Chapter 6: Algebra Play

Let $$x$$ be the number that is chosen. Track each step algebraically.

Step 2 (double it): $$2x$$.
Step 3 (add four): $$2x + 4$$.
Step 4 (divide by two): $$\dfrac{2x + 4}{2} = x + 2$$.
Step 5 (subtract the original): $$(x + 2) - x = 2$$.

Every $$x$$ has cancelled out, leaving just $$2$$. So no matter what starting number you pick, the final value is always $$2$$.

Try it: start with $$7 \to 14 \to 18 \to 9 \to 2$$. Start with $$100 \to 200 \to 204 \to 102 \to 2$$. Every time we get $$2$$.

Replace "Add four" (Step 3) with "Add $$k$$" and rerun the algebra with starting number $$x$$:

Step 2: $$2x$$; Step 3: $$2x + k$$; Step 4: $$\dfrac{2x + k}{2} = x + \dfrac{k}{2}$$; Step 5: $$\left(x + \dfrac{k}{2}\right) - x = \dfrac{k}{2}$$.

So the final value is $$\dfrac{k}{2}$$.

To make the answer $$3$$, set $$\dfrac{k}{2} = 3$$, giving $$k = 6$$: replace Step 3 with "Add six".
To make the answer $$5$$, set $$\dfrac{k}{2} = 5$$, giving $$k = 10$$: replace Step 3 with "Add ten".

Yes. Any sequence of operations whose net algebraic effect on the starting number $$x$$ is a constant will work. Design the steps by first choosing the constant you want, then arranging multiplications, additions, divisions and a final subtraction of $$x$$ (or a multiple of $$x$$) so that everything with $$x$$ cancels.

Here is one example:

1. Think of a number: $$x$$.
2. Multiply by $$3$$: $$3x$$.
3. Add $$21$$: $$3x + 21$$.
4. Divide by $$3$$: $$x + 7$$.
5. Multiply by $$2$$: $$2x + 14$$.
6. Subtract twice the original number: $$2x + 14 - 2x = 14$$.

No matter which starting number is used, the answer is $$14$$.

Let the month be $$M$$ and the day be $$D$$. Track the trick algebraically.

Multiply month by $$5$$: $$5M$$.
Add $$6$$: $$5M + 6$$.
Multiply by $$4$$: $$20M + 24$$.
Add $$9$$: $$20M + 33$$.
Multiply by $$5$$: $$100M + 165$$.
Add the day: $$100M + 165 + D$$.

So the final answer is $$100M + D + 165$$. Rearranging, $$100M + D = \text{final answer} - 165$$.

For Mukta's answer $$291$$: $$291 - 165 = 126 = 100 \times 1 + 26$$. Because the day $$D$$ is at most $$31$$ (a two-digit number), its digits occupy the ones and tens places, while everything to the left of them gives the month.

So $$M = 1$$ (January) and $$D = 26$$, i.e. the date is $$26/01$$.

Using the formula from the previous problem, $$\text{final answer} = 100M + D + 165$$, so $$100M + D = \text{final answer} - 165$$.

Here $$1390 - 165 = 1225 = 100 \times 12 + 25$$.

So $$M = 12$$ (December) and $$D = 25$$. The date is $$25/12$$ — Christmas Day.

Yes. Suppose we replace "Add $$6$$" (Step 3) with "Add $$p$$" and "Add $$9$$" (Step 5) with "Add $$q$$". Track the algebra:

Multiply month by $$5$$: $$5M$$; add $$p$$: $$5M + p$$; multiply by $$4$$: $$20M + 4p$$; add $$q$$: $$20M + 4p + q$$; multiply by $$5$$: $$100M + 20p + 5q$$; add the day: $$100M + D + 20p + 5q$$.

The constant part is $$20p + 5q$$. Subtracting it from the final answer gives back $$100M + D$$, from which the day (last two digits) and month (remaining digits) can be read off.

For the original trick, $$p = 6$$ and $$q = 9$$, so the constant is $$20 \times 6 + 5 \times 9 = 120 + 45 = 165$$ — which is exactly what Shubham subtracted.

For instance, if you change to $$p = 5, q = 7$$, subtract $$20 \times 5 + 5 \times 7 = 135$$ instead of $$165$$.

The recipe: design the steps so that after subtracting the original number (or a multiple of it), all the $$x$$'s cancel and only a constant remains.

Example trick. Ask a friend to:

1. Think of a number: $$x$$.
2. Multiply it by $$4$$: $$4x$$.
3. Add $$28$$: $$4x + 28$$.
4. Divide by $$4$$: $$x + 7$$.
5. Subtract the original number: $$(x + 7) - x = 7$$.

The answer is always $$7$$. By choosing the constant in Step 3 to be any multiple of $$4$$, you can make the trick end in any value you like.

A slightly trickier example. Think of $$x \to +5 \to x + 5 \to \times 3 \to 3x + 15 \to -3x \to 15 \to /3 \to 5$$. Final answer: always $$5$$.

Work from the top downwards, using "top $$=$$ sum of the two cells below" reversed as "missing cell $$=$$ top $$-$$ known cell below".

Second row. The top is $$10 = 4 + \square$$, so the right cell of the second row is $$10 - 4 = 6$$.

Bottom row. The left cell of the second row is $$4 = 1 + \square$$, so the middle bottom cell is $$4 - 1 = 3$$. The right cell of the second row is $$6 = 3 + \square$$, so the right bottom cell is $$6 - 3 = 3$$.

Completed pyramid: top $$10$$; second row $$4, \; 6$$; bottom row $$1, \; 3, \; 3$$.

Because the top depends on all three bottom entries, the cleanest starting point is to use a letter for the missing bottom cell and set up an equation.

Let the missing middle bottom cell be $$c$$.

Second row: $$12 + c$$ (left) and $$c + 8$$ (right).
Top: $$(12 + c) + (c + 8) = 20 + 2c$$.

Given the top is $$60$$:

So the middle bottom cell is $$20$$. The second row becomes $$12 + 20 = 32$$ and $$20 + 8 = 28$$.

Completed pyramid: top $$60$$; second row $$32, \; 28$$; bottom row $$12, \; 20, \; 8$$.

Start with the simplest case: a two-row pyramid whose bottom cells are $$a$$ and $$b$$. The top cell is the sum of the two cells below it, so

For a two-row pyramid the top is just the plain sum of the two bottom entries. Each bottom entry contributes with coefficient $$1$$.

Let the bottom row be $$a, \; b, \; c$$.

Middle row: $$a + b$$ and $$b + c$$.
Top: $$(a + b) + (b + c) = a + 2b + c$$.

So for a three-row pyramid, the top equals $$a + 2b + c$$. The coefficients of the bottom entries are $$1, \; 2, \; 1$$, which is the second row of Pascal's triangle: the middle entry counts twice because it feeds both cells of the middle row.

Let the bottom row be $$a, \; b, \; c, \; d$$. Fill upwards.

Third row: $$a + b, \; b + c, \; c + d$$.
Second row: $$(a + b) + (b + c) = a + 2b + c$$ and $$(b + c) + (c + d) = b + 2c + d$$.
Top: $$(a + 2b + c) + (b + 2c + d) = a + 3b + 3c + d$$.

So the topmost cell equals

The coefficients $$1, \; 3, \; 3, \; 1$$ are the third row of Pascal's triangle.

The first three Virahāṅka-Fibonacci (VF) numbers are $$1, \; 2, \; 3$$. Put them in the bottom row.

Middle row: $$1 + 2 = 3$$ and $$2 + 3 = 5$$.
Top: $$3 + 5 = 8$$.

The numbers that appear in the grid are $$1, \; 2, \; 3, \; 3, \; 5, \; 8$$. Each distinct value here — $$1, \; 2, \; 3, \; 5, \; 8$$ — is itself one of the first few VF numbers.

Why? Because the VF rule is "each term is the sum of the two before it", the sum $$1 + 2$$ is the next VF number ($$3$$), and the sum $$2 + 3$$ is the one after that ($$5$$), and $$3 + 5 = 8$$ is the one after that.

Denote the VF sequence by $$V_1, V_2, V_3, \ldots$$ with $$V_k = V_{k-1} + V_{k-2}$$, so consecutive VF numbers sum to the next one.

Bottom row: $$V_1, V_2, \ldots, V_n$$.
Row 2: $$V_1 + V_2 = V_3, \; V_2 + V_3 = V_4, \; \ldots, \; V_{n-1} + V_n = V_{n+1}$$, giving $$V_3, V_4, \ldots, V_{n+1}$$.
Row 3: similarly $$V_5, V_6, \ldots, V_{n+2}$$.
…
Row $$k$$ (from the bottom): $$V_{2k-1}, V_{2k}, \ldots, V_{n+k-1}$$.

So every entry in the pyramid is a VF number. The topmost cell (row $$n$$) has just one entry, namely

In words: the top of an $$n$$-row VF pyramid is the $$(2n - 1)$$th VF number.

Yes. Because a calendar has $$7$$ columns, the four numbers of any $$2 \times 2$$ grid are related in a fixed way. Let the top-left entry be $$a$$. Then

The sum of the four cells is

If your friend tells you the sum $$S$$, you solve $$4a + 16 = S$$ to get $$a = \dfrac{S - 16}{4}$$, and then the other three numbers are $$a + 1, \; a + 7, \; a + 8$$.

Check with the example: $$S = 40 \Rightarrow a = \dfrac{40 - 16}{4} = 6$$, so the grid is $$6, 7, 13, 14$$. ✓

Using $$\text{sum} = 4a + 16$$ with sum $$36$$:

The other three numbers are $$a + 1 = 6, \; a + 7 = 12, \; a + 8 = 13$$. So the grid is

Here are a couple of examples.

Trick 1 — $$3 \times 3$$ grid. If the top-left entry of a $$3 \times 3$$ calendar grid is $$a$$, the nine numbers are

Their sum is $$9a + 72 = 9(a + 8)$$. Notice $$a + 8$$ is the middle entry! So if your friend reports the sum $$S$$, the middle number is $$\dfrac{S}{9}$$, from which the whole grid can be reconstructed.

Trick 2 — a row of 4 consecutive days. If the leftmost is $$a$$, the four numbers are $$a, a+1, a+2, a+3$$, summing to $$4a + 6$$. From the sum $$S$$, $$a = \dfrac{S - 6}{4}$$ gives the whole row.

Trick 3 — a plus (+) shape. The five cells (a middle cell and its four orthogonal neighbours) are $$a - 7, \; a - 1, \; a, \; a + 1, \; a + 7$$, summing to $$5a$$ — five times the middle entry. Divide by $$5$$ to reveal the middle.

There are $$3! = 6$$ ways to place the three digits. List them and compute:

The largest product is $$32 \times 5 = 160$$.

Group the six options by their multiplier (the one-digit factor). For a fixed multiplier, whichever two-digit multiplicand is larger gives the larger product.

• Multiplier $$2$$: compare $$35 \times 2$$ and $$53 \times 2$$. Since $$53 > 35$$, keep $$53 \times 2$$.
• Multiplier $$3$$: compare $$25 \times 3$$ and $$52 \times 3$$. Since $$52 > 25$$, keep $$52 \times 3$$.
• Multiplier $$5$$: compare $$23 \times 5$$ and $$32 \times 5$$. Since $$32 > 23$$, keep $$32 \times 5$$.

So the largest is among $$53 \times 2, \; 52 \times 3, \; 32 \times 5$$.

Clearly $$52 \times 3 = 156 > 106 = 53 \times 2$$. So compare $$52 \times 3$$ with $$32 \times 5$$ by expanding:

The first terms are the same, and the second term is larger for $$32 \times 5$$. So the largest of all six products is

Let the three distinct digits be $$p < q < r$$. There are six arrangements. Group by multiplier and use "larger tens digit wins for a fixed multiplier" (because increasing the tens digit by $$1$$ adds $$10 \times \text{multiplier}$$ to the product, whereas increasing the units digit only adds the multiplier itself):

• Multiplier $$p$$: best is $$rq \times p$$.
• Multiplier $$q$$: best is $$rp \times q$$.
• Multiplier $$r$$: best is $$qp \times r$$.

Now compare these three candidates using algebra.

Compare $$rp \times q$$ with $$rq \times p$$:

So $$rp \times q > rq \times p$$, which eliminates the multiplier-$$p$$ option.

Compare $$qp \times r$$ with $$rp \times q$$:

The first term $$10qr$$ is common. Since $$r > q$$ and $$p > 0$$, we have $$pr > pq$$, so

Therefore the largest product is

i.e. the largest digit $$r$$ is the multiplier and the other two digits are arranged in decreasing order ($$q$$ in the tens place, $$p$$ in the units place) to form the multiplicand. This confirms the rule holds for all choices of three distinct digits.

Apply the rule from Intext 21: the largest digit is the multiplier and the other two digits are placed in decreasing order in the multiplicand.

Digits in increasing order: $$p = 1, \; q = 3, \; r = 7$$. So the largest product is $$qp \times r = 31 \times 7$$.

Sanity check against the other candidates: $$71 \times 3 = 213$$ and $$73 \times 1 = 73$$, both smaller than $$217$$.

Apply the rule: largest digit as the multiplier, other two digits in decreasing order as the multiplicand.

Digits in increasing order: $$p = 3, \; q = 5, \; r = 9$$. So the largest product is $$qp \times r = 53 \times 9$$.

Sanity check: $$93 \times 5 = 465$$ and $$95 \times 3 = 285$$, both smaller than $$477$$.

Yes, there is never any remainder. To see why, use algebra.

Let the two-digit number have tens digit $$a$$ and units digit $$b$$, so it equals $$10a + b$$. The reverse has tens digit $$b$$ and units digit $$a$$, so it equals $$10b + a$$.

Assume $$b > a$$ (so the reverse is larger). Their difference is

The result is exactly $$9$$ times $$(b - a)$$, so dividing by $$9$$ leaves no remainder.

Try more examples: $$83 - 38 = 45 = 9 \times 5$$; $$92 - 29 = 63 = 9 \times 7$$.

Now the original $$ab = 10a + b$$ is larger than the reverse $$ba = 10b + a$$. Subtract the smaller from the larger:

Again the difference is a multiple of $$9$$, so dividing by $$9$$ leaves no remainder. The quotient is $$a - b$$ instead of $$b - a$$; both are positive because we always subtract the smaller from the larger.

Example: $$74 - 47 = 27 = 9 \times 3$$ and here $$a = 7, b = 4$$, so $$a - b = 3$$. ✓

Let the tens digit of the original number be $$a$$ and the units digit be $$b$$. Then the original is $$10a + b$$ and the reverse is $$10b + a$$.

The difference (larger minus smaller) is $$9 \lvert b - a \rvert$$, so dividing by $$9$$ gives the quotient

In words, the quotient is simply the (positive) difference of the two digits of the original number.

Examples:

• $$47 \to 74 - 47 = 27, \; 27 \div 9 = 3$$. Digits are $$4$$ and $$7$$, difference $$3$$. ✓
• $$92 \to 92 - 29 = 63, \; 63 \div 9 = 7$$. Digits are $$9$$ and $$2$$, difference $$7$$. ✓
• $$83 \to 83 - 38 = 45, \; 45 \div 9 = 5$$. Digits are $$8$$ and $$3$$, difference $$5$$. ✓

Let the original number have tens digit $$a$$ and units digit $$b$$, so it equals $$10a + b$$. The reverse is $$10b + a$$. Their sum is

Because the sum is $$11 \times (a + b)$$, it is always divisible by $$11$$. So yes, the claim holds for every two-digit starting number.

Check with the examples:

• $$31 + 13 = 44 = 11 \times 4$$ and $$a + b = 3 + 1 = 4$$. ✓
• $$28 + 82 = 110 = 11 \times 10$$ and $$a + b = 2 + 8 = 10$$. ✓
• $$12 + 21 = 33 = 11 \times 3$$ and $$a + b = 1 + 2 = 3$$. ✓

As a bonus, the quotient after dividing by $$11$$ is $$a + b$$: the sum of the digits.

Write out the three cyclic numbers in expanded form:

Adding them:

Divisibility by $$37$$. Notice that $$111 = 3 \times 37$$. So

which is clearly a multiple of $$37$$.

Divisibility by $$3$$. The same expression $$3 \times 37 \times (a + b + c)$$ is also a multiple of $$3$$. So yes, the sum is always divisible by $$3$$ as well.

(In fact the sum is always a multiple of $$111$$.)

The final quotient always comes out to be the original three-digit number $$abc$$ itself. Let us see why.

Write $$abc$$ (as a 3-digit number) $$= N = 100a + 10b + c$$. Then

Now factor $$1001$$:

So

Dividing successively by $$7$$, then by $$11$$, then by $$13$$ therefore gives

i.e. the original 3-digit number.

Try: $$abc = 273$$. Then $$abcabc = 273273$$. Compute $$273273 \div 7 = 39039$$, then $$39039 \div 11 = 3549$$, then $$3549 \div 13 = 273$$. ✓

Let the starting number of flowers be $$x$$ and let him place $$k$$ flowers at each shrine.

Pond 1 & shrine 1. After dipping, he has $$2x$$; he places $$k$$ and is left with $$2x - k$$.

Pond 2 & shrine 2. After dipping the remaining, he has $$2(2x - k) = 4x - 2k$$; he places $$k$$ and is left with $$4x - 3k$$.

Pond 3 & shrine 3. After dipping again, he has $$2(4x - 3k) = 8x - 6k$$; he places all of these at shrine 3, and this must equal $$k$$ (the common number placed at every shrine).

For $$x$$ and $$k$$ to be whole numbers, $$x$$ must be a multiple of $$7$$. The smallest positive solution is $$x = 7$$ and $$k = 8$$.

Verify with $$x = 7, k = 8$$: $$7 \to 14 \to 6 \to 12 \to 4 \to 8 \to 8$$ placed at shrine 3. ✓

So he started with $$7$$ flowers and placed $$8$$ flowers at each shrine.

Method 1: using algebra.

Let the number of horses be $$h$$ and the number of hens be $$H$$. Each horse has $$4$$ legs and each hen has $$2$$ legs, so

Divide the leg equation by $$2$$: $$2h + H = 75$$. Subtract the head equation:

Then $$H = 55 - 20 = 35$$.

Method 2: without letter-numbers (following the hint).

Suppose all $$55$$ animals were hens. Then the total legs would be $$55 \times 2 = 110$$. But the actual leg count is $$150$$, which is $$150 - 110 = 40$$ more.

Every time we swap one hen for one horse, the head count stays the same but the leg count goes up by $$4 - 2 = 2$$. To add $$40$$ legs, we need $$40 \div 2 = 20$$ swaps.

So there are $$20$$ horses and $$55 - 20 = 35$$ hens.

Let the daughter's present age be $$d$$ years. Then the mother's present age is $$5d$$ years.

In $$6$$ years the daughter will be $$d + 6$$ and the mother will be $$5d + 6$$. The mother will then be $$3$$ times her daughter's age:

Expand the right side:

Subtract $$3d$$ from both sides:

So the daughter is $$6$$ years old now (and the mother is $$30$$).

Check: in $$6$$ years the daughter will be $$12$$ and the mother $$36$$, and $$36 = 3 \times 12$$. ✓

Let Gauri have $$g$$ cows. Then Naina has $$2g$$ cows (from the first statement).

If Naina gives $$3$$ cows to Gauri, Naina is left with $$2g - 3$$ and Gauri now has $$g + 3$$. These two numbers are equal:

Subtract $$g$$ from both sides:

So Gauri has $$6$$ cows and Naina has $$2 \times 6 = 12$$ cows.

Check: after the exchange, Gauri would have $$6 + 3 = 9$$ and Naina would have $$12 - 3 = 9$$. ✓

Compute each fraction.

Every fraction equals $$\dfrac{1}{3}$$. Let us see why using algebra.

The $$n$$th fraction has, in its numerator, the first $$n$$ odd numbers $$1, 3, 5, \ldots, (2n - 1)$$, and, in its denominator, the next $$n$$ odd numbers $$(2n + 1), (2n + 3), \ldots, (4n - 1)$$.

Sum of the first $$n$$ odd numbers. A well-known identity:

So the numerator equals $$n^2$$.

Denominator. The denominator is the sum of the first $$2n$$ odd numbers minus the sum of the first $$n$$ odd numbers:

Therefore each fraction is

So the observation is that the pattern continues forever: every fraction of this form equals $$\dfrac{1}{3}$$.