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

Long Answer Questions

Question 1. Any four-digit number can be written as \( \overline{dcba} \), where a, b, c and d are its units, tens, hundreds and thousands digits.
(i) Write \( \overline{dcba} \) as a sum of place values.
(ii) Using this form, explain why such a number is divisible by 10 only when a = 0, and divisible by 5 only when a is 0 or 5.
(iii) Explain why, for divisibility by 4, it is enough to look at the last two digits. Apply your reasoning to 7364 and 9350, and state for each whether it is divisible by 4, by 5 and by 10.
Answer:
(i) \( \overline{dcba} \) = 1000d + 100c + 10b + a.
(ii) 1000d + 100c + 10b = 10(100d + 10c + b) is a multiple of 10 (and so of 5) whatever the digits are.
So the number is a multiple of 10 only when the units digit a is a multiple of 10, that is, a = 0; and a multiple of 5 only when a is a multiple of 5, that is, a = 0 or 5.
(iii) 1000d + 100c = 100(10d + c), and 100 = 4 × 25, so this part is always a multiple of 4. Only the last two digits, 10b + a, decide divisibility by 4.
7364: last two digits 64 = 4 × 16, so 7364 is divisible by 4 (7364 = 4 × 1841). Its units digit is 4, so it is not divisible by 5 or 10.
9350: last two digits 50 is not a multiple of 4, so 9350 is not divisible by 4. Its units digit is 0, so it is divisible by both 5 and 10.
In simple words: The higher place values are multiples of 10 and of 100, so the units digit decides 5 and 10, and the last two digits decide 4.

Teacher's Note:
1. Split the number into a part that is surely a multiple and a leftover part.
2. 100 is a multiple of 4, but 10 is not; that is why two digits are needed.

Question 2. Anshu is exploring sums of consecutive numbers.
(i) Write 45 as a sum of two or more consecutive natural numbers in three different ways.
(ii) Taking the middle number as n, show with algebra that the sum of any three consecutive numbers is a multiple of 3.
(iii) Taking the smallest number as n, show that the sum of four consecutive numbers can never be a multiple of 4. State the remainder such a sum always leaves on division by 4.
Answer:
(i) 45 = 22 + 23 = 14 + 15 + 16 = 7 + 8 + 9 + 10 + 11.
(ii) The numbers are n − 1, n, n + 1, and (n − 1) + n + (n + 1) = 3n, which always has 3 as a factor. Example: 14 + 15 + 16 = 45 = 3 × 15.
(iii) The numbers are n, n + 1, n + 2, n + 3, with sum 4n + 6 = 4(n + 1) + 2.
This is 2 more than a multiple of 4, so it is never a multiple of 4; it always leaves remainder 2. Example: 6 + 7 + 8 + 9 = 30 = 4 × 7 + 2.
In simple words: Three numbers in a row add to 3 times the middle one, while four in a row always leave remainder 2 on division by 4.

Teacher's Note:
1. Other answers to (i) such as 5 + 6 + 7 + 8 + 9 + 10 are also correct.
2. Pick the letter for the middle number when the count is odd.

Teacher's Note:
1. 56790 is a multiple of 9 because its root is 9.
2. Remember the one adjustment: root 9 means remainder 0.

Question 4. Decide whether each statement below is always true, sometimes true or never true. In each case justify your decision with algebra, and give an example and a non-example wherever they help.
(i) If 7 exactly divides two numbers, then it exactly divides their sum.
(ii) If a number is divisible by 20, then it is divisible by 10.
(iii) The sum of a multiple of 4 and a multiple of 6 is a multiple of 12.
Answer:
(i) Always true. The numbers are 7m and 7n, and 7m + 7n = 7(m + n). Example: 21 + 56 = 77 = 7 × 11. In general, if a divides M and N, it divides M + N and M − N.
(ii) Always true. A multiple of 20 is 20m = 10(2m), which has 10 as a factor; a number divisible by k is divisible by every factor of k. Example: 140 = 20 × 7 = 10 × 14.
(iii) Sometimes true. 4a + 6b = 2(2a + 3b) is always even, but 12 need not be a factor.
Example: 12 + 12 = 24 = 12 × 2, a multiple of 12. Non-example: 4 + 6 = 10, not a multiple of 12.
In simple words: Sums of multiples of 7 stay multiples of 7, multiples of 20 are multiples of 10, but 4-multiples plus 6-multiples only sometimes give 12-multiples.

Teacher's Note:
1. 'Always' needs algebra; 'sometimes' needs an example and a non-example.
2. Factor out the common number to show divisibility.

Question 5. This question is about why the shortcut for 9 works.
(i) Write each of 9, 99 and 999 as “one less than a place value”, that is, complete 10 = 9 + 1, 100 = __ + 1, 1000 = __ + 1.
(ii) Using these, split the four-digit number \( \overline{dcba} \) = 1000d + 100c + 10b + a into a part that is clearly a multiple of 9 and a part that is the sum of the digits. Hence justify the test: a number is divisible by 9 if and only if the sum of its digits is divisible by 9.
(iii) Use the test on 6471 and on 6473. For the one that is not a multiple of 9, state the remainder it leaves.
Answer:
(i) 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1.
(ii) \( \overline{dcba} \) = (999 + 1)d + (99 + 1)c + (9 + 1)b + a = \( \underbrace{(999d + 99c + 9b)}_{\text{a multiple of 9}} + \underbrace{(d + c + b + a)}_{\text{the digit sum}} \).
999d + 99c + 9b = 9(111d + 11c + b) is a multiple of 9 whatever the digits are.
So the number and its digit sum leave the same remainder on division by 9; in particular the number is divisible by 9 if and only if its digit sum is.
(iii) 6471: 6 + 4 + 7 + 1 = 18, a multiple of 9, so 6471 is divisible by 9 (6471 = 9 × 719).
6473: 6 + 4 + 7 + 3 = 20 → 2 + 0 = 2, so 6473 is not a multiple of 9 and leaves remainder 2.
In simple words: Each place value is a string of nines plus 1, so the number is a multiple of 9 plus its digit sum.

Teacher's Note:
1. The proof shows the remainders match, not only divisibility.
2. 6473 is 2 more than 6471, which agrees with remainder 2.

Question 6. This question is about why the shortcut for 11 works.
(i) Show that 10 is one less than a multiple of 11, 100 is one more than a multiple of 11, and 1000 is one less than a multiple of 11, by writing each as 11 × __ ± 1.
(ii) Using these, show that for the four-digit number \( \overline{dcba} \) the remainder on division by 11 is decided by a − b + c − d.
(iii) Apply the rule to 8217 and to 6394. State which one is a multiple of 11, and give the remainder for the other.
Answer:
(i) 10 = 11 × 1 − 1, 100 = 11 × 9 + 1, 1000 = 11 × 91 − 1. The place values are alternately one less and one more than a multiple of 11.
(ii) \( \overline{dcba} \) = (11 × 91 − 1)d + (11 × 9 + 1)c + (11 × 1 − 1)b + a = 11(91d + 9c + b) + (a − b + c − d).
The first part is a multiple of 11, so the remainder on division by 11 is decided entirely by a − b + c − d, the alternating sum from the units digit.
(iii) 8217: 7 − 1 + 2 − 8 = 0, so 8217 is a multiple of 11 (8217 = 11 × 747).
6394: 4 − 9 + 3 − 6 = −8, so 6394 is 8 short of a multiple of 11, that is, 11 − 8 = 3 more than one. Its remainder is 3.
In simple words: Place values are alternately 1 below and 1 above multiples of 11, so the alternating digit sum decides the remainder.

Teacher's Note:
1. The ‘−1’ place values give the minus signs in the rule.
2. Check: 11 × 581 = 6391, and 6394 − 6391 = 3.

Question 7. Shortcuts for composite divisors.
(i) Explain why a number is divisible by 6 exactly when it is divisible by both 2 and by 3. Test 4728 for divisibility by 6.
(ii) Nikhil says, “24 = 4 × 6, so I will check divisibility by 4 and by 6.” Show with the number 1836 that his method fails.
(iii) Using prime factorisation, explain why checking divisibility by 3 and by 8 does work for 24, and use it to show that 4728 is a multiple of 24. State the general rule involving the LCM.
Answer:
(i) 6 = 2 × 3, and 2 and 3 have no common factor other than 1, so a number having both 2 and 3 as factors has 2 × 3 = 6 as a factor (and every multiple of 6 is a multiple of 2 and 3).
4728 ends in 8, so it is divisible by 2; 4 + 7 + 2 + 8 = 21 is a multiple of 3. So 4728 is divisible by 6 (4728 = 6 × 788).
(ii) 1836: last two digits 36 = 4 × 9, so it is divisible by 4; it is even with digit sum 18, so it is divisible by 6.
But 1836 = 24 × 76 + 12, so 1836 is not divisible by 24; Nikhil's method fails.
(iii) 24 = 2³ × 3. Now 4 = 2² and 6 = 2 × 3 together guarantee only 2² × 3 = 12, since the factor 2 is shared. But 8 = 2³ and 3 together give exactly 2³ × 3 = 24.
For 4728: 728 = 8 × 91, so it is divisible by 8; digit sum 21 shows it is divisible by 3. Hence 4728 is a multiple of 24 (4728 = 24 × 197).
General rule: if a number is divisible by k and by m, it is divisible by LCM(k, m). Here LCM(4, 6) = 12, but LCM(3, 8) = 24.
In simple words: Two tests together guarantee only their LCM: 4 and 6 give 12, while 3 and 8 give 24.

Teacher's Note:
1. Choose factors with no common factor.
2. For 8, test the last three digits.

Question 8. Solve the two cryptarithms below. In each, every letter stands for exactly one digit, different letters stand for different digits, and no number begins with 0. Show the reasoning that fixes each digit, not just the answer.
(i) KM × 9 = NMK
(ii) TU × 8 = VUT
Answer:
(i) In place-value form, 9(10K + M) = 100N + 10M + K, so 90K + 9M = 100N + 10M + K, which gives 89K = 100N + M.
So M is the remainder when 89K is divided by 100, and it must be a single digit. For K = 1 to 9, 89K = 89, 178, 267, 356, 445, 534, 623, 712, 801; only 801 (K = 9) leaves a single-digit remainder: 801 = 100 × 8 + 1.
Hence K = 9, N = 8, M = 1, and the puzzle reads 91 × 9 = 819.
(ii) 8(10T + U) = 100V + 10U + T gives 80T + 8U = 100V + 10U + T, that is, 79T = 100V + 2U.
So 79T must leave an even remainder of at most 18 on division by 100. For T = 1 to 9, 79T = 79, 158, 237, 316, 395, 474, 553, 632, 711, with remainders 79, 58, 37, 16, 95, 74, 53, 32, 11.
Only T = 4 works: 2U = 16, so U = 8 and V = 3. The puzzle reads 48 × 8 = 384.
In simple words: Writing each side with place values gives 89K = 100N + M and 79T = 100V + 2U, which allow only 91 × 9 = 819 and 48 × 8 = 384.

Teacher's Note:
1. Turn the cryptarithm into one equation, then test the few possible values.
2. Check that all letters stand for different digits.

Teacher's Note:
1. List the expressions in a fixed order so none is missed.
2. Zero is an even number.

Question 10. Let \( \overline{ab} \) stand for a two-digit number with digits a and b, and \( \overline{ba} \) for the number formed by reversing them.
(i) Show with algebra that \( \overline{ab} \) + \( \overline{ba} \) is always a multiple of 11, and that \( \overline{ab} \) − \( \overline{ba} \) is always a multiple of 9. Verify both results using the number 73.
(ii) For a three-digit number, show that \( \overline{abc} \) − \( \overline{cba} \) is always a multiple of 99. Verify using 825.
(iii) Sreelatha claims, “If I reverse the digits of a multiple of 9, the new number is still a multiple of 9.” Use the digit-sum test to decide whether her claim is correct, and say whether any other rearrangement of the digits would also work.
Answer:
(i) \( \overline{ab} \) + \( \overline{ba} \) = (10a + b) + (10b + a) = 11a + 11b = 11(a + b).
\( \overline{ab} \) − \( \overline{ba} \) = (10a + b) − (10b + a) = 9a − 9b = 9(a − b).
Check with 73: 73 + 37 = 110 = 11 × 10 and 73 − 37 = 36 = 9 × 4.
(ii) \( \overline{abc} \) − \( \overline{cba} \) = (100a + 10b + c) − (100c + 10b + a) = 99a − 99c = 99(a − c). The middle digit cancels.
Check with 825: 825 − 528 = 297 = 99 × 3.
(iii) Her claim is correct. Reversing does not change which digits are present, so the digit sum stays the same; by the test for 9, the reversed number is again a multiple of 9. Example: 4752 and 2574 both have digit sum 18 and both are multiples of 9.
For the same reason any rearrangement of the digits gives a multiple of 9, as long as the new number does not begin with 0.
In simple words: Reversal sums and differences factor as 11, 9 and 99 times something, and rearranging digits keeps the digit sum, so multiples of 9 stay multiples of 9.

Teacher's Note:
1. Write each number with place values before adding or subtracting.
2. Divisibility by 9 depends only on the digits, not their order.

Case Study Based Questions

Question 11. Trays of nine at the sweet shop. Vimala runs a sweet shop. Every laddoo she makes is packed into trays that hold exactly 9 laddoos, and a tray is sent out only when it is completely full. On Monday her team made 4536 laddoos and on Tuesday they made 2814 laddoos. Vimala does not want to divide by 9 each time, so her son reminds her of the shortcut she learnt in school: add up the digits of the number, and keep adding until a single digit is left. That single digit tells her how many laddoos will be left over after all the full trays are packed.
(i) State the shortcut for checking whether a number is divisible by 9.
(ii) Show, using the shortcut, that Monday’s 4536 laddoos fill whole trays exactly, and find how many trays are filled.
(iii) How many laddoos will be left over on Tuesday after all full trays are packed?
(iv) What is the smallest number of extra laddoos Vimala must make on Tuesday so that nothing is left over?
Answer:
(i) A number is divisible by 9 if the sum of its digits is divisible by 9. Adding the digits again and again until one digit is left gives the remainder on division by 9 (9 means no remainder).
(ii) 4 + 5 + 3 + 6 = 18, a multiple of 9, so 4536 is divisible by 9. Number of trays = 4536 ÷ 9 = 504.
(iii) 2 + 8 + 1 + 4 = 15 and 1 + 5 = 6, so 6 laddoos are left over.
(iv) She needs 9 − 6 = 3 more laddoos, making 2817, a multiple of 9.
In simple words: Monday's 4536 fills 504 trays exactly; Tuesday leaves 6 over, so 3 more laddoos complete a tray.

Teacher's Note:
1. The leftover count is the digital root, unless it is 9.
2. Extra needed = 9 − leftover.

Question 12. The torn bill. Rohit buys stationery for his school and the shopkeeper hands him a printed bill. On the way home the bill tears and one digit of the total gets smudged. All that can be read is the amount ₹2 1 ▫ 4, a four-digit total whose tens digit is missing. Rohit remembers two things the shopkeeper said: the total was shared equally by 3 classes with nothing left over, and the same total could also be shared equally by 6 groups with nothing left over.
(i) Using the digit-sum shortcut, list all digits that can go in the smudged place so that the total is a multiple of 3.
(ii) Explain, without any calculation, why the total is certain to be an even number.
(iii) Hence state which of your answers in (i) also make the total a multiple of 6.
(iv) For which digit is the total also a multiple of 9? Find that total and the quotient on dividing it by 9.
Answer:
(i) Digit sum = 2 + 1 + ▫ + 4 = 7 + ▫. For a multiple of 3 it must be 9, 12 or 15, so the digit is 2, 5 or 8.
(ii) The units digit is 4, which is even, so the total is even whatever the smudged digit is (the other place values 2000, 100 and 10 × ▫ are all multiples of 10).
(iii) A number divisible by both 2 and 3 is divisible by 6. The totals 2124, 2154 and 2184 are all even multiples of 3, so all three digits 2, 5 and 8 work.
(iv) For a multiple of 9, 7 + ▫ = 9, so the digit is 2; the total is ₹2124, and 2124 ÷ 9 = 236.
In simple words: The missing digit is 2, 5 or 8 for divisibility by 3 and by 6, and only 2 makes ₹2124, a multiple of 9.

Teacher's Note:
1. The units digit alone settles evenness.
2. 7 + ▫ = 18 would need ▫ = 11, which is not a digit.

Question 13. Bus ticket serial numbers. The tickets in a conductor’s book are numbered one after another in order. At the first stop he tears off four tickets in a row for a family, and later he notices that the four serial numbers on them add up to 2186. At the next stop he tears off five tickets in a row for a group of students, and those five serial numbers add up to 2190. The conductor is curious whether he can work backwards from a total to the actual ticket numbers, without checking the book at all.
(i) If the smallest of the four serial numbers is n, write their sum in terms of n.
(ii) Find the four serial numbers torn off at the first stop.
(iii) The conductor claims that four consecutive serial numbers can sometimes add up to an odd total. Is he right? Give a reason.
(iv) Find the middle serial number of the five tickets torn off at the next stop.
Answer:
(i) The numbers are n, n + 1, n + 2, n + 3, so the sum is 4n + 6.
(ii) 4n + 6 = 2186 gives 4n = 2180 and n = 545. The tickets are 545, 546, 547 and 548.
(iii) He is wrong. 4n + 6 = 2(2n + 3) always has 2 as a factor, so the total is always even.
(iv) Five consecutive numbers with middle number m add up to 5m. 5m = 2190 gives m = 438 (the tickets are 436 to 440).
In simple words: Four tickets in a row add to 4n + 6, giving 545 to 548, and five in a row add to 5 times the middle one, 438.

Teacher's Note:
1. Check (ii): 545 + 546 + 547 + 548 = 2186.
2. For an odd count of numbers, use the middle one as the letter.

Question 14. Sharing the prize money. A school’s inter-house quiz carries a team prize of ₹7560. The prize has to be shared equally among the members of the winning team, and only whole rupees may be given out. Before the result is announced, the teacher checks which team sizes would allow an exact sharing. She uses the shortcuts she teaches in class rather than dividing each time: the last digit for 5, the last two digits for 4, the digit sum for 9, and the alternating digit sum for 11.
(i) Can a team of 5 share the prize exactly? Give the shortcut you used.
(ii) Can a team of 9 share the prize exactly? If yes, state each member’s share.
(iii) Can a team of 4 share the prize exactly? Give the shortcut you used.
(iv) For a team of 11, how many rupees would be left over?
Answer:
(i) Yes. The units digit of 7560 is 0, so it is divisible by 5; each member gets ₹1512.
(ii) Yes. 7 + 5 + 6 + 0 = 18, a multiple of 9. Each share is 7560 ÷ 9 = ₹840.
(iii) Yes. The last two digits form 60 = 4 × 15, so 7560 is divisible by 4; each member gets ₹1890.
(iv) Alternating sum: 0 − 6 + 5 − 7 = −8, so 7560 is 8 short of a multiple of 11, that is, 11 − 8 = 3 more than one. ₹3 would be left over (7560 = 11 × 687 + 3).
In simple words: 7560 shares exactly among 5, 9 or 4 people, but a team of 11 leaves ₹3 over.

Teacher's Note:
1. Each shortcut matches one divisor.
2. Turn a negative alternating sum into a remainder by adding 11.

Question 15. A library’s check digit. A college library gives every book a six-digit accession code. The last digit is a check digit: it is chosen so that the whole six-digit code is divisible by 11. If a clerk types a code that is not divisible by 11, the computer refuses it at once, so most typing slips are caught before the record is saved. A new book has been given the code 9 2 4 6 3 ▫, and the librarian must now work out the check digit. Later the same day, a clerk enters the code 924683 for this book and the computer rejects it.
(i) State the shortcut for checking divisibility by 11.
(ii) Find the check digit that completes the new book’s code.
(iii) Show that 924683 is rejected, and state the remainder it leaves on division by 11.
(iv) The clerk’s error was to interchange the last two digits of the correct code. Explain, using the alternating sum, why interchanging two neighbouring digits almost always changes the result of the check.
Answer:
(i) Put alternating ‘+’ and ‘−’ signs before the digits, starting with ‘+’ at the units digit, and add. If the result is 0 or a multiple of 11, the number is divisible by 11.
(ii) For 9 2 4 6 3 ▫: ▫ − 3 + 6 − 4 + 2 − 9 = ▫ − 8. For this to be 0 with a single digit, ▫ = 8. The code is 924638 = 11 × 84058.
(iii) For 924683: 3 − 8 + 6 − 4 + 2 − 9 = −10, not a multiple of 11, so the code is rejected. Being 10 short of a multiple of 11, it leaves remainder 1.
(iv) Two neighbouring digits x and y always carry opposite signs in the alternating sum. Swapping them changes the sum by 2(x − y) (or its negative), which is 0 only when x = y.
So for unequal digits the check value changes and the slip is caught; it can escape only if 2(x − y) is a multiple of 11, which is impossible for two different digits. Here x − y = 8 − 3 = 5, a change of 10, from 0 to −10.
In simple words: The check digit is 8; swapping two neighbouring different digits shifts the alternating sum by twice their difference, so the error shows up.

Teacher's Note:
1. Check: 11 × 84058 = 924638.
2. The difference of two different digits is at most 9, so 2(x − y) can never be ±11 or ±22.

Question 16. Counting the diyas. A potter has made a tray of diyas for a fair. He counts them in different ways. Arranging them in rows of 4, three diyas are left over. Arranging them in rows of 5, four are left over. Arranging them in rows of 6, five are left over. He tells his daughter, “I have fewer than one hundred diyas, and if you think about what is missing each time rather than what is left over, you will find the number at once.”
(i) Write an algebraic expression for all numbers that leave a remainder of 3 when divided by 4.
(ii) Explain what the potter means by thinking about what is missing, and show that the number of diyas is 1 less than a multiple of 4, of 5 and of 6.
(iii) Find the number of diyas.
(iv) What is the next number of diyas that would behave in exactly the same way?
Answer:
(i) Such numbers are 3 more than a multiple of 4: 4k + 3 (also written 4k − 1).
(ii) 3 left out of a row of 4 means he is 1 diya short of another full row; 4 out of 5 and 5 out of 6 also mean 1 short. So if N is the number of diyas, N + 1 is a multiple of 4, of 5 and of 6.
(iii) N + 1 is a multiple of LCM(4, 5, 6) = 60. As N < 100, N + 1 = 60 and N = 59. Check: 59 = 4 × 14 + 3 = 5 × 11 + 4 = 6 × 9 + 5.
(iv) The next possibility is N + 1 = 120, giving 119 diyas.
In simple words: One more diya would fill every arrangement, so the count is one less than 60, that is 59; the next such count is 119.

Teacher's Note:
1. 'Remainder one less than the divisor' means 'one short'.
2. Use the LCM to find common multiples.

Question 17. The nine-check at the till. Before calculators were common, shopkeepers checked long additions using digital roots, a method described by Aryabhata II in his work Mahāsiddhānta. You replace each number by its digital root, add the roots, take the digital root of that sum, and compare it with the digital root of your answer. If the two differ, the answer is certainly wrong. Meera adds three bills, ₹4892, ₹3756 and ₹2135, and writes the total as ₹10783. Her assistant then copies the total into the register as ₹10738.
(i) Find the digital roots of 4892, 3756 and 2135.
(ii) Carry out the nine-check on Meera’s total of 10783 and say whether it passes.
(iii) Carry out the same check on the assistant’s figure of 10738.
(iv) What does part (iii) tell you about the kind of mistake this check can never detect? Explain briefly.
Answer:
(i) 4892 → 23 → 5; 3756 → 21 → 3; 2135 → 11 → 2.
(ii) Sum of roots: 5 + 3 + 2 = 10 → 1. Digital root of 10783: 1 + 0 + 7 + 8 + 3 = 19 → 10 → 1. They agree, so the total passes (and 4892 + 3756 + 2135 = 10783 is indeed correct).
(iii) 10738: 1 + 0 + 7 + 3 + 8 = 19 → 10 → 1. This also matches, so the wrong figure passes the check too.
(iv) The check depends only on the digit sum, and interchanging digits does not change the digit sum. So the check can never detect a mistake that only swaps digits (here 8 and 3). It can prove an answer wrong, but never prove it right.
In simple words: Both 10783 and 10738 pass the nine-check, because swapping digits keeps the digit sum the same.

Teacher's Note:
1. A passed check is not a proof of correctness.
2. Digit-swap errors need a different check, such as the one for 11.

Question 18. The scoreboard puzzle. At a school quiz, the three teams are called Agni, Bhumi and Cauvery. The scoreboard painter is playful: instead of the real scores he paints a puzzle on the board. Each team’s score is a two-digit number whose two digits are the same, written with the team’s initial, so Agni scored AA, Bhumi scored BB and Cauvery scored CC. The painter writes that the three scores together come to the three-digit total ABC. Different letters stand for different digits, and no number begins with 0.
AA + BB + CC = ABC
(i) Show that AA + BB + CC is always a multiple of 11.
(ii) Explain why A cannot be 3 or more.
(iii) Find the digits A, B and C, and write out the completed sum.
(iv) Which team won, and by how many points?
Answer:
(i) AA = 11A, BB = 11B and CC = 11C, so AA + BB + CC = 11(A + B + C), which always has 11 as a factor.
(ii) The largest possible scores with different digits are 99, 88 and 77, with total 99 + 88 + 77 = 264, less than 300. So the hundreds digit A cannot be 3 or more.
(iii) In place-value form, 11A + 11B + 11C = 100A + 10B + C, which simplifies to B + 10C = 89A.
Since B + 10C is at most 99, A = 1, so B + 10C = 89, giving C = 8 and B = 9. The sum is 11 + 99 + 88 = 198.
(iv) Bhumi won with 99 points, beating Cauvery (88) by 11 points (and Agni, with 11, by 88 points).
In simple words: Place values give B + 10C = 89A, so A = 1, B = 9, C = 8 and 11 + 99 + 88 = 198.

Teacher's Note:
1. A repeated-digit number like AA is 11 × A.
2. Bound the total first to limit A.

Question 19. Boxes of twenty-four. A chocolate unit packs its bars into cartons of 24. The supervisor, Nikhil, says, “Since 24 = 4 × 6, I will simply check whether the day’s output is divisible by 4 and by 6. If it passes both tests, the cartons will come out exactly full.” On Wednesday the output is 1836 bars and on Thursday it is 4728 bars. The quality officer disagrees with Nikhil’s rule and suggests checking divisibility by 3 and by 8 instead.
(i) Show that 1836 passes Nikhil’s test, that is, that it is divisible by both 4 and 6.
(ii) Show that 1836 is nevertheless not divisible by 24, and state how many bars would be left over.
(iii) Using prime factorisation, explain why checking 4 and 6 cannot settle divisibility by 24, while checking 3 and 8 can.
(iv) Apply the officer’s test to Thursday’s output of 4728 and state the number of full cartons.
Answer:
(i) The last two digits form 36 = 4 × 9, so 1836 is divisible by 4. It is even and its digit sum 1 + 8 + 3 + 6 = 18 is a multiple of 3, so it is divisible by 6. Nikhil’s test is passed.
(ii) 24 × 76 = 1824 and 1836 − 1824 = 12, so 1836 is not divisible by 24; 12 bars are left over. (Quick check: 836 is not a multiple of 8, so 1836 is not divisible by 8, a factor of 24.)
(iii) 24 = 2³ × 3 needs three factors of 2 and one 3. 4 = 2² and 6 = 2 × 3 share a factor 2, so together they guarantee only 2² × 3 = 12 = LCM(4, 6). Indeed 1836 = 2² × 3³ × 17 has only two 2s.
8 = 2³ and 3 have no common factor, so LCM(3, 8) = 24, and passing both tests does guarantee divisibility by 24.
(iv) 4728: digit sum 21, a multiple of 3; last three digits 728 = 8 × 91, a multiple of 8. So 4728 is divisible by 24, giving 4728 ÷ 24 = 197 full cartons.
In simple words: 4 and 6 together only guarantee 12, so 1836 leaves 12 bars over; 3 and 8 guarantee 24, and 4728 makes 197 cartons.

Teacher's Note:
1. Shared prime factors are the reason a pair of tests can fail.
2. The test for 8 uses the last three digits.

Question 20. The crate code. A fruit-packing unit stamps a five-digit code on every crate. To be valid, the code must be a multiple of 45. Today’s stamping machine has failed twice, so the code on the crate reads 3 ▫ 7 5 ▫, where the missing digit in the thousands place is called p and the missing units digit is called q. The supervisor must find every code that could possibly be valid, because the crates will later be sorted in order of their codes.
(i) Since 45 = 9 × 5, state the two separate tests the code must pass, and explain why passing both is enough.
(ii) Use the test for 5 to find the possible values of q.
(iii) For each value of q, use the digit-sum test to find the matching value of p, and hence write down every valid code.
(iv) Exactly one of the valid codes is also a multiple of 90. Identify it, give a reason, and state the quotient on dividing it by 90.
Answer:
(i) The code must pass the test for 5 (units digit 0 or 5) and the test for 9 (digit sum a multiple of 9). As 9 and 5 have no common factor, LCM(9, 5) = 45, so passing both means divisible by 45.
(ii) The units digit is q, so q = 0 or 5.
(iii) Digit sum = 3 + p + 7 + 5 + q = 15 + p + q.
If q = 0: 15 + p = 18, so p = 3, giving 33750 = 45 × 750.
If q = 5: 20 + p = 27, so p = 7, giving 37575 = 45 × 835. These are the only valid codes.
(iv) 90 = 45 × 2, so a valid code is a multiple of 90 only if it is also even. 37575 is odd, while 33750 is even, so the code is 33750, and 33750 ÷ 90 = 375.
In simple words: The tests for 5 and 9 give only 33750 and 37575, and the even one, 33750, is the multiple of 90.

Teacher's Note:
1. For each q the digit sum allows only one digit p.
2. A multiple of 90 must end in 0.