CBSE Class 8 Mathematics Number Play Short Answer Questions with Answers

Short Answer Questions

Question 1. The sum of four consecutive numbers is 54.
(i) Write the four numbers.
(ii) If the smallest of them is n, write the sum in terms of n.
Answer:
(i) The numbers are 12, 13, 14 and 15, since 12 + 13 + 14 + 15 = 54.
(ii) The numbers are n, n + 1, n + 2, n + 3, so the sum is 4n + 6.
Check: 4n + 6 = 54 gives 4n = 48 and n = 12, matching part (i).
In simple words: Four numbers in a row starting at n add to 4n + 6, and 4n + 6 = 54 gives 12, 13, 14, 15.

Teacher's Note:
1. Solving 4n + 6 = 54 is a neat way to find part (i).
2. Always check by adding the four numbers.

Question 2. Express 30 as a sum of two or more consecutive natural numbers in three different ways.
Answer:
30 = 9 + 10 + 11 (three numbers)
30 = 6 + 7 + 8 + 9 (four numbers)
30 = 4 + 5 + 6 + 7 + 8 (five numbers)
30 cannot be written as a sum of two consecutive numbers, because such a sum is always odd.
In simple words: 30 = 9 + 10 + 11 = 6 + 7 + 8 + 9 = 4 + 5 + 6 + 7 + 8.

Teacher's Note:
1. For three numbers, divide by 3 to get the middle one.
2. Two consecutive numbers always add to an odd number.

Question 3. Without dividing, decide whether each number below is divisible by 9. Give the reason in each case.
(i) 5283
(ii) 72104
Answer:
(i) 5 + 2 + 8 + 3 = 18, a multiple of 9, so 5283 is divisible by 9.
(ii) 7 + 2 + 1 + 0 + 4 = 14, not a multiple of 9, so 72104 is not divisible by 9. As 1 + 4 = 5, it leaves remainder 5.
In simple words: Digit sum 18 means divisible by 9; digit sum 14 means not divisible.

Teacher's Note:
1. The digit-sum test decides both cases.
2. The repeated digit sum also gives the remainder.

Question 4. Use the shortcut for 11 to decide whether 61809 is divisible by 11. Show the alternating sum you use.
Answer:
Starting with ‘+’ at the units digit: 9 − 0 + 8 − 1 + 6 = 22.
22 is a multiple of 11, so 61809 is divisible by 11. Indeed 61809 = 11 × 5619.
In simple words: The alternating sum is 22, a multiple of 11, so 61809 divides exactly by 11.

Teacher's Note:
1. The alternating sum may be 11, 22 or any multiple of 11, not just 0.
2. Write each sign clearly to avoid slips.

Question 5. Find the digital root of 487356. What does this digital root tell you about the remainder when 487356 is divided by 9?
Answer:
4 + 8 + 7 + 3 + 5 + 6 = 33 and 3 + 3 = 6, so the digital root is 6.
Each place value is 1 more than a multiple of 9, so the digital root is the remainder on division by 9. Hence 487356 leaves remainder 6.
In simple words: The digits add to 33, then 6, so dividing by 9 leaves 6.

Teacher's Note:
1. Digital root and remainder on division by 9 are the same (9 means 0).
2. Check: 9 × 54150 = 487350, leaving 6.

Question 6. A three-digit number is written as 100c + 10b + a, where a, b and c are its units, tens and hundreds digits. Use this form to explain why such a number is divisible by 10 only when a = 0.
Answer:
100c + 10b + a = 10(10c + b) + a.
The part 10(10c + b) is a multiple of 10 whatever b and c are.
So the number is a multiple of 10 only if a is a multiple of 10; as a is a single digit, a = 0.
In simple words: All of the number except the units digit is already a multiple of 10, so the units digit must be 0.

Teacher's Note:
1. Take 10 out as a common factor of the first two terms.
2. The same idea gives the tests for 2 and 5.

Question 7. Solve the cryptarithm below. Each letter stands for exactly one digit, different letters stand for different digits, and a number does not begin with 0.
MN + N = NM
Answer:
In place-value form, MN = 10M + N and NM = 10N + M.
10M + N + N = 10N + M gives 9M = 8N.
As M and N are single digits and 9 does not divide 8, N must be a multiple of 9, so N = 9 and M = 8.
So M = 8, N = 9, and the sum is 89 + 9 = 98.
In simple words: Writing MN and NM with place values gives 9M = 8N, so M = 8, N = 9 and 89 + 9 = 98.

Teacher's Note:
1. Place-value form turns a cryptarithm into a simple equation.
2. N = 0 is not allowed, since NM cannot begin with 0.

Question 8. Is the statement “If a number is divisible by 24, then it is also divisible by 6” always true, sometimes true or never true? Justify your answer using algebra.
Answer:
The statement is always true.
A number divisible by 24 is 24m for some integer m, and 24m = 6 × 4m = 6(4m), which has 6 as a factor.
General rule: if a number is divisible by k, it is divisible by every factor of k, and 6 is a factor of 24.
In simple words: Any multiple of 24 is 6 × (4 × something), so it is always a multiple of 6.

Teacher's Note:
1. The reverse is not true: 6 is not divisible by 24.
2. Write 24 as a product containing 6.

Question 9. Show using algebra that the sum of any five consecutive numbers is a multiple of 5. Can the sum of four consecutive numbers ever be a multiple of 4? Give a reason.
Answer:
Take the middle number as n: (n − 2) + (n − 1) + n + (n + 1) + (n + 2) = 5n, a multiple of 5.
For four consecutive numbers starting at n, the sum is 4n + 6 = 4(n + 1) + 2.
It always leaves remainder 2 on division by 4, so it is never a multiple of 4.
In simple words: Five numbers in a row add to 5 times the middle one, but four in a row always leave remainder 2 when divided by 4.

Teacher's Note:
1. Choosing the middle number as n makes the ±1 and ±2 cancel.
2. Example: 3 + 4 + 5 + 6 = 18, remainder 2 on division by 4.

Question 10. Two even numbers each leave a remainder of 2 when divided by 4. Show with algebra that their sum is always a multiple of 4, and illustrate with one example.
Answer:
Such numbers have the form 4p + 2 and 4q + 2.
(4p + 2) + (4q + 2) = 4p + 4q + 4 = 4(p + q + 1), a multiple of 4.
The two remainders of 2 join to make one more group of 4. Example: 6 + 10 = 16 = 4 × 4.
In simple words: Two leftover 2s make a full 4, so the sum is a multiple of 4, like 6 + 10 = 16.

Teacher's Note:
1. Picture the numbers as rows of 4 dots with 2 extra.
2. Pick any example such as 2 + 14 = 16 to check.

Question 11. In the four-digit number \( \overline{4y16} \), the letter y stands for a digit.
(i) Find y if the number is a multiple of 9.
(ii) List all values of y for which the number is a multiple of 3.
Answer:
Digit sum = 4 + y + 1 + 6 = 11 + y, which lies between 11 and 20.
(i) 11 + y must be 18, so y = 7; the number is 4716.
(ii) 11 + y must be 12, 15 or 18, so y = 1, 4 or 7.
In simple words: The digits add to 11 + y, which is 18 for y = 7 and a multiple of 3 for y = 1, 4, 7.

Teacher's Note:
1. Every multiple of 9 is a multiple of 3, so 7 appears in both answers.
2. Find the range of the digit sum first.

Question 12. Without doing long division, find the remainder when 8356 is divided by
(i) 9
(ii) 11.
Answer:
(i) 8 + 3 + 5 + 6 = 22 and 2 + 2 = 4, so the remainder is 4.
(ii) Alternating sum from the units digit: 6 − 5 + 3 − 8 = −4. So 8356 is 4 short of a multiple of 11, that is, 11 − 4 = 7 more than one. The remainder is 7.
In simple words: The digit sum gives remainder 4 for 9, and the alternating sum −4 gives remainder 7 for 11.

Teacher's Note:
1. A negative alternating sum needs one extra step.
2. Check: 11 × 759 = 8349 and 8356 − 8349 = 7.

Question 13. Write an algebraic expression that describes every number leaving a remainder of 4 when divided by 7. Using it, find the smallest three-digit number of this kind.
Answer:
Such numbers are 4 more than a multiple of 7: 7k + 4, where k = 0, 1, 2, 3, …
For a three-digit number, 7k + 4 ≥ 100, so 7k ≥ 96 and k ≥ 13.7…, giving k = 14.
The smallest such number is 7 × 14 + 4 = 102.
In simple words: The numbers are 7k + 4, and the first one with three digits is 102.

Teacher's Note:
1. k must be a whole number, so round 13.7 up.
2. Check: 102 = 7 × 14 + 4.

Question 14. Tanvi picks three numbers, each of which leaves a remainder of 3 when divided by 9. Show that the sum of her three numbers is always a multiple of 9.
Answer:
The numbers can be written as 9a + 3, 9b + 3 and 9c + 3.
Their sum is 9a + 9b + 9c + 9 = 9(a + b + c + 1), which has 9 as a factor.
Example: 12 + 21 + 39 = 72 = 9 × 8.
In simple words: The three leftover 3s add up to 9, so the sum is always a multiple of 9.

Teacher's Note:
1. Three remainders of 3 make one full 9.
2. This is like the 'pairs to make fours' idea.

Question 15. For which of these expressions is the value an even number for every integer value of the letters? Justify each answer, giving a non-example where needed.
(i) 6m − 4n
(ii) 5p + 3p
(iii) 3a + 3b
Answer:
(i) 6m − 4n = 2(3m − 2n) has 2 as a factor, so it is always even.
(ii) 5p + 3p = 8p = 2(4p), so it is always even.
(iii) 3a + 3b = 3(a + b) is even only when a + b is even. a = 2, b = 4 gives 18 (even), but a = 1, b = 2 gives 9 (odd). So it is not always even.
In simple words: The first two always have 2 as a factor; 3a + 3b can be odd, as 3 + 6 = 9 shows.

Teacher's Note:
1. Combine like terms first, as in 5p + 3p = 8p.
2. One odd value is enough to rule out 'always even'.

Question 16. Use the shortcuts for 2 and for 3 to decide which of 3174 and 2936 is divisible by 6. Explain why checking 2 and 3 is enough for 6.
Answer:
3174 ends in 4, so it is divisible by 2; 3 + 1 + 7 + 4 = 15 is a multiple of 3, so it is divisible by 3. Hence 3174 is divisible by 6 (3174 = 6 × 529).
2936 is even, but 2 + 9 + 3 + 6 = 20 is not a multiple of 3, so 2936 is not divisible by 6.
2 and 3 have no common factor, so a number with both as factors has 2 × 3 = 6 as a factor.
In simple words: 3174 passes both tests for 2 and 3, so it is divisible by 6; 2936 fails the test for 3.

Teacher's Note:
1. Both tests must be passed.
2. LCM(2, 3) = 6, which is why the pair works.

Question 17. Write the three-digit number \( \overline{cba} \) in expanded place-value form. Using the fact that 9 and 99 are multiples of 3, justify the shortcut “a number is divisible by 3 if the sum of its digits is divisible by 3”.
Answer:
\( \overline{cba} \) = 100c + 10b + a = (99 + 1)c + (9 + 1)b + a = (99c + 9b) + (c + b + a).
99c + 9b = 3(33c + 3b) is a multiple of 3 whatever the digits are.
So the number and its digit sum c + b + a leave the same remainder on division by 3; the number is divisible by 3 exactly when its digit sum is.
Example: 582 = (99 × 5 + 9 × 8) + (5 + 8 + 2) and 15 is a multiple of 3, so 582 is too.
In simple words: Apart from the digit sum, the number is made of 99s and 9s, which are multiples of 3.

Teacher's Note:
1. Split each place value as a multiple of 9 plus 1.
2. The same proof works for 9.

Question 18. Find all pairs of digits a and b for which the four-digit number \( \overline{5a3b} \) is a multiple of 36. Show the two conditions you use.
Answer:
36 = 4 × 9, and 4 and 9 have no common factor, so the number must pass both tests.
Test for 4: the last two digits 3b must be a multiple of 4, so 3b = 32 or 36, giving b = 2 or 6.
Test for 9: 5 + a + 3 + b = 8 + a + b must be a multiple of 9.
If b = 2: 10 + a is a multiple of 9, so a = 8. If b = 6: 14 + a is a multiple of 9, so a = 4.
The pairs are (a, b) = (8, 2) and (4, 6): 5832 = 36 × 162 and 5436 = 36 × 151.
In simple words: The test for 4 gives b = 2 or 6, and the digit-sum test then gives a = 8 or 4.

Teacher's Note:
1. Use the test that limits the choices most first.
2. Check each final number by dividing by 36.

Question 19. Solve the cryptarithm below, where each letter stands for one digit and different letters stand for different digits.
PQ × P = QRQ
Answer:
QRQ is a three-digit number, so Q ≠ 0. The units digit of Q × P is Q, so Q × (P − 1) ends in 0.
P = 1 gives PQ × 1 = PQ, only two digits, so P ≠ 1. If P = 6, Q is even: 62 × 6 = 372, 64 × 6 = 384, 68 × 6 = 408, none of the form QRQ.
Otherwise Q = 5 and P is odd (P ≠ 5): 35 × 3 = 105, 95 × 9 = 855 do not fit, but 75 × 7 = 525 does.
So P = 7, Q = 5, R = 2, and the cryptarithm is 75 × 7 = 525.
In simple words: The units digit forces Q = 5 (the case P = 6 fails), and only 75 × 7 = 525 has the pattern QRQ.

Teacher's Note:
1. List every case allowed by the units digit before testing.
2. Check the final product: 75 × 7 = 525.

Question 20. A six-digit number is made by writing a three-digit number twice over, as in 437437. Show with algebra that every such number is divisible by 11.
Answer:
Let the three-digit number be N. Writing it twice puts the first copy in the thousands places, so the six-digit number is 1000N + N = 1001N.
1001 = 11 × 91, so the number is 11 × 91N, which is always divisible by 11.
Example: 437437 = 437 × 1001 = 11 × 39767.
In simple words: Repeating a three-digit number multiplies it by 1001, and 1001 = 11 × 91.

Teacher's Note:
1. 1001 = 7 × 11 × 13, so such numbers are also divisible by 7 and 13.
2. Spot the 1001 by writing 1000N + N.