NCERT Solutions – Number Play
Exercise 5.1
1. The sum of four consecutive numbers is 34. What are these numbers?
Let the numbers be x, x+1, x+2, x+3
x + (x+1) + (x+2) + (x+3) = 34
4x + 6 = 34
4x = 28 → x = 7
Numbers are: 7, 8, 9, 10
2. Suppose p is the greatest of five consecutive numbers.
The numbers are:
p−4, p−3, p−2, p−1, p
3. Determine whether each statement is always true, sometimes true, or never true:
(i) The sum of two even numbers is a multiple of 3.
Example: 2+4=6 (multiple of 3), but 2+6=8 (not multiple of 3)
Answer: Sometimes True
(ii) If a number is not divisible by 18, then it is also not divisible by 9.
Example: 9 is divisible by 9 but not by 18
Answer: False (Sometimes True)
(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.
Example: 1 and 5 → both not divisible by 6, but sum = 6
Answer: Sometimes True
(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.
6a + 9b = 3(2a + 3b)
Answer: Always True
(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
Example: 6 + 3 = 9 (true), but 12 + 3 = 15 (not multiple of 9)
Answer: Sometimes True
4. Numbers leaving remainder 2 when divided by 3 and 4:
Form: 12k + 2
Examples: 2, 14, 26, 38, …
5. Pebbles problem:
Conditions:
- n ≡ 1 (mod 3)
- n ≡ 1 (mod 2)
- n ≡ 1 (mod 5)
- n ≡ 0 (mod 7)
- n > 100
Solution: n = 301
6. Tathagat’s claim:
Numbers of form: 6k + 2
Sum of three such numbers:
(6a+2) + (6b+2) + (6c+2) = 6(a+b+c+1)
Answer: Always True
7. Remainders when divided by 7:
Given:
- 661 ≡ 3 (mod 7)
- 4779 ≡ 5 (mod 7)
(i) 4779 + 661
(4779 + 661) ≡ (5 + 3) mod 7 = 8 mod 7 = 1
Answer: Remainder = 1
Algebraic form:
4779 = 7a + 5
661 = 7b + 3
Sum = 7(a + b) + 8 = 7(a + b + 1) + 1
(ii) 4779 − 661
(4779 − 661) ≡ (5 − 3) mod 7 = 2
Answer: Remainder = 2
Algebraic form:
Difference = (7a + 5) − (7b + 3) = 7(a − b) + 2
8. Find the smallest number with given conditions:
Conditions:
- n ≡ 2 (mod 3)
- n ≡ 3 (mod 4)
- n ≡ 4 (mod 5)
Observe pattern:
n + 1 is divisible by 3, 4, and 5
LCM of (3, 4, 5) = 60
So, n + 1 = 60 ⇒ n = 59
Answer: 59
Verification:
- 59 ÷ 3 → remainder 2
- 59 ÷ 4 → remainder 3
- 59 ÷ 5 → remainder 4
Conclusion:
59 is the smallest such number because it is just one less than the LCM of 3, 4, and 5.