CBSE Class 8 Mathematics Number Play Very Short Answer Questions with Answers

Very Short Answer Questions

Question 1. State the shortcut for checking whether a given number is divisible by 9.
Answer:
A number is divisible by 9 if and only if the sum of its digits is divisible by 9. The digits may be added again and again until a single digit is left.

Teacher's Note:
1. The final single digit is 9 for every multiple of 9.
2. This works because each place value is 1 more than a multiple of 9.

Question 2. Find the digital root of 742615.
Answer:
7 + 4 + 2 + 6 + 1 + 5 = 25 and 2 + 5 = 7.
The digital root is 7.

Teacher's Note:
1. Keep adding until one digit remains.
2. It is also the remainder on division by 9.

Question 3. Hemant writes the four consecutive numbers 3, 4, 5, 6 in that order and puts a plus or a minus sign between each pair of them. What can you say about the parity of every value he can obtain?
Answer:
Every value is even. Changing one sign changes the value by twice a number, an even change, so all the expressions 3 ± 4 ± 5 ± 6 have the same parity as 3 + 4 + 5 + 6 = 18, which is even.

Teacher's Note:
1. Find one value and all the others share its parity.
2. For example, 3 − 4 + 5 − 6 = −2 is even too.

Question 4. Is the expression 3g + 5h even for all integer values of g and h? Give one pair of values to support your answer.
Answer:
No. For g = 2, h = 1 it gives 6 + 5 = 11, which is odd. (For g = h = 1 it gives 8, so it is only sometimes even.)

Teacher's Note:
1. One counter-example disproves an 'always' claim.
2. 2 cannot be taken out as a common factor of 3g + 5h.

Question 5. Write the general form of a three-digit number whose hundreds, tens and units digits are c, b and a respectively.
Answer:
The number is 100c + 10b + a.

Teacher's Note:
1. Each digit is multiplied by its place value.
2. Do not write cba as a product c × b × a.

Question 6. Find the smallest digit d for which the number \( \overline{4d7} \) is a multiple of 3.
Answer:
Digit sum = 4 + d + 7 = 11 + d, which must be a multiple of 3.
The smallest digit is d = 1, giving 417 (digit sum 12).

Teacher's Note:
1. Try d = 0 first: 11 is not a multiple of 3.
2. The next digits that work are 4 and 7.

Question 7. State the shortcut given in this chapter for checking whether a number is divisible by 11.
Answer:
Place alternating plus and minus 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.

Teacher's Note:
1. The result may be negative, and that is fine.
2. This is called the alternating sum of the digits.

Question 8. Using the shortcut for 11, check whether 6853 is divisible by 11.
Answer:
Alternating sum: 3 − 5 + 8 − 6 = 0.
So 6853 is divisible by 11 (6853 = 11 × 623).

Teacher's Note:
1. An alternating sum of 0 counts as a multiple of 11.
2. Start from the units digit, not from the left.

Question 9. Without dividing, find the remainder when 452117 is divided by 9.
Answer:
4 + 5 + 2 + 1 + 1 + 7 = 20 and 2 + 0 = 2.
The remainder is 2.

Teacher's Note:
1. The digital root is the remainder on division by 9.
2. The only exception is a digital root of 9, which means remainder 0.

Question 10. The digital root of a number is 4. What remainder does the number leave when it is divided by 9?
Answer:
The digital root is the remainder on division by 9, so the remainder is 4.

Teacher's Note:
1. Example: 13 has digital root 4, and 13 = 9 + 4.
2. Only a digital root of 9 means remainder 0.

Question 11. For which divisors is it enough to look only at the units digit of a number to decide divisibility?
Answer:
For 2, 5 and 10. Every place value above the units place is a multiple of 10, and so of 2 and 5, so only the units digit decides.

Teacher's Note:
1. For 4 and 8 you need the last two and last three digits.
2. For 3, 9 and 11 all the digits matter.

Question 12. Name the Indian mathematician and his work in which the method of finding a digital root is mentioned.
Answer:
Aryabhata II (about 950 CE) described it in his work Mahāsiddhānta.

Teacher's Note:
1. He used digital roots to check arithmetic.
2. Do not confuse him with the earlier Aryabhata (about 499 CE).

Question 13. Using the place-value form … + 1000d + 100c + 10b + a, explain why a number is divisible by 10 only when the units digit a is 0.
Answer:
Every term except a (10b, 100c, 1000d, …) is a multiple of 10.
So the whole number is a multiple of 10 only when the leftover part a is a multiple of 10; as a is a single digit, a = 0.

Teacher's Note:
1. Split the number into a 'multiple of 10' part and the units digit.
2. A single digit is a multiple of 10 only if it is 0.

Question 14. Find the remainder when 745296 is divided by 11.
Answer:
Alternating sum from the units digit: 6 − 9 + 2 − 5 + 4 − 7 = −9.
So 745296 is 9 short of a multiple of 11, that is, 11 − 9 = 2 more than one. The remainder is 2.

Teacher's Note:
1. A negative result −r means the remainder is 11 − r.
2. Check: 11 × 67754 = 745294, and 745296 − 745294 = 2.

Question 15. Write an algebraic expression that describes all the numbers which leave a remainder of 2 when divided by 3 and also a remainder of 2 when divided by 4.
Answer:
Such a number is 2 more than a common multiple of 3 and 4, that is, of 12.
So it is 12n + 2 (2, 14, 26, 38, …).

Teacher's Note:
1. The same remainder for both divisors means the LCM can be used.
2. LCM(3, 4) = 12.

Question 16. Solve the cryptarithm EF × 4 = GFE, where different letters stand for different digits.
Answer:
The product has three digits, so E ≥ 3; and 4 × F ends in E, so E is even: 4, 6 or 8.
Trying each: 41 × 4 = 164, 46 × 4 = 184, 64 × 4 = 256, 69 × 4 = 276, 87 × 4 = 348 fail, but 82 × 4 = 328 has the form GFE.
So E = 8, F = 2, G = 3: 82 × 4 = 328.

Teacher's Note:
1. The product has three digits, so E is at least 3.
2. Always check the final sum by multiplying.

Question 17. Write one number which shows that checking divisibility by 4 and by 6 is not enough to decide divisibility by 24.
Answer:
12 is divisible by both 4 and 6, but not by 24.
Checking 4 and 6 only guarantees a multiple of LCM(4, 6) = 12.

Teacher's Note:
1. 36 or 60 would also work.
2. 4 and 6 share the factor 2, so their LCM is less than 24.

Question 18. Find the digit a for which the three-digit number \( \overline{7a4} \) is a multiple of 11.
Answer:
Alternating sum: 4 − a + 7 = 11 − a, which must be 0 or a multiple of 11.
So a = 0 and the number is 704 = 11 × 64.

Teacher's Note:
1. For a digit, 11 − a lies between 2 and 11.
2. Only 11 itself is a multiple of 11 in that range.

Question 19. Show, using algebra, that the sum of four consecutive numbers is always even but is never a multiple of 4.
Answer:
n + (n + 1) + (n + 2) + (n + 3) = 4n + 6 = 2(2n + 3), so it is always even.
Also 4n + 6 = 4(n + 1) + 2 leaves remainder 2 on division by 4, so it is never a multiple of 4.

Teacher's Note:
1. Write the sum two ways: as 2 × something and as 4 × something + 2.
2. Example: 1 + 2 + 3 + 4 = 10.

Question 20. When divided by 8, one number leaves a remainder of 5 and another leaves a remainder of 7. What remainder will their sum leave when it is divided by 8?
Answer:
The numbers are 8p + 5 and 8q + 7; their sum is 8p + 8q + 12 = 8(p + q + 1) + 4.
The remainder is 4.

Teacher's Note:
1. Add the remainders: 5 + 7 = 12, then take out one more 8.
2. Example: 13 + 15 = 28 = 8 × 3 + 4.

Question 21. Solve the cryptarithm AB + AB + AB = CBB, where different letters stand for different digits.
Answer:
3 × B must end in B, so B = 0 or 5. B = 0 would need 30A = 100C, which no digits satisfy, so B = 5.
Then 3 × A5 = C55; trying values, 85 × 3 = 255 works.
So A = 8, B = 5, C = 2: 85 + 85 + 85 = 255.

Teacher's Note:
1. Adding a number three times is multiplying it by 3.
2. Start from the units column.

Question 22. How many multiples of 9 lie between 500 and 560?
Answer:
The multiples are 504, 513, 522, 531, 540, 549 and 558.
There are 7 of them.

Teacher's Note:
1. The first one has digit sum 9: 5 + 0 + 4.
2. Then keep adding 9 until you pass 560.

Question 23. Two even numbers each leave a remainder of 2 when divided by 4. Is their sum a multiple of 4? Justify your answer using algebra.
Answer:
Yes. The numbers are 4p + 2 and 4q + 2, and (4p + 2) + (4q + 2) = 4(p + q + 1), a multiple of 4.

Teacher's Note:
1. The two remainders of 2 make one more 4.
2. Example: 6 + 10 = 16.

Question 24. Write the smallest three-digit number that is divisible by both 9 and 10.
Answer:
It must be a multiple of LCM(9, 10) = 90.
The smallest three-digit multiple of 90 is 180.

Teacher's Note:
1. 90 itself has only two digits.
2. Check: 1 + 8 + 0 = 9 and it ends in 0.

Question 25. Find the smallest number greater than 1 that leaves a remainder of 1 when divided by 2, by 3 and by 4, and is also divisible by 5.
Answer:
The number is 1 more than a common multiple of 2, 3 and 4, that is, 1 more than a multiple of 12: 13, 25, 37, …
The first of these divisible by 5 is 25 (25 = 12 × 2 + 1 = 5 × 5).

Teacher's Note:
1. LCM(2, 3, 4) = 12.
2. List 12k + 1 and look for one ending in 0 or 5.

Question 26. Find the digital root of the number 27a + 45b + 17 for all whole numbers a and b.
Answer:
27a + 45b + 17 = 9(3a + 5b) + 17, and 17 = 9 + 8.
So the remainder on division by 9 is 8, and the digital root is always 8.

Teacher's Note:
1. Multiples of 9 do not affect the digital root.
2. Check with a = b = 0: 17 → 1 + 7 = 8.

Question 27. Sunita claims, “Some multiples of 7 stop being multiples of 7 when they are tripled.” Is her claim correct? Justify using algebra.
Answer:
No, her claim is wrong. A multiple of 7 is 7k, and 3 × 7k = 7(3k), which is again a multiple of 7.

Teacher's Note:
1. Every multiple of a multiple of 7 is a multiple of 7.
2. Example: 14 × 3 = 42 = 7 × 6.

Question 28. List all the pairs of digits (a, b) for which the four-digit number \( \overline{2a6b} \) is divisible by 44.
Answer:
44 = 4 × 11. Test for 4: the last two digits 6b must be a multiple of 4, so b = 0, 4 or 8.
Test for 11: b − 6 + a − 2 = a + b − 8 must be 0 or a multiple of 11, so a + b = 8.
Pairs: (0, 8), (4, 4) and (8, 0), giving 2068, 2464 and 2860.

Teacher's Note:
1. 4 and 11 have no common factor, so both tests together give 44.
2. a + b = 19 is impossible here, as b is at most 8.

Question 29. Find three consecutive numbers in which the first is a multiple of 3, the second is a multiple of 4 and the third is a multiple of 5.
Answer:
3, 4 and 5 work. (Such triples repeat every LCM(3, 4, 5) = 60 numbers, so 63, 64, 65 is the next set.)

Teacher's Note:
1. Small numbers are worth trying first.
2. Adding 60 keeps each number a multiple of the same divisor.

Question 30. Explain why the product of any four consecutive integers is always a multiple of 24.
Answer:
Among any four consecutive integers, one is a multiple of 4, another is a different even number, and one is a multiple of 3.
So the product has the factors 4 × 2 × 3 = 24 and is always a multiple of 24.

Teacher's Note:
1. Example: 5 × 6 × 7 × 8 = 1680 = 24 × 70.
2. Two even numbers appear in any four consecutive integers.