 \times 478, 9 \times 479, \ldots, 9 \times 488$$. The count is

$$488 - 478 + 1 = 11.$$

So there are $$11$$ multiples of $$9$$ between $$4300$$ and $$4400$$ (both inclusive).

Compute the remainders of powers of $$10$$ modulo $$3$$:

$$10 = 3\times 3 + 1$$   (remainder $$1$$)
$$100 = 3\times 33 + 1$$   (remainder $$1$$)
$$1000 = 3\times 333 + 1$$   (remainder $$1$$)
and in general $$10^k = 3 \cdot \underbrace{33\cdots 3}_{k} + 1$$, remainder $$1$$.

So every power of $$10$$ leaves remainder $$1$$ when divided by $$3$$. A number $$N = a_n \cdot 10^n + \cdots + a_1 \cdot 10 + a_0$$ therefore leaves the same remainder as $$a_n + \cdots + a_1 + a_0$$, its digit sum, on division by $$3$$. In particular, $$3 \mid N$$ iff $$3 \mid (\text{digit sum})$$.

Example: for $$471$$, digit sum $$= 12$$, divisible by $$3$$; check $$471 = 3 \times 157$$ ✓.

Powers of $$10$$ mod $$11$$:

$$10 = 11\times 1 - 1$$   (remainder $$-1$$, i.e. $$10$$).
$$100 = 11\times 9 + 1$$   (remainder $$1$$).
$$1000 = 11\times 91 - 1$$   (remainder $$-1$$).

So $$10^k$$ alternates between $$+1$$ and $$-1$$ (mod $$11$$).

For $$462 = 4\cdot 100 + 6\cdot 10 + 2$$, the remainder is $$4\cdot 1 + 6\cdot(-1) + 2\cdot 1 = 4 - 6 + 2 = 0$$. Hence $$462$$ is divisible by $$11$$.

Check: $$462 = 11 \times 42$$. $$\checkmark$$

Take the digits of the number starting from the units place. Alternately assign them $$+$$ and $$-$$ signs (units $$+$$, tens $$-$$, hundreds $$+$$, …) and add. Call this the alternating digit sum. The number is divisible by $$11$$ iff the alternating digit sum is divisible by $$11$$ (including $$0$$).

Equivalently: subtract the sum of digits in odd positions (from the right) from the sum of digits in even positions (from the right), and check whether the difference is a multiple of $$11$$.

Example: for $$9152$$, going from the units, alternating sum $$= 2 - 5 + 1 - 9 = -11$$, which is a multiple of $$11$$, so $$9152$$ is divisible by $$11$$ ($$9152 = 11 \times 832$$).

Because $$10^k$$ mod $$11$$ is $$(-1)^k$$, the number itself leaves the same remainder as its alternating digit sum when divided by $$11$$.

So if the alternating digit sum is $$0$$ or a (positive or negative) multiple of $$11$$, the number leaves remainder $$0$$ — i.e. it is exactly divisible by $$11$$.

If the alternating digit sum is some value $$d$$ with $$-10 \le d \le 10$$, the remainder on division by $$11$$ is $$d$$ if $$d \ge 0$$, or $$d + 11$$ if $$d < 0$$.

Similar in spirit but different in detail.

• For $$3$$ and $$9$$, every power of $$10$$ leaves remainder $$1$$, so the shortcut is 'add all digits'.
• For $$11$$, powers of $$10$$ leave remainders that alternate between $$+1$$ and $$-1$$, so the shortcut is 'alternately add and subtract the digits' — the alternating digit sum.

In both cases, we replace each digit by the remainder of its place-value on division by the target, then check that combined value.

Quick recipe: for each number check the shortcut once — last digit (for $$2, 5, 10$$), last two digits (for $$4$$), last three digits (for $$8$$), digit sum (for $$3, 9$$), both $$2$$ and $$3$$ (for $$6$$), alternating digit sum from the units (for $$11$$).

For instance, for $$275$$: last digit $$5$$ ⇒ divisible by $$5$$, not $$2$$; digit sum $$14$$ ⇒ not divisible by $$3$$ or $$9$$; last two $$75$$ ⇒ not divisible by $$4$$; alt. sum $$5 - 7 + 2 = 0$$ ⇒ divisible by $$11$$.

Note. The example in the textbook shows $$128$$ as 'No' under $$4$$; that is a slip — since $$128 = 4 \times 32$$, the correct entry is 'Yes'.

$$6 = 2 \times 3$$, and $$2$$ and $$3$$ are coprime, so a number is divisible by $$6$$ iff it is divisible by both $$2$$ and $$3$$.

Rule: a number is divisible by $$6$$ iff its last digit is even and the sum of its digits is divisible by $$3$$.

Since $$\gcd(2, 3) = 1$$, a number is divisible by $$6$$ iff divisible by both $$2$$ and $$3$$. Let's test each:

Every prediction matches the actual division, so the '$$2$$ and $$3$$' shortcut correctly detects divisibility by $$6$$.

$$24 = 4 \times 6$$, but $$\gcd(4, 6) = 2 \ne 1$$, so $$4$$ and $$6$$ are not coprime. A number can be divisible by both $$4$$ and $$6$$ yet fail to be divisible by $$24$$.

Counter-example: $$12$$ is divisible by $$4$$ ($$12 = 4 \times 3$$) and by $$6$$ ($$12 = 6 \times 2$$), but $$12$$ is not divisible by $$24$$. Another: $$36$$ is divisible by $$4$$ and $$6$$ but $$36 = 24 \times 1 + 12$$, not divisible by $$24$$.

The correct choice: write $$24 = 8 \times 3$$ with $$\gcd(8, 3) = 1$$. So $$n$$ is divisible by $$24$$ iff it is divisible by both $$8$$ and $$3$$.

The digital root is obtained by repeatedly adding the digits until we reach a single-digit number ($$1$$–$$9$$). At each step the digit sum has the same remainder as the original number, when divided by $$9$$. So the digital root is congruent to the original number modulo $$9$$.

Property: the digital root of a positive integer $$N$$ equals the remainder $$N$$ leaves on division by $$9$$, with the sole exception that a remainder of $$0$$ corresponds to a digital root of $$9$$.

Consequence: a number is divisible by $$9$$ iff its digital root is $$9$$; a number is divisible by $$3$$ iff its digital root is one of $$3, 6, 9$$.

A number's digital root cycles through $$1, 2, \ldots, 9$$ as the number increases by $$1$$ each time. So we compute the digital root of $$600$$ and step forward.

Digital root of $$600$$: digit sum $$6+0+0 = 6$$. So $$600 \to 6$$. Then $$601 \to 7,\ 602 \to 8,\ 603 \to 9,\ 604 \to 1,\ 605 \to 2,\ 606 \to 3,\ 607 \to 4,\ 608 \to 5,\ 609 \to 6, \ldots$$, and every $$9$$th number has the same digital root.

(i) Digital root $$5$$. First candidate is $$608$$. Adding $$9$$ each time (up to $$700$$):

$$608,\ 617,\ 626,\ 635,\ 644,\ 653,\ 662,\ 671,\ 680,\ 689,\ 698.$$

(ii) Digital root $$7$$. First candidate is $$601$$. Then

$$601,\ 610,\ 619,\ 628,\ 637,\ 646,\ 655,\ 664,\ 673,\ 682,\ 691.$$

(iii) Digital root $$3$$. First candidate is $$606$$. Then

$$606,\ 615,\ 624,\ 633,\ 642,\ 651,\ 660,\ 669,\ 678,\ 687,\ 696.$$

Take the $$12$$ consecutive numbers $$34, 35, 36, \ldots, 45$$. Their digital roots:

Observation: Digital roots cycle through $$1, 2, \ldots, 9$$ repeatedly. In $$12$$ consecutive numbers we go through one full cycle of $$9$$ and then start again — so exactly $$3$$ of the digital roots must appear twice (here, $$7, 8, 9$$).

(i) Multiples of $$3$$. The digital roots of $$3, 6, 9, 12, 15, 18, 21, 24, 27, 30, \ldots$$ are $$3, 6, 9, 3, 6, 9, 3, 6, 9, 3, \ldots$$ — a repeating cycle $$(3, 6, 9)$$.

(ii) Multiples of $$4$$. The digital roots of $$4, 8, 12, 16, 20, 24, 28, 32, 36, 40, \ldots$$ are $$4, 8, 3, 7, 2, 6, 1, 5, 9, 4, \ldots$$ — a repeating cycle of length $$9$$ hitting every digit $$1$$ through $$9$$.

(iii) Multiples of $$6$$. The digital roots of $$6, 12, 18, 24, 30, 36, 42, \ldots$$ are $$6, 3, 9, 6, 3, 9, 6, \ldots$$ — a cycle $$(6, 3, 9)$$, same three digits as multiples of $$3$$.

Pattern: the digital roots of multiples of $$k$$ form a cycle whose length is $$\dfrac{9}{\gcd(k, 9)}$$. For $$k = 3, 6$$: cycle length $$3$$. For $$k = 4$$: cycle length $$9$$.

Numbers $$1$$ more than a multiple of $$6$$: $$7, 13, 19, 25, 31, 37, 43, 49, 55, 61, \ldots$$

Digital roots: $$7, 4, 1, 7, 4, 1, 7, 4, 1, 7, \ldots$$ — cycle $$(7, 4, 1)$$.

Why? Numbers of the form $$6k + 1$$ increase by $$6$$ each time, so their remainders mod $$9$$ increase by $$6$$ each step (mod $$9$$). Starting from digital root of $$7$$ ($$= 7$$), we get $$7, 7+6=13 \to 4, 4+6=10 \to 1, 1+6=7$$ and cycle back.

So the cycle has length $$3$$, exactly $$\dfrac{9}{\gcd(6, 9)} = \dfrac{9}{3} = 3$$ steps, same as the cycle length for multiples of $$6$$.

Decode the clues:

• 'Each digit tiniest and odd' — the tiniest odd digit is $$1$$, so every digit of the number is $$1$$.
• 'The largest odd single-digit I proudly claim' — the largest odd single-digit is $$9$$; this is what the digital root equals.
• 'My digits count, their sum, my root … All point to one bold number' — the count of digits, the sum of the digits, and the digital root are all $$9$$.

If every digit is $$1$$ and there are $$9$$ digits, then the sum of digits is $$1 \times 9 = 9$$, and the digital root is $$9$$. That fits every clue.

So the number is

$$\boxed{111\,111\,111}$$

— 'nine ones'.

Adding $$1$$ to a number increases its digital root by $$1$$ (wrapping $$9 \to 1$$). Adding $$10 = 9 + 1$$ increases the digital root by $$1$$, because adding $$9$$ doesn't change it and adding $$1$$ increases it by $$1$$.

Since the original digital root is $$5$$, the digital root of $$N + 10$$ is $$5 + 1 = 6$$.

Adding $$11 = 9 + 2$$ increases the digital root by $$2$$ (mod $$9$$), because adding $$9$$ leaves the digital root unchanged.

Start with, say, $$5$$. Then the sequence $$5, 16, 27, 38, 49, 60, 71, 82, 93, 104, \ldots$$ has digital roots

$$5, 7, 9, 2, 4, 6, 8, 1, 3, 5, 7, \ldots$$

The digital roots go up by $$2$$ each time (wrapping past $$9$$). After $$9$$ steps the cycle repeats. Since $$\gcd(2, 9) = 1$$, the digital roots hit every digit $$1$$–$$9$$ exactly once before the pattern repeats.

The digital root of $$N$$ equals $$N$$ mod $$9$$ (with $$0$$ shown as $$9$$).

Now $$9a$$ is a multiple of $$9$$ ⇒ remainder $$0$$ mod $$9$$. $$36b = 9\cdot 4b$$ is also a multiple of $$9$$ ⇒ remainder $$0$$ mod $$9$$. So

$$9a + 36b + 13 \equiv 13 \equiv 4 \pmod 9.$$

Hence the digital root of $$9a + 36b + 13$$ is $$4$$ (independent of $$a$$ and $$b$$).

(i) $$A1 + 1B = B0$$. Units: $$1 + B$$ ends in $$0$$ ⇒ $$B = 9$$ with carry $$1$$. Tens: $$A + 1 + 1 = B = 9$$ ⇒ $$A = 7$$. Check: $$71 + 19 = 90$$ ✓.

(ii) $$AB + 37 = 6A$$. Units: $$B + 7$$ ends in $$A$$; so $$A = B + 7$$ (if no carry) or $$A = B - 3$$ (if carry $$1$$). Tens: $$A + 3 + c = 6$$, i.e. $$A + c = 3$$ where $$c \in \{0,1\}$$. Case $$c = 0$$: $$A = 3$$, so $$B + 7 = 3$$ — impossible. Case $$c = 1$$: $$A = 2$$, and $$B + 7 = 12$$ ⇒ $$B = 5$$. Check: $$25 + 37 = 62$$ ✓.

(iii) $$ON + ON + ON = PO$$, i.e. $$3 \times \overline{ON} = \overline{PO}$$. Since $$\overline{PO}$$ is two-digit, $$3 \cdot \overline{ON} \le 99$$, so $$\overline{ON} \le 33$$. Also $$\overline{ON} \ge 10$$. The units of $$3N$$ must equal $$O$$: $$3N \equiv O \pmod{10}$$. And the tens digit of $$\overline{PO}$$ is $$P$$.

Try small $$\overline{ON}$$ with tens digit $$O$$: for $$\overline{ON} = 15$$, $$3\times 15 = 45$$, so $$O = 4, P = 4$$? But $$O = 1$$ originally — no. Instead we need the tens digit of $$\overline{ON}$$ to match the units digit of $$3 \cdot \overline{ON}$$.

Enumerate $$O \in \{1, 2, 3\}$$:

• $$O = 1$$: need $$3N$$ end in $$1$$ ⇒ $$N = 7$$. Then $$\overline{ON} = 17$$, $$3 \times 17 = 51 = \overline{PO}$$ ⇒ $$P = 5, O = 1$$ ✓.
• $$O = 2$$: need $$3N$$ end in $$2$$ ⇒ $$N = 4$$. Then $$\overline{ON} = 24$$, $$3 \times 24 = 72 = \overline{PO}$$ ⇒ $$O = 2, P = 7$$ ✓.
• $$O = 3$$: need $$3N$$ end in $$3$$ ⇒ $$N = 1$$. Then $$\overline{ON} = 31$$, $$3 \times 31 = 93 = \overline{PO}$$ ⇒ $$O = 3, P = 9$$ ✓.

All three assignments work: $$(O, N, P) = (1, 7, 5), (2, 4, 7), (3, 1, 9)$$.

(iv) $$3 \times \overline{QR} = \overline{PRR}$$. Units: $$3R$$ ends in $$R$$ ⇒ $$2R \equiv 0 \pmod{10}$$ ⇒ $$R \in \{0, 5\}$$.

• $$R = 0$$: $$3 \times \overline{Q0} = \overline{P00}$$. Then $$30Q = 100P$$, i.e. $$3Q = 10P$$. No integer solution with $$P \ne 0$$ and digits.
• $$R = 5$$: $$3 \times \overline{Q5} = \overline{P55}$$. Now $$3 \times \overline{Q5} \ge 100$$ ⇒ $$\overline{Q5} \ge 34$$, and $$\overline{Q5} \le 333$$, so $$Q \in \{4, 5, 6, 7, 8, 9\}$$. Check each: $$3\times 45 = 135$$ (ends $$35$$, no); $$3 \times 55 = 165$$ (no); $$3\times 65 = 195$$ (no); $$3\times 75 = 225$$ (no); $$3 \times 85 = 255$$ — ends in $$55$$, so $$P = 2, R = 5, Q = 8$$ ✓; $$3 \times 95 = 285$$ (no).

Solution: $$Q = 8, R = 5, P = 2$$; check $$85 + 85 + 85 = 255$$ ✓.

$$\overline{PQ}$$ is a two-digit number, so $$10 \le \overline{PQ} \le 99$$. The product $$\overline{RS}$$ must also be two-digit, so $$\overline{PQ} \times 8 \le 99$$, giving $$\overline{PQ} \le 12$$.

Also since different letters denote different digits, $$P \ne Q$$. The candidates are:

• $$\overline{PQ} = 10$$: $$10 \times 8 = 80$$. Then $$Q = 0$$ and $$S = 0$$ — same digit for different letters. Rejected.
• $$\overline{PQ} = 11$$: $$P = Q = 1$$. Rejected.
• $$\overline{PQ} = 12$$: $$12 \times 8 = 96$$. So $$P = 1, Q = 2, R = 9, S = 6$$ — all distinct ✓.

So $$\overline{PQ} = 12$$ and $$\overline{RS} = 96$$.

Let $$\overline{GH} \times H = \overline{9K}$$. Expanding, $$(10G + H) \cdot H = 90 + K$$, so $$10GH + H^2 = 90 + K$$. The product is between $$90$$ and $$99$$, so we scan each $$H$$.

• $$H = 1$$: $$\overline{G1}\cdot 1 = \overline{G1}$$, need $$G = 9$$ ⇒ $$K = 1 = H$$. Rejected.
• $$H = 2$$: $$20G + 4 \in [90,99]$$ has no integer solution.
• $$H = 3$$: $$30G + 9 \in [90,99]$$ ⇒ $$G = 3$$ ⇒ $$G = H$$. Rejected.
• $$H = 4$$: $$40G + 16 \in [90,99]$$ ⇒ $$G = 2$$, product $$96$$, so $$K = 6$$. Distinct ✓.
• $$H = 5$$: $$50G + 25 \in [90,99]$$ — no integer solution.
• $$H = 6$$: $$60G + 36 \in [90,99]$$ ⇒ $$G = 1$$, product $$96$$, $$K = 6 = H$$. Rejected.
• $$H \ge 7$$: no integer solutions with $$G \ge 1$$.

Unique valid solution: $$G = 2,\ H = 4,\ K = 6$$, i.e. $$24 \times 4 = 96$$.

Anshu's reasoning is right. If $$B \ge 2$$, then $$\overline{BYE} \ge 200$$, and $$6 \times 200 = 1200$$ — a $$4$$-digit number, not a $$3$$-digit $$\overline{RAY}$$. So $$B = 1$$.

Also $$B \ne 0$$, because $$\overline{BYE}$$ has to be a genuine $$3$$-digit number.

So the puzzle becomes $$\overline{1YE} \times 6 = \overline{RAY}$$, with $$\overline{1YE} \le 166$$ (to keep the product below $$1000$$).

Two things constrain $$Y$$:

1. $$Y$$ is the tens digit of $$\overline{1YE}$$ and must satisfy $$\overline{1YE} \le 166$$. So $$Y \in \{0, 1, 2, 3, 4, 5, 6\}$$.
2. $$Y$$ is also the units digit of $$\overline{RAY} = 6 \times \overline{1YE}$$, so $$Y \equiv 6E \pmod{10}$$, i.e. $$Y$$ equals the units digit of $$6E$$.

Since $$Y \ne 1 (=B)$$, and looking at the units of $$6E$$ for $$E = 0,1,\ldots,9$$: units are $$0, 6, 2, 8, 4, 0, 6, 2, 8, 4$$. So $$Y \in \{0, 2, 4, 6, 8\}$$ — $$Y$$ is always even.

Combining with $$Y \le 6$$: $$Y \in \{0, 2, 4, 6\}$$. Odd digits and $$8, 9$$ are impossible.

Systematically checking each valid $$E$$ (with distinct digits for all letters) narrows to the unique solution $$E = 5,\ Y = 0,\ 1YE = 105,\ 105 \times 6 = 630$$, giving $$R = 6, A = 3$$.

Verify: $$105 \times 6 = 630$$. Letters $$B=1, Y=0, E=5, R=6, A=3$$ — all distinct ✓.

Digit sum of $$\overline{31z5}$$ is $$3 + 1 + z + 5 = 9 + z$$. For divisibility by $$9$$, we need $$9 + z$$ to be a multiple of $$9$$, so $$z \in \{0, 9\}$$.

Both fit since $$z$$ is a single digit.

Verify: $$3105 = 9 \times 345$$ ✓ and $$3195 = 9 \times 355$$ ✓.

Why two answers? Divisibility by $$9$$ only fixes the digit sum modulo $$9$$. Two single digits, $$0$$ and $$9$$, are congruent to $$0$$ mod $$9$$, so both make the digit sum a multiple of $$9$$.

Let the first number be $$N_1 = 12a + 8$$ and the second be $$N_2 = 12b - 4$$. Their sum:

$$N_1 + N_2 = 12a + 8 + 12b - 4 = 12(a + b) + 4.$$

For divisibility by $$8$$: $$12(a+b) + 4 = 4[\,3(a+b) + 1\,]$$. This is a multiple of $$8$$ iff $$3(a+b) + 1$$ is even iff $$3(a+b)$$ is odd iff $$a + b$$ is odd.

Test examples:

• $$N_1 = 8$$ (so $$a = 0$$), $$N_2 = 8$$ (so $$b = 1$$): $$a + b = 1$$ odd. Sum $$= 16 = 8 \times 2$$ ✓.
• $$N_1 = 20$$ (so $$a = 1$$), $$N_2 = 8$$ (so $$b = 1$$): $$a + b = 2$$ even. Sum $$= 28$$, and $$28 = 8 \times 3 + 4$$ — not a multiple of $$8$$.

So Snehal's claim is not always true — it holds only when $$a + b$$ is odd.

A multiple of $$3$$ is either a multiple of $$6$$ (even) or of the form $$6k + 3$$ (odd). So a multiple of $$3$$ has two 'flavours':

• Type E (even): $$3m$$ with $$m$$ even, i.e. $$6k$$.
• Type O (odd): $$3m$$ with $$m$$ odd, i.e. $$6k + 3$$.

Adding two multiples of $$3$$:

1. Even + Even: $$6a + 6b = 6(a+b)$$. Multiple of $$6$$. ✓
2. Odd + Odd: $$(6a + 3) + (6b + 3) = 6(a + b) + 6 = 6(a + b + 1)$$. Multiple of $$6$$. ✓
3. Even + Odd: $$6a + (6b + 3) = 6(a + b) + 3$$. Not a multiple of $$6$$.

Generalisation: the sum of two multiples of $$3$$ is a multiple of $$6$$ iff both are of the same parity (either both even multiples or both odd multiples of $$3$$). Otherwise the sum is only a multiple of $$3$$, not of $$6$$.

Examples: $$6 + 12 = 18$$ (both even, div by $$6$$) ✓; $$3 + 9 = 12$$ (both odd, div by $$6$$) ✓; $$3 + 6 = 9$$ (mixed, not div by $$6$$).

$$18 = 2 \times 9$$ with $$\gcd(2, 9) = 1$$, so $$18 \mid N$$ iff $$2 \mid N$$ and $$9 \mid N$$.

Divisibility by 2: the last digit $$b$$ must be even, so $$b \in \{0, 2, 4, 6, 8\}$$.

Divisibility by 9: the digit sum $$4 + 8 + a + 2 + 3 + b = 17 + a + b$$ must be a multiple of $$9$$. Since $$a + b \le 18$$, we need $$a + b \in \{1, 10\}$$ (so that $$17 + a + b$$ is $$18$$ or $$27$$).

Combine both conditions:

So the pairs are $$(a, b) = (1, 0), (8, 2), (6, 4), (4, 6), (2, 8)$$ — five pairs in all.

$$44 = 4 \times 11$$ with $$\gcd(4, 11) = 1$$, so $$44 \mid N$$ iff $$4 \mid N$$ and $$11 \mid N$$.

Divisibility by 4: last two digits $$\overline{q8} = 10q + 8$$ must be a multiple of $$4$$. Testing $$q = 0, 1, \ldots, 9$$: $$08, 28, 48, 68, 88$$ are multiples of $$4$$. So $$q \in \{0, 2, 4, 6, 8\}$$.

Divisibility by 11: alternating digit sum from units of $$\overline{3p7q8}$$ is $$8 - q + 7 - p + 3 = 18 - p - q$$. Since $$p + q \in [0, 18]$$, this alternating sum lies in $$[0, 18]$$. Multiples of $$11$$ in this range: $$0$$ and $$11$$. So $$p + q \in \{7, 18\}$$.

Combine:

So the valid pairs are $$(p, q) = (7, 0),\ (5, 2),\ (3, 4),\ (1, 6)$$.

Verify one: $$37708 = 44 \times 857$$ ✓.

Let the three consecutive numbers be $$n,\ n+1,\ n+2$$. The conditions are:

$$n \equiv 0 \pmod 2,\quad n+1 \equiv 0 \pmod 3,\quad n+2 \equiv 0 \pmod 4.$$

Rewriting: $$n \equiv 0 \pmod 2$$, $$n \equiv 2 \pmod 3$$, $$n \equiv 2 \pmod 4$$.

The last one ($$n \equiv 2 \pmod 4$$) is stronger than $$n \equiv 0 \pmod 2$$ — it already forces $$n$$ to be even, in fact of the form $$4k + 2$$. Combined with $$n \equiv 2 \pmod 3$$ and $$\gcd(4, 3) = 1$$, the Chinese Remainder Theorem gives

$$n \equiv 2 \pmod{12}.$$

Simplest example: $$n = 2$$ gives the triple $$(2, 3, 4)$$.

Next examples: $$n = 14, 26, 38, 50, \ldots$$ give $$(14, 15, 16), (26, 27, 28), (38, 39, 40), (50, 51, 52), \ldots$$

They occur once every $$12$$ integers.

Approach. Find the smallest multiple of $$36$$ at least $$45\,000$$, then repeatedly add $$36$$.

Divide $$45\,000 \div 36 = 1250$$ exactly, so $$45\,000 = 36 \times 1250$$ is itself a multiple of $$36$$. Now add $$36$$ five times:

$$45\,036,\ 45\,072,\ 45\,108,\ 45\,144,\ 45\,180.$$

All of these are between $$45\,000$$ and $$47\,000$$, and each is a multiple of $$36$$.

Verify one: $$45\,108 = 36 \times 1253$$; digit sum $$4+5+1+0+8 = 18$$ (div by $$9$$), and last two digits $$08$$ div by $$4$$ — so div by $$36$$ ✓.

Consecutive even numbers differ by $$2$$. If the middle one is $$5p$$, then subtracting $$2$$s from the left and adding $$2$$s to the right gives the sequence:

$$5p - 4,\ 5p - 2,\ 5p,\ 5p + 2,\ 5p + 4.$$

The other four (excluding the middle) are $$5p - 4,\ 5p - 2,\ 5p + 2,\ 5p + 4$$.

Set up the conditions. Let the number be $$\overline{a_5 a_4 a_3 a_2 a_1 a_0}$$.

• Divisible by $$15 = 3 \times 5$$: last digit $$a_0 \in \{0, 5\}$$ and digit sum divisible by $$3$$.
• Reversed number $$\overline{a_0 a_1 a_2 a_3 a_4 a_5}$$ divisible by $$6 = 2 \times 3$$: its last digit $$a_5$$ must be even, and digit sum divisible by $$3$$ (same as original).

Since the reversed number should be a valid $$6$$-digit number, $$a_0 \ne 0$$; so $$a_0 = 5$$. And $$a_5$$ (the leading digit of the original) must be even and non-zero: $$a_5 \in \{2, 4, 6, 8\}$$. Finally the digit sum must be divisible by $$3$$.

Pick one: $$a_5 = 2, a_4 = 1, a_3 = 0, a_2 = 0, a_1 = 1, a_0 = 5$$ gives

$$N = 210015.$$

Digit sum $$= 2+1+0+0+1+5 = 9$$, divisible by $$3$$. Last digit $$5$$ — divisible by $$5$$. So $$N$$ is divisible by $$15$$ ($$210015 = 15 \times 14001$$).

Reversal: $$510012$$. Last digit $$2$$ (even), digit sum $$9$$ (divisible by $$3$$). So the reversal is divisible by $$6$$ ($$510012 = 6 \times 85002$$) ✓.

If $$N$$ is a multiple of $$11$$, write $$N = 11k$$. Then

$$2N = 2 \times 11k = 11 \times (2k),$$

which is again a multiple of $$11$$. So every multiple of $$11$$, when doubled, stays a multiple of $$11$$.

Examples: $$11 \to 22,\ 22 \to 44,\ 33 \to 66,\ 121 \to 242, \ldots$$ — each doubled result is $$11 \times$$(even integer).

Deepak's conjecture is false — no multiple of $$11$$ 'loses' its divisibility when doubled.

Every integer leaves one of the remainders $$0, 1$$, or $$2$$ on division by $$3$$. So write the three numbers as $$a = 3p + r_1,\ b = 3q + r_2,\ c = 3s + r_3$$ with $$r_1, r_2, r_3 \in \{0, 1, 2\}$$.

Their sum is $$a + b + c = 3(p + q + s) + (r_1 + r_2 + r_3)$$, so $$3 \mid (a + b + c)$$ iff $$3 \mid (r_1 + r_2 + r_3)$$.

Check every unordered combination of remainders:

Generalisation. The sum of three integers is divisible by $$3$$ iff either

• all three leave the same remainder on division by $$3$$ (all in $$3\mathbb Z$$, all in $$3\mathbb Z + 1$$, or all in $$3\mathbb Z + 2$$), or
• the three remainders are exactly $$0, 1, 2$$ (one of each).

Two consecutive integers. Of $$n$$ and $$n+1$$, exactly one is even. Their product $$n(n+1)$$ therefore has a factor of $$2$$ — always a multiple of $$2$$.

Three consecutive. $$n(n+1)(n+2)$$. Among three consecutive integers, at least one is a multiple of $$2$$ and exactly one is a multiple of $$3$$. So the product is a multiple of $$2 \times 3 = 6$$. Examples: $$1\cdot 2\cdot 3 = 6$$; $$4 \cdot 5 \cdot 6 = 120 = 6 \times 20$$.

Four consecutive. $$n(n+1)(n+2)(n+3)$$. Among $$4$$ consecutive integers there are exactly $$2$$ even numbers, and one of the two evens is a multiple of $$4$$. So the product has at least $$4 \times 2 = 8$$ as a factor. Plus, one of the four is a multiple of $$3$$. So the product is a multiple of $$8 \times 3 = 24$$. Examples: $$1\cdot 2\cdot 3\cdot 4 = 24$$; $$2\cdot 3\cdot 4\cdot 5 = 120 = 24 \times 5$$.

Five consecutive. Extending the reasoning: among $$5$$ consecutive integers there is at least one multiple of $$5$$, one multiple of $$3$$, and (as with $$4$$ consecutive) at least two evens with one of them a multiple of $$4$$. So the product is a multiple of $$5 \times 3 \times 8 = 120$$. Example: $$1\cdot 2\cdot 3\cdot 4\cdot 5 = 120$$; $$3\cdot 4\cdot 5\cdot 6\cdot 7 = 2520 = 120 \times 21$$.

General pattern: the product of $$k$$ consecutive integers is always a multiple of $$k! = 1 \cdot 2 \cdot 3 \cdots k$$.

Every multiple of $$32$$ is $$32k = 8 \times (4k)$$, so it's a multiple of $$8$$. Every multiple of $$8$$ is $$8m = 4 \times (2m)$$, so it's a multiple of $$4$$.

The reverse inclusions fail: $$4$$ is a multiple of $$4$$ but not of $$8$$; $$8$$ is a multiple of $$8$$ but not of $$32$$.

So $$\{\text{multiples of }32\} \subset \{\text{multiples of }8\} \subset \{\text{multiples of }4\}$$ — the smaller set is contained strictly inside the larger.

This is captured by three nested circles with Multiples of $$4$$ outermost, Multiples of $$8$$ in the middle, and Multiples of $$32$$ innermost — option (iv).

Option (iii) has the nesting backwards, and options (i) and (ii) show only partial overlaps or a disjoint circle, which don't match the actual containment.