CBSE Class 8 Mathematics Algebra Play Long Answer and Case Study Questions with Answers

Long Answer Questions

Question 1. Study this ‘Think of a Number’ trick and answer the parts that follow.
Step 1: Think of a number. Step 2: Add 7. Step 3: Multiply the result by 2. Step 4: Subtract 6. Step 5: Divide by 2. Step 6: Subtract the number you first thought of.
(i) Carry out the trick with the starting numbers 6 and 11, showing the value after every step.
(ii) Taking the starting number as x, write the expression after each step and explain with algebra why the trick always ends at the same value.
(iii) Change exactly one step of the trick so that the answer is always 9, and justify your change using algebra.
Answer:
(i) Starting with 6: 6 → 13 → 26 → 20 → 10 → 4.
Starting with 11: 11 → 18 → 36 → 30 → 15 → 4. Both end at 4.
(ii) Let the chosen number be x.
Add 7: x + 7
Multiply by 2: 2x + 14
Subtract 6: 2x + 8
Divide by 2: x + 4
Subtract the original number: (x + 4) − x = 4
x stands for any number and it cancels at the last step, so the answer is 4 for every starting number, not only for the two tested above.
(iii) Change “add 7” to “add 12”. Then x → x + 12 → 2x + 24 → 2x + 18 → x + 9 → 9.
Check with 6: 18, 36, 30, 15, 9. In general, if step 2 is “add k”, the trick ends at k − 3, because 2(x + k) − 6 = 2x + 2k − 6, which halves to x + k − 3.
In simple words: The starting number cancels at the end, so the trick always gives 4; adding 12 instead of 7 makes it give 9.

Teacher's Note:
1. Examples show the pattern; algebra proves it.
2. Another valid change is ‘subtract 6’ → ‘add 4’, which also gives 9.

Question 2. In a number pyramid, each brick holds the sum of the two bricks directly below it. Study the pyramid shown and answer the parts that follow.
(i) Copy the pyramid and fill in all the missing bricks.
(ii) Taking the bottom row of a three-row pyramid as a, b and c, build the pyramid with letter-numbers and write the expression for the top brick.
(iii) Using that expression, find the top brick for the bottom row 12, 15, 11 without building the pyramid, and state which brick of the bottom row has the greatest effect on the top.
Answer:
(i) Bottom row 9, 6, 7. Middle row: 9 + 6 = 15 and 6 + 7 = 13. Top: 15 + 13 = 28.

(ii) With bottom row a, b, c: middle row a + b and b + c; top (a + b) + (b + c) = a + 2b + c.
Check with the pyramid above: 9 + 2 × 6 + 7 = 28, which matches.
(iii) Top = 12 + 2 × 15 + 11 = 12 + 30 + 11 = 53.
The middle brick has the greatest effect, because its coefficient is 2 while each end brick has coefficient 1. Raising the middle number by 1 raises the top by 2, but raising an end number by 1 raises the top by only 1. This is because two paths lead from the middle brick to the top, and only one from each end brick.
In simple words: The top of a three-row pyramid is a + 2b + c, so the middle brick counts double; here the tops are 28 and 53.

Teacher's Note:
1. Always check a formula against a pyramid you have filled.
2. Counting paths to the top explains the coefficients.

Question 3. On the August 2025 calendar page, a friend outlines a 2 × 2 square of four dates and tells you only their sum.
(i) Taking the top-left date as a, write the other three dates in terms of a, and show that the sum of the four dates is 4a + 16.
(ii) Find the four dates if the sum reported is 76.
(iii) Your friend now claims that the sum of her four dates is 90. Show by algebra that she must be mistaken, and state the test that any correct sum must pass.
Answer:
(i) Dates in a row go up by 1, and the date directly below is 7 more. So the square is: top row a and a + 1; bottom row a + 7 and a + 8.
Sum = a + (a + 1) + (a + 7) + (a + 8) = 4a + 16.
(ii) 4a + 16 = 76, so 4a = 60 and a = 15. The dates are 15, 16, 22 and 23.
(iii) 4a + 16 = 90 gives 4a = 74 and a = 18.5, which is not a whole number, so no date can be the top-left corner.
Test: 4a + 16 = 4(a + 4), so a correct sum must always be a multiple of 4. 90 is not, so the friend has added wrongly. (The sum must also be large or small enough to keep all four dates on the page.)
In simple words: The sum is 4a + 16, a multiple of 4; 76 gives 15, 16, 22, 23, while 90 is impossible.

Teacher's Note:
1. Check: 15 + 16 + 22 + 23 = 76.
2. A multiple-of-4 test catches wrong sums at once.

Question 4. The digits 3, 4 and 8 are to be placed in the boxes of □□ × □, each digit being used exactly once.
(i) Write all six possible products and evaluate them.
(ii) Name the arrangement that gives the largest product and the one that gives the smallest.
(iii) State the rule for placing three different digits to get the largest product, and explain with the expansion of qp × r and rp × q why the largest digit must be the multiplier.
Answer:
(i) 48 × 3 = 144; 84 × 3 = 252; 38 × 4 = 152; 83 × 4 = 332; 34 × 8 = 272; 43 × 8 = 344.
(ii) Largest: 43 × 8 = 344. Smallest: 48 × 3 = 144.
(iii) Rule: use the largest digit as the multiplier and write the other two in decreasing order to make the two-digit number.
Let the digits be p < q < r. The two strongest candidates are qp × r and rp × q:
qp × r = (10q + p) × r = 10qr + pr
rp × q = (10r + p) × q = 10qr + pq
The first terms are equal (both 10qr). In the second terms, pr > pq because r > q. Hence qp × r is larger, so the biggest digit must multiply. Here p = 3, q = 4, r = 8, giving 43 × 8 = 344.
In simple words: Of the six products, 43 × 8 = 344 is largest, because the biggest digit as multiplier gives the bigger leftover term.

Teacher's Note:
1. Pair products that share the same multiplier to compare quickly.
2. The expansion turns a guess into a proof.

Question 5. In the date trick, a friend chooses a date, multiplies the month by 5, adds 6, multiplies by 4, adds 9, multiplies by 5 and finally adds the day.
(i) Taking the month as M and the day as D, work through the steps and show that the number announced is always 100M + 165 + D.
(ii) A friend announces 479. Find the date she had chosen and explain how the last two digits are read.
(iii) The step “add 6” is now replaced by “add 8”, all other steps staying the same. Find the new number that must be subtracted from the announced value, and use it to decode the announced value 1020.
Answer:
(i) Multiply the 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
(ii) 479 − 165 = 314, so 100M + D = 314. A day is at most 31, so D takes only the last two digits: D = 14 and M = 3. The date is 14 March.
(iii) With “add 8”: 5M → 5M + 8 → 20M + 32 → 20M + 41 → 100M + 205, then + D.
So the number is 100M + 205 + D, and now 205 must be subtracted.
1020 − 205 = 815, giving M = 8 and D = 15, that is 15 August.
In simple words: The steps always give 100M + 165 + D; 479 decodes to 14 March, and with ‘add 8’ we subtract 205, so 1020 means 15 August.

Teacher's Note:
1. Multiply every term of the expression at each step.
2. The fixed number is (6 × 4 + 9) × 5 = 165, or (8 × 4 + 9) × 5 = 205.

Question 6. Mukta’s divisibility trick runs like this: choose a two-digit number whose digits are different, reverse the digits, subtract the smaller number from the larger and divide the difference by 9.
(i) Try the trick with 62 and with 95, showing the difference and the quotient in each case.
(ii) Let the chosen number have tens digit a and units digit b, with b > a. Using the expanded forms 10a + b and 10b + a, explain with algebra why the difference is always divisible by 9.
(iii) Work out what happens when a > b, and state what the quotient tells you about the digits.
Answer:
(i) 62: reversed 26; difference 62 − 26 = 36; 36 ÷ 9 = 4.
95: reversed 59; difference 95 − 59 = 36; 36 ÷ 9 = 4.
(ii) The number is 10a + b and the reverse is 10b + a. Since b > a, the larger is 10b + a, so
difference = (10b + a) − (10a + b) = 10b − b − 10a + a = 9b − 9a = 9(b − a).
As b − a is a whole number, the difference is 9 times a whole number, so it is always divisible by 9.
(iii) Now 10a + b is larger, and the difference is (10a + b) − (10b + a) = 9(a − b), again a multiple of 9.
In both cases the quotient is the difference of the two digits. That is why 62 and 95 both gave 4: their digits differ by 4. (If the digits were equal, the difference would be 0, still divisible by 9.)
In simple words: The difference is always 9 times the gap between the digits, so 9 always divides it and the quotient is that gap.

Teacher's Note:
1. Subtract larger minus smaller so the difference is positive.
2. The quotient tells you the digit difference.

Question 7. Invent a new ‘Think of a Number’ trick of your own and present it as follows.
(i) Write the steps of a trick of at least four steps that always ends with the answer 6, whatever number is chosen to start with.
(ii) Demonstrate your trick on two different starting numbers, showing every step.
(iii) Using a letter-number for the starting number, justify with algebra that your trick must always give 6.
(iv) Explain the single change you would make so that the trick always ends with 20 instead.
Answer:
(i) One possible trick: 1. Think of a number. 2. Add the number that comes just after it. 3. Add 11. 4. Divide by 2. 5. Subtract the number you first thought of.
(ii) Start with 7: 7 + 8 = 15; 15 + 11 = 26; 26 ÷ 2 = 13; 13 − 7 = 6.
Start with 20: 20 + 21 = 41; 41 + 11 = 52; 52 ÷ 2 = 26; 26 − 20 = 6.
(iii) Let the number be x; the next number is x + 1.
Add it: x + (x + 1) = 2x + 1
Add 11: 2x + 12
Divide by 2: x + 6
Subtract the original number: (x + 6) − x = 6. Since x cancels, the answer must be 6 for every start.
(iv) Replace “add 11” by “add 39”. Then 2x + 1 + 39 = 2x + 40, which halves to x + 20, and subtracting x leaves 20. In general, “add 2k − 1” at step 3 makes the trick end at k.
In simple words: The trick reaches x + 6 before the starting number is taken away, so it always ends at 6; adding 39 instead of 11 makes it end at 20.

Teacher's Note:
1. Any trick works if the letter-number cancels at the end.
2. Other correct tricks are also acceptable if justified.

Teacher's Note:
1. Comparing rows that differ by one shape removes the other shapes at one stroke.
2. Always substitute back into every row.

Question 9. Take any three-digit number abc with digits a, b and c. Cycle the digits to form bca and cab, and add the three numbers.
(i) Carry this out for 471 and write the total.
(ii) Using the expanded forms 100a + 10b + c, 100b + 10c + a and 100c + 10a + b, prove with algebra that the total is always divisible by 37 and also by 3.
(iii) Find the total when the digits add up to 18, and explain why two three-digit numbers made of the same three digits always give the same total.
Answer:
(i) The three numbers are 471, 714 and 147. 471 + 714 + 147 = 1332.
(ii) Adding the expanded forms, each digit appears once in the hundreds, once in the tens and once in the units place:
Total = 100(a + b + c) + 10(a + b + c) + (a + b + c) = 111(a + b + c) = 3 × 37 × (a + b + c).
So the total is always a multiple of 37 and of 3, whatever the digits. For 471: a + b + c = 12, 111 × 12 = 1332, and 1332 ÷ 37 = 36, 1332 ÷ 3 = 444.
(iii) Total = 111 × 18 = 1998.
The total depends only on a + b + c, not on the order of the digits. So any two numbers built from the same three digits give the same total; for example 471 and 174 both lead to 1332.
In simple words: The total is always 111 × (sum of digits), so it is divisible by 37 and 3 and depends only on the digit sum.

Teacher's Note:
1. 111 = 3 × 37 is the key factorisation.
2. Cycling the digits does not change the digit sum.

Question 10. A traveller meets a spirit beside a magic well. Each time the traveller walks once round the well, the coins in his bag become three times as many, but he must then pay the spirit 27 coins. He walks round three times and finds that after the third payment he has no coins left at all.
(i) Taking the number of coins he started with as x, write an expression for the coins left after each round and find how many coins he had at the start.
(ii) If the spirit instead charges k coins per round, find the condition on k for the traveller’s coins to increase after a round when he is carrying x coins.
(iii) Explain how the spirit, who magically knows the number of coins in the bag, should fix the charge so that the traveller is left with nothing after a single round.
Answer:
(i) Let the traveller start with x coins.
After round 1: 3x − 27
After round 2: 3(3x − 27) − 27 = 9x − 108
After round 3: 3(9x − 108) − 27 = 27x − 351
He is left with nothing, so 27x − 351 = 0, giving x = 13. He started with 13 coins.
Check: 13 → 39 − 27 = 12 → 36 − 27 = 9 → 27 − 27 = 0.
(ii) With x coins and a charge of k, he ends the round with 3x − k. For this to be more than x: 3x − k > x, that is 2x > k, so k < 2x.
The charge must be less than twice the coins he carries at the start of the round. With 13 coins it had to be below 26, and 27 was just too much, so his money fell instead of rising.
(iii) The spirit should charge exactly 3x coins, three times the number in the bag. Then 3x − 3x = 0, and the traveller is left with nothing after one round.
In simple words: Working the three rounds gives 27x − 351 = 0, so he began with 13 coins; his coins grow only if the charge is less than 2x.

Teacher's Note:
1. Use brackets when the whole amount is tripled.
2. Check the answer by running the rounds forwards.

Case Study Based Questions

Question 11. The magician at the school fair. At her school’s annual fair, Veena sets up a stall called “I Can Read Your Mind”. A visitor pays two rupees, thinks of any number and is asked to follow five instructions: add 5 to the number, double the result, subtract 4, divide by 2, and finally take away the number first thought of. Veena never asks what the visitor chose, yet she announces the final answer before the visitor can speak, and she is right every single time. Ravi, who is in Class 8, watches twenty visitors try it and decides to find out with algebra what Veena knows.
(i) Ravi starts with 9. Write the value after each of the five instructions.
(ii) If the chosen number is x, write the expression obtained just after the step “double the result”.
(iii) Show with algebra that the final answer is the same for every starting number, and write that answer.
(iv) Veena wants the answer to be 8 instead. Which single instruction should she change, and how?
Answer:
(i) 9 → 9 + 5 = 14 → 14 × 2 = 28 → 28 − 4 = 24 → 24 ÷ 2 = 12 → 12 − 9 = 3.
(ii) After “add 5” the expression is x + 5, so after doubling it is 2x + 10.
(iii) Subtract 4: 2x + 6; divide by 2: x + 3; take away the chosen number: (x + 3) − x = 3. The letter-number cancels, so the answer is always 3.
(iv) Change “add 5” to “add 10”. Then 2x + 20 − 4 = 2x + 16, which halves to x + 8, and taking away x leaves 8.
In simple words: Every visitor reaches x + 3 before taking away x, so the answer is always 3; adding 10 instead of 5 makes it 8.

Teacher's Note:
1. Veena’s secret is that the chosen number cancels.
2. Another single change: ‘subtract 4’ → ‘add 6’ also gives 8.

Question 12. The number-pyramid game. In a Class 8 mathematics period, the teacher draws an empty three-row number pyramid on the board and explains the rule: each brick must hold the sum of the two bricks directly below it. Teams fill in only the bottom row; the fastest team to announce the correct top brick wins a point. Team A writes 6, 11, 4 in the bottom row, and Team B writes the same three numbers but in the reverse order, 4, 11, 6. Both teams shout out the same top number, which surprises the class, and the teacher asks them to explain the coincidence using letter-numbers.
(i) Write the two bricks of the middle row for Team A.
(ii) Write the top brick for Team A.
(iii) Taking the bottom row as a, b and c, write the expression for the top brick.
(iv) Using that expression, explain why Team B gets the same top number as Team A.
Answer:
(i) Middle row: 6 + 11 = 17 and 11 + 4 = 15.
(ii) Top brick: 17 + 15 = 32.
(iii) Middle row a + b and b + c, so the top is a + 2b + c.
(iv) In a + 2b + c the two end bricks have the same coefficient 1, so swapping them leaves the value unchanged. Team B: 4 + 22 + 6 = 32, the same as Team A’s 6 + 22 + 4 = 32.
In simple words: The top is a + 2b + c, and swapping the end numbers a and c does not change it, so both teams get 32.

Teacher's Note:
1. The middle number stays in the middle when the row is reversed.
2. Symmetric coefficients explain the coincidence.

Teacher's Note:
1. Check (i) with the formula: 4 × 13 + 16 = 68.
2. A total that is not a multiple of 4 is always wrong.

Question 14. Two friends test a shortcut. Amina tells Rohit about a shortcut she has read: “Take any two-digit number, reverse its digits to get a second number, and subtract the smaller from the larger. The answer is always a multiple of 9.” Rohit is doubtful, so the two of them spend a lunch break testing it. Amina tries 72, Rohit tries 94, and both of them get a difference that 9 divides exactly. Rohit then deliberately chooses a number with two equal digits to try to break the shortcut, while Amina writes the two-digit number as 10a + b to see whether the shortcut can be proved for every case.
(i) Carry out Amina’s test with 72: write the reversed number, the difference, and the quotient on dividing by 9.
(ii) Carry out Rohit’s test with 94 in the same way.
(iii) Both friends get the same quotient. Explain what this quotient is, in terms of the digits of the chosen number.
(iv) Rohit chooses 55. What difference does he get, and does the shortcut still hold? Give a reason.
Answer:
(i) 72 reversed is 27; 72 − 27 = 45; 45 ÷ 9 = 5.
(ii) 94 reversed is 49; 94 − 49 = 45; 45 ÷ 9 = 5.
(iii) The quotient is the difference between the two digits: 7 − 2 = 5 and 9 − 4 = 5. With the number 10a + b, the difference is (10a + b) − (10b + a) = 9(a − b), so dividing by 9 leaves a − b.
(iv) 55 − 55 = 0, and 0 is divisible by 9 (quotient 0). This fits 9(a − b) with a = b, so the shortcut still holds; Rohit’s choice does not break it.
In simple words: The difference is always 9 times the gap between the digits, so 72 and 94 both give 5, and 55 gives 0, still a multiple of 9.

Teacher's Note:
1. 0 is a multiple of every number.
2. Algebra covers the special cases too.

Question 15. The newspaper puzzle column. A Sunday newspaper runs a puzzle column titled “I Know Your Birthday”. Readers are told to take the date of their birthday, multiply the month number by 5, add 6, multiply by 4, add 9, multiply by 5, and finally add the day of the month. Readers then post only this final number to the column, and the next Sunday the column prints their birthdays correctly. The puzzle setter explains that the whole trick rests on one algebraic expression: the number a reader posts is always 100M + 165 + D, where M is the month number and D is the day.
(i) A reader whose birthday is 28 June follows the steps. Show the working and write the number she posts.
(ii) Another reader posts 1270. Find his birthday.
(iii) One reader posts 1231. Show that this number cannot have come from any real date.
(iv) What is the largest number any reader can possibly post, and which date gives it?
Answer:
(i) M = 6, D = 28: 6 × 5 = 30; 30 + 6 = 36; 36 × 4 = 144; 144 + 9 = 153; 153 × 5 = 765; 765 + 28 = 793. She posts 793.
(ii) 1270 − 165 = 1105, so M = 11 and D = 05. His birthday is 5 November.
(iii) 1231 − 165 = 1066, so 100M + D = 1066. The last two digits give D = 66 and the digits before them give M = 10. No month has a 66th day, so no date can give 1231; the reader must have made a mistake.
(iv) The largest value comes from M = 12 and D = 31: 1200 + 165 + 31 = 1396, from 31 December.
In simple words: Subtracting 165 gives 100M + D; 1270 means 5 November, 1231 leads to an impossible day 66, and 31 December gives the largest number 1396.

Teacher's Note:
1. Check (i) with the formula: 600 + 165 + 28 = 793.
2. A decoded day above 31 or a month above 12 shows an error.

Teacher's Note:
1. Each row is a small equation.
2. Use each new value in the next row.

Question 17. The largest product game. The mathematics club meets every Friday. This week the teacher hands each team three number cards, 2, 6 and 9, and a frame of the shape □□ × □. Each card must be used exactly once, and the team whose arrangement gives the largest product wins. One team argues that the biggest digit should go into the tens place of the two-digit number, another team says the biggest digit should be used as the single-digit multiplier, and a third team decides to list every possible arrangement before answering.
(i) How many different arrangements of the three cards are possible in the frame?
(ii) Work out 96 × 2 and 62 × 9, and say which team’s reasoning they support.
(iii) Write the arrangement that gives the largest product, with its value.
(iv) State the general rule for three different digits p < q < r.
Answer:
(i) The first box can be filled in 3 ways and the next in 2 ways, leaving 1 card for the last box, so there are 3 × 2 = 6 arrangements.
(ii) 96 × 2 = 192 and 62 × 9 = 558. They support the second team: the largest digit should be the single-digit multiplier, not the tens digit.
(iii) The largest product is 62 × 9 = 558. The next best, 92 × 6 = 552, falls short by 6.
(iv) For digits p < q < r, the largest product is qp × r: the largest digit r is the multiplier and the other two are written in decreasing order.
In simple words: Of the six arrangements, 62 × 9 = 558 is the largest, so the biggest digit should be the multiplier.

Teacher's Note:
1. Listing all six cases is a safe check.
2. The rule saves listing every time.

Question 18. Counting at the village farm. Lata helps at her grandmother’s farm, where only goats and ducks are kept. One morning her grandmother asks her to find out how many of each are there without separating the animals. Lata counts the heads and gets 48. She then counts the legs, carefully, and gets 134. Her younger brother begins by imagining that every animal in the shed is a duck, and then corrects his count; Lata prefers to use a letter-number for the number of goats and to write a single equation.
(i) If all 48 animals were ducks, how many legs would there be?
(ii) By how much does this fall short of 134, and how many extra legs does each goat contribute?
(iii) Find the number of goats and the number of ducks.
(iv) Write Lata’s equation using x for the number of goats, and solve it to check your answer.
Answer:
(i) 48 × 2 = 96 legs.
(ii) 134 − 96 = 38 legs short. Each goat has 4 legs instead of 2, so each goat adds 2 extra legs.
(iii) Goats = 38 ÷ 2 = 19, so ducks = 48 − 19 = 29.
(iv) With x goats there are 48 − x ducks, so 4x + 2(48 − x) = 134. This gives 2x + 96 = 134, so 2x = 38 and x = 19.
Check: 19 × 4 + 29 × 2 = 76 + 58 = 134 legs and 19 + 29 = 48 heads.
In simple words: All ducks would give 96 legs; the 38 extra legs come 2 per goat, so there are 19 goats and 29 ducks.

Teacher's Note:
1. The brother’s method and Lata’s equation give the same answer.
2. Always check both the heads and the legs.

Question 19. A pyramid on a worksheet. A worksheet gives Class 8 students the partly filled four-row number pyramid shown below. The rule is the usual one: every brick is the sum of the two bricks directly below it. Three of the four bottom bricks and the top brick are printed; everything else has to be worked out. Meera says the puzzle cannot be done without guessing, but her partner Faiz points out that the top brick can be written in terms of the four bottom bricks, which turns the puzzle into a single equation.
(i) Taking the bottom row as a, b, c, d, write the expression for the top brick.
(ii) Form an equation and find the missing number in the bottom row.
(iii) Write the complete third row and second row of the pyramid.
(iv) The last two bottom bricks are now interchanged, so that the bottom row reads 4, 11, 2, 7. Find the new top brick and explain why the top changes here although interchanging the two end bricks of a three-row pyramid does not change its top.
Answer:
(i) The row above the bottom is a + b, b + c, c + d; the next is a + 2b + c, b + 2c + d; so the top is a + 3b + 3c + d.
(ii) With a = 4, c = 7, d = 2 and top 60: 4 + 3b + 21 + 2 = 60, so 3b + 27 = 60, 3b = 33 and b = 11. The missing bottom brick is 11.
(iii) Row above the bottom: 4 + 11 = 15, 11 + 7 = 18, 7 + 2 = 9. Next row: 15 + 18 = 33 and 18 + 9 = 27. Top: 33 + 27 = 60, as printed.

(iv) New top = 4 + 3 × 11 + 3 × 2 + 7 = 4 + 33 + 6 + 7 = 50.
In a four-row pyramid the coefficients are 1, 3, 3, 1, so moving 7 from a coefficient-3 place to a coefficient-1 place (and 2 the other way) changes the total. In a three-row pyramid the coefficients are 1, 2, 1, and the two end bricks carry the same coefficient, so interchanging them changes nothing.
In simple words: The top is a + 3b + 3c + d, which gives the missing brick 11; swapping 7 and 2 moves them to places with different weights, so the top becomes 50.

Teacher's Note:
1. Swapping changes the top only when the two places have different coefficients.
2. Fill the pyramid fully to check the equation.

Question 20. The repeating-number trick at the fair. At a mathematics fair, a stall holder makes this offer: “Write down any three-digit number. Now write the same three digits again to its right, so that a six-digit number is formed. Divide your six-digit number by 7; there will be no remainder. Divide the answer by 11; again no remainder. Divide that answer by 13. I promise you a surprise.” A Class 8 student writes 347, forms 347347 and follows the instructions. She then wonders whether the stall holder has chosen 7, 11 and 13 for some special reason, and whether the same idea would work with a two-digit number written twice.
(i) Carry out the three divisions on 347347 and write the three answers.
(ii) Taking the three-digit number as n, write the six-digit number in terms of n and explain the surprise.
(iii) Explain why exactly the divisors 7, 11 and 13 are used in the trick.
(iv) A two-digit number is written twice to form a four-digit number, such as 5252. State a number that always divides it exactly, and justify your answer with algebra.
Answer:
(i) 347347 ÷ 7 = 49621; 49621 ÷ 11 = 4511; 4511 ÷ 13 = 347.
(ii) Writing the digits again shifts the first copy three places, so the six-digit number is n × 1000 + n = n × 1001. Since 1001 = 7 × 11 × 13, dividing by 7, then 11, then 13 removes the factor 1001 and returns n. The surprise: the original number comes back.
(iii) Because 7 × 11 × 13 = 1001 exactly. These are the prime factors of 1001, so each division leaves no remainder, and together they undo the multiplication by 1001.
(iv) 101 always divides it. If the two-digit number is m, writing it twice gives m × 100 + m = m × 101, so 101 is a factor and the quotient is the original number: 5252 ÷ 101 = 52.
In simple words: Writing a number twice multiplies it by 1001 (or by 101 for two digits), so dividing by those factors gives the number back.

Teacher's Note:
1. The multiplier is 1 followed by zeros and a 1: 101, 1001, 10001 …
2. 1001 = 7 × 11 × 13 explains the choice of divisors.