Why does abcabc always divide evenly by 7, 11, and 13?

The 6-digit number abcabc always equals the 3-digit number abc multiplied by 1001, and 1001 factors exactly as 7 x 11 x 13, so dividing by all three in turn returns the original 3-digit number.

Class 8 Maths NCERT Solutions Part 2 Chapter 6: Algebra Play (Ganita Prakash, Part 2)

Key Concepts & Formulae at a Glance

Think of a Number tricks can be decoded by replacing the chosen number with a letter, such as \(x\), and simplifying the resulting expression step by step.
Number pyramid rule: each cell equals the sum of the two cells directly below it. For a 3-row pyramid with bottom row \(a,b,c\): top \(=a+2b+c\). For a 4-row pyramid with bottom row \(a,b,c,d\): top \(=a+3b+3c+d\).
Calendar grid (2×2): if the top-left number is \(a\), the four numbers are \(a,\,a+1,\,a+7,\,a+8\), and their sum is \(4a+16\).
Largest product rule: to make the biggest product from three given digits arranged as (2-digit) × (1-digit), put the largest digit as the 1-digit multiplier, and arrange the other two digits in decreasing order as the 2-digit multiplicand.
Reverse-and-subtract trick: for a 2-digit number \(ab\) (digits \(a,b\)), the difference between it and its digit-reversal \(ba\) is always a multiple of 9, namely \(9(b-a)\) or \(9(a-b)\).
Reverse-and-add trick: the sum of a 2-digit number and its digit-reversal is always a multiple of 11, namely \(11(a+b)\).
3-digit cyclic-sum trick: for a 3-digit number \(abc\), the sum \(abc+bca+cab\) always equals \(111(a+b+c)\), which is always divisible by both 37 and 3.
6-digit repeat trick: a number formed by repeating a 3-digit number, \(abcabc\), always equals \(abc\times1001=abc\times7\times11\times13\).

\[\text{3-row pyramid: } a+2b+c \qquad \text{4-row pyramid: } a+3b+3c+d \qquad \text{Calendar grid: } 4a+16\]

6.2 Thinking about 'Think of a Number' Tricks

In Grade 7, we learned about 'Think of a Number' tricks. Here is one: think of a number, double it, add four, divide by two, then subtract the original number. Using algebra, if the chosen number is \(x\): double it gives \(2x\); add four gives \(2x+4\); divide by 2 gives \(x+2\); subtract the original number gives \(x+2-x=2\). No matter the starting number, the result is always 2.

MTMath Talk — I predict you get 2. Am I right? Try it out with different starting numbers. Do you always end up with the same value, 2? Why?

Trying a few starting numbers confirms the prediction every time:

Start (x)	Double (2x)	Add 4	÷ 2	Subtract x
3	6	10	5	2
10	20	24	12	2
−5	−10	−6	−3	2

Algebraically, the working \(x\to2x\to2x+4\to x+2\to(x+2)-x=2\) shows the original letter-number \(x\) cancels out completely in the last step, leaving only the constant 2.

Yes — the answer is always 2, for every starting number, because the algebra shows \(x\) cancels out and only the constant term survives.

MTMath Talk — How would you change this game to make the final answer 3? What about 5?

The final constant is always half of whatever is added in the "add" step, since that step is the only one contributing a constant (doubling and subtracting \(x\) always cancel to leave just that constant divided by 2).

To get 3: double the number, add 6 instead of 4, divide by 2, subtract the original number. Check: \(x\to2x\to2x+6\to x+3\to(x+3)-x=3\).

To get 5: double the number, add 10, divide by 2, subtract the original number. Check: \(x\to2x\to2x+10\to x+5\to(x+5)-x=5\).

Replace "add four" with "add six" to always get 3, or with "add ten" to always get 5 — in general, "add \(2k\)" always gives a final answer of \(k\).

MTMath Talk — Can you come up with more complicated steps that always lead to the same final value?

Yes. For example: think of a number \(x\); multiply by 3; add 12; divide by 3; subtract the original number. Working through: \(x\to3x\to3x+12\to x+4\to(x+4)-x=4\). This always gives 4, however complicated the steps look, because the multiply/divide steps are designed to exactly cancel and leave only a fixed constant.

Example: "multiply by 3, add 12, divide by 3, subtract the original number" always gives 4 — any set of steps works as long as the final subtraction removes every trace of \(x\).

Let us now look at a different trick of this type — a calendar date-guessing trick. Shubham asks Mukta to think of a date, multiply the month by 5, add 6, multiply by 4, add 9, multiply by 5, then add the day. Mukta thinks of 26/01 (Republic Day) and reports her final answer as 291.

MTMath Talk — How did Shubham figure out the date chosen by Mukta?

Let the month be \(M\) and the day be \(D\). Following the steps in algebra:

Step	Expression
Multiply \(M\) by 5	\(5M\)
Add 6	\(5M+6\)
Multiply by 4	\(20M+24\)
Add 9	\(20M+33\)
Multiply by 5	\(100M+165\)
Add the day	\(100M+165+D\)

Mukta's answer was 291, so \(291=100M+165+D\), giving \(291-165=126=100M+D\). Since \(D\) is a day within a month, it is at most 31 and needs only 2 digits, so the last 2 digits of 126 are \(D\), and what comes before is \(M\): here \(M=1\) and \(D=26\).

Shubham subtracts 165 from the answer; the last two digits of the result give the day, and whatever remains before that gives the month. Here that is the 26th of January.

MTMath Talk — Mukta thinks of another date, follows the same steps, and reports her answer as 1390. What date did Mukta start with this time?

Subtracting 165 from 1390 gives \(1390-165=1225\). The last two digits are the day (25) and what remains is the month (12).

1225 → month 12, day 25 — the date she thought of was 25th December.

Figure it Out — Set 1

Three date-finding questions using the calendar trick from page 137 of the textbook.

1Find the date if the final answer is 1269.

Subtracting 165: \(1269-165=1104\). The last two digits give the day and the remaining digits give the month: month \(=11\), day \(=04\).

The date is 4th November (04/11).

2Find the date if the final answer is 394.

Subtracting 165: \(394-165=229\). The last two digits give the day and the remaining digit gives the month: month \(=2\), day \(=29\).

The date is 29th February (29/02).

3Find the date if the final answer is 296.

Subtracting 165: \(296-165=131\). The last two digits give the day and the remaining digit gives the month: month \(=1\), day \(=31\).

The date is 31st January (31/01).

MTMath Talk — Can you change the steps in this trick and still find the original date? Instead of subtracting 165 from the final answer, you might have to subtract some other number.

Yes — any set of steps of the form "multiply \(M\) by some number, do some arithmetic, multiply by another number, then add \(D\)" works, as long as the final constant added to \(100M+D\) is known so it can be subtracted off at the end. For example, using "multiply the month by 4, add 8, multiply by 5, add the day" gives \(20M+40+D\), which is not of the form \(100M+\text{const}+D\), so the multipliers must be chosen so the month's coefficient becomes 100 (a power of ten with two zeros, since the day needs 2 digits) — e.g. any pair of multipliers whose product is 100, such as "multiply by 4, then multiply by 25", or "multiply by 20, then multiply by 5". Whatever fixed constant results from the "add" steps just needs to be subtracted at the end.

Yes — keep the two multipliers' product equal to 100 (so the month lands in the hundreds place while the day, at most 2 digits, sits in the last two places), and subtract whatever constant the "add" steps contribute.

MTMath Talk — Try to devise your own 'Think of a Number' trick.

Example: think of a number \(x\); add 5; multiply by 2; subtract 4; divide by 2; subtract the original number. Working through: \(x\to x+5\to2x+10\to2x+6\to x+3\to(x+3)-x=3\). This always gives 3, whatever number is chosen.

Any trick works as long as the operations reduce, after full simplification, to \(x+k\) followed by "subtract \(x\)" — the answer will always be the constant \(k\).

6.3 Number Pyramids

In a number pyramid, each number is the sum of the two numbers directly below it. For example, with bottom row 1, 9, 4: the row above is \(1+9=10\) and \(9+4=13\), and the top is \(10+13=23\).

MTMath Talk — Use the same rule to fill these pyramids: bottom rows 6, 2; then 3, 4, 3; then 5, 4, 5, 0.

Bottom 6, 2: top \(=6+2=8\).

Bottom 3, 4, 3: middle row \(=3+4=7\) and \(4+3=7\); top \(=7+7=14\).

Bottom 5, 4, 5, 0: row of 3 \(=5+4,\ 4+5,\ 5+0=9,9,5\); row of 2 \(=9+9,\ 9+5=18,14\); top \(=18+14=32\).

Tops: 8; 14; and 32, found by repeatedly adding adjacent pairs from the bottom upward.

MTMath Talk — How do we fill this pyramid: top 10, second row 4 and a blank, bottom row 1 and two blanks?

Working from the known cells: the bottom-left is 1 and the cell above-left is 4, so the bottom's second cell must be \(4-1=3\) (since \(1+3=4\)). The top is 10 and the known second-row cell is 4, so the other second-row cell is \(10-4=6\) (since \(4+6=10\)). Finally, the bottom's third cell must satisfy \(3+{?}=6\), so it is \(6-3=3\).

Bottom row: 1, 3, 3. Middle row: 4, 6. Top: 10.

What about filling in the numbers in this pyramid: top 60, second row both blank, bottom row 12, blank, 8? Let us fill the empty boxes with letter-numbers \(a\), \(b\) (second row) and \(c\) (bottom middle).

From the pyramid rule: \(a+b=60\), \(12+c=a\), \(c+8=b\). Adding the last two: \((12+c)+(c+8)=a+b=60\), so \(20+2c=60\), giving \(2c=40\) and \(c=20\). Once \(c\) is known, \(a=12+20=32\) and \(b=20+8=28\).

MTMath Talk — Fill the following pyramids: (a) top 50, one middle cell 22, bottom 4, blank, 6, blank; (b) top 40, one middle cell 9, bottom 5, blank, 7, blank; (c) top 35, one middle cell 7, bottom 3, 5, blank, blank.

Each pyramid has a bottom row of 4 numbers, a row of 3 above it, a row of 2 above that, and the top. Working from the known clue in the row of 3 back down to the bottom row, then forward up to the top, gives a consistent solution for all three:

(a) Bottom: 4, 4, 6, 16 → row of 3: 8, 10, 22 → row of 2: 18, 32 → top: 50.

(b) Bottom: 5, 4, 7, 2 → row of 3: 9, 11, 9 → row of 2: 20, 20 → top: 40.

(c) Bottom: 3, 5, 2, 11 → row of 3: 8, 7, 13 → row of 2: 15, 20 → top: 35.

If the exact positions of the given/blank cells in your printed copy differ slightly from the layout above, re-apply the same method: use the pyramid rule to work outward from whichever cells are known, solving one unknown at a time.

Tops confirmed as 50, 40, and 35 respectively, each consistent with the given middle-row clue.

What is the relationship between the numbers in the bottom row and the number at the top? Let us start with the simplest pyramid, with bottom row \(a,b\): the top is simply \(a+b\).

What about a pyramid with three rows? Using letter-numbers \(a,b,c\) for the bottom row, the middle row is \(a+b,\ b+c\), and the top row is \((a+b)+(b+c)=a+2b+c\).

a+2b+c

a+b

b+c

Figure it Out — Set 2

Six questions from pages 140–141 of the textbook, on number pyramids and the Virahanka-Fibonacci sequence.

1Without building the entire pyramid, find the number in the topmost row given the bottom row in each case: (i) 4, 13, 8 (ii) 7, 11, 3 (iii) 10, 14, 25.

Using the 3-row pyramid formula, top \(=a+2b+c\):

(i) \(4+2(13)+8=4+26+8=38\).

(ii) \(7+2(11)+3=7+22+3=32\).

(iii) \(10+2(14)+25=10+28+25=63\).

(i) 38 (ii) 32 (iii) 63.

2Write an expression for the topmost row of a pyramid with 4 rows in terms of the values in the bottom row.

Let the bottom row be \(a,b,c,d\). The row above is \(a+b,\ b+c,\ c+d\). The row above that is \((a+b)+(b+c)=a+2b+c\) and \((b+c)+(c+d)=b+2c+d\). The top is the sum of these two: \((a+2b+c)+(b+2c+d)\).

\[\text{Top}=a+3b+3c+d\]

Top \(=a+3b+3c+d\), where \(a,b,c,d\) is the bottom row.

3Without building the entire pyramid, find the number in the topmost row given the bottom row in each case: (i) 8, 19, 21, 13 (ii) 7, 18, 19, 6 (iii) 9, 7, 5, 11.

Using the 4-row pyramid formula, top \(=a+3b+3c+d\):

(i) \(8+3(19)+3(21)+13=8+57+63+13=141\).

(ii) \(7+3(18)+3(19)+6=7+54+57+6=124\).

(iii) \(9+3(7)+3(5)+11=9+21+15+11=56\).

(i) 141 (ii) 124 (iii) 56.

4If the first three Virahanka-Fibonacci numbers are written in the bottom row of a number pyramid with three rows, fill in the rest of the pyramid. What numbers appear in the grid? What is the number at the top? Are they all Virahanka-Fibonacci numbers?

The Virahanka-Fibonacci sequence is 1, 2, 3, 5, 8, 13, 21, ..., where each number is the sum of the two before it. The first three terms are 1, 2, 3.

Bottom row: 1, 2, 3. Middle row: \(1+2=3\), \(2+3=5\). Top: \(3+5=8\).

The numbers appearing in the grid are 1, 2, 3, 3, 5, 8 (with 3 repeated once), and the top is 8.

Every single number in the pyramid — 1, 2, 3, 3, 5, and 8 — is itself a Virahanka-Fibonacci number. The top of the pyramid, 8, is the 5th term of the sequence.

5What can you say about the numbers in the pyramid and the number at the top in these cases? (i) The first four Virahanka-Fibonacci numbers in a four-row pyramid. (ii) The first 29 Virahanka-Fibonacci numbers in a 29-row pyramid.

(i) Bottom row: 1, 2, 3, 5. Row of 3: \(1+2,\ 2+3,\ 3+5=3,5,8\). Row of 2: \(3+5,\ 5+8=8,13\). Top: \(8+13=21\).

Every number appearing in this pyramid (1, 2, 3, 5, 3, 5, 8, 8, 13, 21) is again a Virahanka-Fibonacci number, and the top, 21, is the 7th term of the sequence — that is, term number \(2(4)-1=7\).

(ii) The same pattern continues: with the first 29 Virahanka-Fibonacci numbers on the bottom row of a 29-row pyramid, every number that appears anywhere in the pyramid will again be a Virahanka-Fibonacci number, and the number at the top will be the \(2(29)-1=57\)th term of the sequence — an extremely large Fibonacci number.

(i) Top \(=21=\)the 7th Virahanka-Fibonacci number. (ii) Top \(=\)the 57th Virahanka-Fibonacci number; every entry in either pyramid is itself a term of the sequence.

6If the bottom row of an \(n\) row pyramid contains the first \(n\) Virahanka-Fibonacci numbers, what can we say about the numbers in the pyramid? What can we say about the number at the top?

Every number that appears anywhere inside such a pyramid is itself a Virahanka-Fibonacci number. This happens because each entry in the pyramid is a sum of Fibonacci numbers with binomial (Pascal's triangle) coefficients, and Fibonacci-type sequences satisfy the identity \(\sum_{k=0}^{m}\binom{m}{k}F(k+1)=F(2m+1)\) (checked above for \(m=2,3\), giving tops 8 and 21).

Every number in the pyramid is a Virahanka-Fibonacci number, and the number at the top is always the \((2n-1)\)th term of the Virahanka-Fibonacci sequence.

6.4 Fun with Grids

Calendar Magic

A friend picks a 2 × 2 grid from a calendar page, adds the 4 numbers in the grid, and reports the sum. For example, picking 6, 7, 13, 14 from August 2025 gives \(6+7+13+14=40\).

MTMath Talk — Can we find the 4 numbers in the grid from just knowing this sum?

Let \(a\) be the top-left number of a 2×2 calendar grid. Since dates in the next column are one more, and dates in the row below are 7 more (a week later), the grid is:

a+1

a+7

a+8

Adding all four numbers: \(a+(a+1)+(a+7)+(a+8)=4a+16\).

Suppose the sum is 36: \(4a+16=36\), so \(4a=20\) (subtracting 16 from both sides), giving \(a=5\) (dividing both sides by 4). The other three numbers are \(a+1=6\), \(a+7=12\), \(a+8=13\).

Yes — from the sum, solve \(4a+16=\text{sum}\) for \(a\), then the grid is \(a,\ a+1,\ a+7,\ a+8\). For sum 36, the grid is 5, 6, 12, 13.

MTMath Talk — Create your own calendar trick. For instance, choose a grid of a different size and shape.

Example: a 1×3 horizontal strip of three consecutive dates in the same row, say \(a,\ a+1,\ a+2\). Their sum is \(3a+3\). Given the sum, solve for \(a\) by subtracting 3 and dividing by 3.

Another example: a 3×3 grid centred on a date \(a\) (with the grid extending one column left/right and one row above/below). By symmetry, the sum of all 9 dates is exactly \(9a\), so the sum divided by 9 gives the centre date directly.

Any grid shape works, as long as every cell can be expressed as the corner (or centre) value plus a fixed offset — then the total sum is a known multiple of that value plus a constant, which can always be solved backwards.

Algebra Grids

In these grids, shapes represent numbers, and in each row the last column is the sum of the values to its left.

🟦

🔴

🟦

Row 1: 🟦 + 🟦 + 🟦 \(=27\), so 🟦 \(=9\). Row 2: 🔴 + 🔴 + 🟦 \(=19\), so \(2\times\)🔴\(+9=19\), giving \(2\times\)🔴\(=10\), so 🔴 \(=5\).

MTMath Talk — In the following grids, find the values of the shapes and fill in the empty squares.

Grid 1 (square \(=s\), circle \(=c\)): Row 1: \(s+s+c=27\) → \(2s+c=27\). Row 2: \(c+c+s=21\) → \(2c+s=21\). Row 3 (to find): \(c+s+c=2c+s\).

From Row 1: \(c=27-2s\). Substituting into Row 2: \(2(27-2s)+s=21\Rightarrow54-4s+s=21\Rightarrow54-3s=21\Rightarrow3s=33\Rightarrow s=11\). Then \(c=27-2(11)=5\).

Row 3 total: \(2c+s=2(5)+11=21\).

Square \(=11\), circle \(=5\), and the empty Row 3 total is 21.

Grid 2 (blue circle \(=b\), purple diamond \(=d\)): Row 1: \(b+d+d=18\) → \(b+2d=18\). Row 2: \(d+b+b=15\) → \(d+2b=15\).

From Row 1: \(b=18-2d\). Substituting into Row 2: \(d+2(18-2d)=15\Rightarrow d+36-4d=15\Rightarrow-3d=-21\Rightarrow d=7\). Then \(b=18-2(7)=4\).

Blue circle \(=4\), purple diamond \(=7\). Use these values with the same row-total rule to fill any further empty rows in your printed copy.

6.5 The Largest Product

MTMath Talk — Fill the digits 2, 3, and 5 in □□ × □, using each digit once. What is the largest product possible?

There are six ways to place the three digits: \(23\times5,\ 25\times3,\ 32\times5,\ 35\times2,\ 52\times3,\ 53\times2\).

Grouping by the common multiplier: \((35\times2,\,53\times2)\), \((25\times3,\,52\times3)\), \((23\times5,\,32\times5)\). In each pair, the larger multiplicand wins, leaving three products to compare: \(53\times2,\ 52\times3,\ 32\times5\).

\(53\times2=106\) is smaller than \(52\times3=156\), so only \(52\times3\) and \(32\times5\) need comparing. Expanding: \(32\times5=(3\times10\times5)+(2\times5)=150+10\); \(52\times3=(5\times10\times3)+(2\times3)=150+6\). The first terms match, and the second term shows \(32\times5\) is larger.

The largest product is \(32\times5=160\) — put the largest digit (5) as the multiplier, and the other two digits in decreasing order (32) as the multiplicand.

Suppose \(p\lt q\lt r\) are the three digits. As before, there are six possible products, grouped by multiplier: \((qr\times p,\ rq\times p)\), \((pr\times q,\ rp\times q)\), \((pq\times r,\ qp\times r)\). In each pair, the multiplicand with the larger tens digit wins, leaving \(rq\times p,\ rp\times q,\ qp\times r\) to compare.

Since \(q\gt p\), \(rp\times q\) beats \(rq\times p\). Comparing \(qp\times r\) and \(rp\times q\): \(qp\times r=(10\times q\times r)+(p\times r)\) and \(rp\times q=(10\times r\times q)+(p\times q)\). The first terms match; since \(r\gt q\), the second term of \(qp\times r\) is larger.

Largest product = (largest digit) × (other two digits arranged in decreasing order)

Figure it Out — Set 3

Two questions from page 144 of the textbook.

1Fill the digits 1, 3, and 7 in □□ × □ to make the largest product possible.

By the rule, the largest digit (7) is the multiplier, and the other two digits are arranged in decreasing order (31) as the multiplicand: \(31\times7=217\).

Checking all six arrangements confirms this is the maximum: \(13\times7=91,\ 17\times3=51,\ 31\times7=217,\ 37\times1=37,\ 71\times3=213,\ 73\times1=73\).

31 × 7 = 217 is the largest possible product.

2Fill the digits 3, 5, and 9 in □□ × □ to make the largest product possible.

By the rule, the largest digit (9) is the multiplier, and the other two digits are arranged in decreasing order (53) as the multiplicand: \(53\times9=477\).

Checking all six arrangements confirms this is the maximum: \(35\times9=315,\ 39\times5=195,\ 53\times9=477,\ 59\times3=177,\ 93\times5=465,\ 95\times3=285\).

53 × 9 = 477 is the largest possible product.

6.6 Decoding Divisibility Tricks

Mukta shows Shubham a trick: choose a 2-digit number with different digits (don't reveal it), reverse the digits to get another number, find their difference, and divide the result by 9. For example, choosing 47: reversed is 74; the difference is \(74-47=27\); and \(27\div9=3\), with no remainder.

MTMath Talk — If we choose other 2-digit numbers, and follow the steps, will there always be no remainder?

Suppose the two-digit number is \(ab\) (with tens digit \(a\) and units digit \(b\)), so its value is \(10a+b\). Reversed, the new number is \(ba\), with value \(10b+a\).

If \(b\gt a\), then \(ba\gt ab\), and the difference is:

\[(10b+a)-(10a+b)=10b-b-10a+a=9b-9a=9(b-a)\]

This difference is always divisible by 9, since it is exactly 9 times the whole number \(b-a\).

Yes — the difference between any 2-digit number and its digit-reversal is always a multiple of 9, so dividing by 9 always leaves no remainder.

MTMath Talk — Can you work out what happens if a > b?

If \(a\gt b\), then \(ab\gt ba\), and the (positive) difference is instead \((10a+b)-(10b+a)=9a-9b=9(a-b)\), which is still a multiple of 9.

If \(a\gt b\), the difference is \(9(a-b)\) instead of \(9(b-a)\) — either way, the result is always a multiple of 9.

Figure it Out — Set 4

Eleven questions from pages 145–147 of the textbook.

1In the trick given above, what is the quotient when you divide by 9? Is there a relationship between the two numbers and the quotient?

Since the difference is \(9(b-a)\) (or \(9(a-b)\)), dividing by 9 gives exactly \(b-a\) (or \(a-b\)) as the quotient.

The quotient always equals the positive difference between the two digits of the original number. E.g. for 47 and 74, the quotient 3 equals \(7-4\).

2In the trick given above, instead of finding the difference of the two 2-digit numbers, find their sum. Observe that all these numbers are divisible by 11. Is this always true? Justify using algebra.

For a 2-digit number \(ab=10a+b\) and its reversal \(ba=10b+a\), the sum is:

\[(10a+b)+(10b+a)=11a+11b=11(a+b)\]

This is always exactly 11 times the sum of the digits \(a+b\), so it is always divisible by 11 — matching the given examples: \(31+13=44=11\times4\) (digits sum to 4); \(28+82=110=11\times10\) (digits sum to 10); \(12+21=33=11\times3\) (digits sum to 3).

Yes, always true — the sum of a 2-digit number and its reversal always equals \(11(a+b)\), a multiple of 11.

3Consider any 3-digit number abc (100a + 10b + c). Make two other numbers by cycling the digits, yielding bca and cab. Using algebra, justify that the sum is always divisible by 37. Will it also always be divisible by 3?

\(abc=100a+10b+c\), \(bca=100b+10c+a\), \(cab=100c+10a+b\). Adding all three:

\[(100a+10b+c)+(100b+10c+a)+(100c+10a+b)=111a+111b+111c=111(a+b+c)\]

Since \(111=3\times37\), the sum \(111(a+b+c)\) is always a multiple of 37, regardless of the value of \(a+b+c\).

Because \(111=3\times37\) already contains a factor of 3, the sum \(111(a+b+c)=3\times37\times(a+b+c)\) is also always divisible by 3, for every choice of digits — the factor of 3 comes from 111 itself, not from \(a+b+c\) needing to be a multiple of 3.

The cyclic sum is always \(111(a+b+c)\), which is always divisible by both 37 and 3, for every 3-digit number.

4Math Talk — Consider any 3-digit number abc. Make it a 6-digit number by repeating the digits: abcabc. Divide by 7, then 11, then 13. What do you get? Figure out why it works.

The 6-digit number \(abcabc\) equals the 3-digit number \(abc\) shifted three places and added to itself: \(abcabc=abc\times1000+abc=abc\times1001\).

Since \(1001=7\times11\times13\), dividing \(abcabc\) successively by 7, then 11, then 13 removes exactly this factor of 1001, leaving the original 3-digit number \(abc\).

For example, with \(abc=248\): \(248248\div7=35464\), \(35464\div11=3224\), \(3224\div13=248\) — back to the original number.

Dividing by 7, 11, and 13 in turn always gives back the original 3-digit number, because \(abcabc=abc\times1001=abc\times7\times11\times13\).

5There are 3 shrines, each with a magical pond that doubles any flowers dipped in it. A person dips his flowers in each pond in turn, placing an equal number of flowers in each of the 3 shrines, with all remaining flowers placed in the third shrine. How many flowers did he start with? How many flowers did he place in each shrine?

Let the starting number of flowers be \(x\), and let \(k\) be the equal number placed in each shrine.

Pond 1 doubles \(x\) to \(2x\); placing \(k\) in shrine 1 leaves \(2x-k\).

Pond 2 doubles this to \(2(2x-k)=4x-2k\); placing \(k\) in shrine 2 leaves \(4x-3k\).

Pond 3 doubles this to \(2(4x-3k)=8x-6k\), and this entire remaining amount is placed in shrine 3. For shrine 3 to also equal \(k\): \(8x-6k=k\Rightarrow8x=7k\Rightarrow x=\dfrac{7k}{8}\).

For \(x\) to be a whole number, \(k\) must be a multiple of 8. The smallest such case is \(k=8\), giving \(x=7\).

Checking \(x=7,k=8\): pond 1 doubles 7 to 14; place 8 in shrine 1, leaving 6. Pond 2 doubles 6 to 12; place 8 in shrine 2, leaving 4. Pond 3 doubles 4 to 8; place all 8 in shrine 3. Each shrine received exactly 8 flowers.

He started with 7 flowers and placed 8 flowers in each of the 3 shrines.

6Math Talk — A farm has some horses and hens. The total number of heads is 55 and the total number of legs is 150. How many horses and how many hens are on the farm? Can you solve this without letter-numbers?

With letter-numbers: let horses \(=h\) and hens \(=n\). Then \(h+n=55\) and \(4h+2n=150\). Substituting \(n=55-h\): \(4h+2(55-h)=150\Rightarrow4h+110-2h=150\Rightarrow2h=40\Rightarrow h=20\). Then \(n=55-20=35\).

Without letter-numbers: if all 55 animals were hens, there would be \(55\times2=110\) legs. The actual total is 150 legs, a difference of \(150-110=40\). Each horse has 2 more legs than a hen, so this difference of 40 legs must come from the horses: \(40\div2=20\) horses. The remaining \(55-20=35\) animals are hens.

20 horses and 35 hens.

7A mother is 5 times her daughter's age. In 6 years' time, the mother will be 3 times her daughter's age. How old is the daughter now?

Let the daughter's current age be \(d\), so the mother's current age is \(5d\). In 6 years: daughter \(=d+6\), mother \(=5d+6\).

Given the mother will then be 3 times the daughter's age: \(5d+6=3(d+6)=3d+18\). So \(2d=12\), giving \(d=6\).

Check: currently, daughter is 6 and mother is 30. In 6 years, daughter is 12 and mother is 36, and \(36=3\times12\). ✓

The daughter is currently 6 years old.

8Gauri says to Naina, "You have twice as many cows as I do." Naina replies, "That's true, but if I gave you three of my cows, we would each have the same number of cows." How many cows do Gauri and Naina have?

Let Gauri have \(g\) cows, so Naina has \(2g\) cows. If Naina gives 3 cows to Gauri: Gauri has \(g+3\) and Naina has \(2g-3\), and these are equal: \(g+3=2g-3\Rightarrow g=6\).

So Gauri has 6 cows and Naina has \(2(6)=12\) cows. Check: after the transfer, Gauri has \(6+3=9\) and Naina has \(12-3=9\) — equal. ✓

Gauri has 6 cows and Naina has 12 cows.

9A dosa cart has rent ₹5000 per day and each dosa costs ₹10 to make. (i) If 100 dosas are sold a day, what selling price gives a profit of ₹2000? (ii) If customers pay only ₹50 per dosa, how many dosas should be sold to make a profit of ₹2000?

(i) Let the selling price per dosa be \(p\). Selling 100 dosas gives revenue \(100p\), while total cost is rent plus making cost: \(5000+100(10)=6000\). Profit is revenue minus cost: \(100p-6000=2000\Rightarrow100p=8000\Rightarrow p=80\).

(ii) Let the number of dosas sold be \(n\), sold at ₹50 each. Revenue is \(50n\), and cost is \(5000+10n\). Profit: \(50n-(5000+10n)=40n-5000=2000\Rightarrow40n=7000\Rightarrow n=175\).

(i) The selling price should be ₹80 per dosa. (ii) 175 dosas should be sold in a day.

10Evaluate the sequence of fractions: 1/3, (1+3)/(5+7), (1+3+5)/(7+9+11). What do you observe? Explain why.

In the \(k\)-th fraction, the numerator is the sum of the first \(k\) odd numbers, which equals \(k^2\) (a standard identity: \(1+3+\cdots+(2k-1)=k^2\)). The denominator is the sum of the next \(k\) odd numbers after that, which equals the sum of the first \(2k\) odd numbers minus the sum of the first \(k\) odd numbers: \((2k)^2-k^2=4k^2-k^2=3k^2\).

So every fraction in the sequence equals \(\dfrac{k^2}{3k^2}=\dfrac13\). Checking: \(k=1\): \(\dfrac13\). \(k=2\): \(\dfrac{1+3}{5+7}=\dfrac4{12}=\dfrac13\). \(k=3\): \(\dfrac{1+3+5}{7+9+11}=\dfrac9{27}=\dfrac13\).

Every fraction in the sequence equals exactly 1/3, because the numerator (sum of the first \(k\) odd numbers) is always \(k^2\), and the denominator (sum of the next \(k\) odd numbers) is always \(3k^2\).

11Karim and the Genie — In a dream, a genie offers to double the money in Karim's pocket each time he walks around a banyan tree, in exchange for 8 coins per round. After 3 rounds, Karim is left with only 8 coins — exactly what he owes. (i) How many coins did Karim initially have? (ii) For what cost per round should Karim agree to the deal, if he wants to increase his coins? (iii) How should the genie set the cost per round so that it gets all of Karim's coins?

(i) Let the initial number of coins be \(x\), and the cost per round be \(c=8\). After each round, coins double and then \(c\) is subtracted:

Round 1: \(2x-8\). Round 2: \(2(2x-8)-8=4x-24\). Round 3 (doubling only, before the final payment): \(2(4x-24)=8x-48\).

This is the moment of "horror" — the amount right after the third doubling equals 8, the very debt he owes: \(8x-48=8\Rightarrow8x=56\Rightarrow x=7\).

(i) Karim initially had 7 coins.

(ii) With a general cost \(c\) per round, the coins after one round become \(2x-c\). For his coins to increase after a round (rather than stay the same or shrink), we need \(2x-c\gt x\), i.e. \(x\gt c\). As long as the cost per round is less than the number of coins he currently holds, doubling outpaces the fixed cost and his total keeps growing.

(ii) Karim should only agree if the cost per round \(c\) is less than his number of coins \(x\) (i.e. \(c\lt x\)); this keeps his total increasing after every round.

(iii) After 3 rounds, the amount remaining right after the final doubling (before the last payment) is \(8x-6c\) — using \(a_1=2x-c\), \(a_2=2a_1-c=4x-3c\), and the pre-payment amount \(2a_2=8x-6c\). For the genie to claim every remaining coin, this pre-payment amount must exactly equal the cost \(c\) itself: \(8x-6c=c\Rightarrow8x=7c\Rightarrow c=\dfrac{8x}{7}\).

(iii) Knowing Karim's coin count \(x\), the genie should set the cost per round to \(c=\dfrac{8x}{7}\) — this drains Karim to exactly zero after 3 rounds. (For \(x=7\), this gives \(c=8\), matching the story.)

Summary

Algebra is very useful in modeling and understanding numerical scenarios. Because of this, it occurs in almost all areas of mathematics, science and beyond.
Algebra is an indispensable tool in justifying mathematical statements.
We applied algebra to analyse 'Think of a Number' tricks, number pyramids, grids, ways of forming numbers using given digits to maximise certain products, divisibility tricks, and various other problems.

Frequently Asked Questions

Because replacing the chosen number with a letter, say x, and simplifying the steps algebraically always cancels x out completely, leaving only a fixed constant that doesn't depend on the starting number.

Each cell in a number pyramid equals the sum of the two cells directly below it. For a 3-row pyramid with bottom row a, b, c, the top is a + 2b + c; for a 4-row pyramid with bottom row a, b, c, d, the top is a + 3b + 3c + d.

Place the largest of the three digits as the single-digit multiplier, and arrange the other two digits in decreasing order to form the 2-digit multiplicand.

For a number ab (value 10a+b) and its reversal ba (value 10b+a), the difference simplifies to 9(b−a) or 9(a−b), which is always exactly 9 times a whole number.

The 6-digit number abcabc always equals the 3-digit number abc multiplied by 1001, and 1001 factors exactly as 7 × 11 × 13, so dividing by all three in turn returns the original 3-digit number.

Continue with Ganita Prakash, Part II

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