Class 8 Maths Ganita Prakash Chapter 5 Number Play NCERT Solutions

Figure it Out – Page 122

1. The sum of four consecutive numbers is 34. What are these numbers?
Solution:
Let four consecutive numbers be x, (x + 1), (x + 2) and (x + 3).
x + (x + 1) + (x + 2) + (x + 3) = 34
x + x + 1 + x + 2 + x + 3 = 34
4x + 6 = 34
4x = 34 – 6
4x = 28
x = 28/7 = 7.

So, (x + 1 ) = 7 + 1 = 8
(x + 2) = 7 + 2 = 9
(x + 3) = 7 + 3 = 10
Therefore, the given four consecutive numbers are 7, 8, 9, and 10.

2. Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.
Solution:
If p is the greatest of five consecutive numbers, then the other four numbers are (p – 1), (p – 2), (p – 3), and (p – 4).

3. For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.
(i)  The sum of two even numbers is a multiple of 3.
Solution:
Let the two even numbers be 2a + 2b
Sum = 2a + 2b = 2(a + b)
For 2(a + b) to be a multiple of 3, (a + b) must be multiple of 3.
Example:
2 + 4 = 6 → divisible by 3
2 + 8 = 10 → not divisible by 3 
Conclusion: Sometimes true.

(ii) If a number is not divisible by 18, then it is also not divisible by 9.
Solution:
If a number is divisible by 18, then it is also divisible by 9 because 9 is a factor of 18.
18a ÷ 9 = 2a → divisible by 9.
But if a number is divisible by 9, it is not always divisible by 18.
9b ÷ 18 = b/2 → not divisible by 9.
Example: 9 is divisible by 9 but not divisible by 18.
27 is divisible by 9 but not 18.
Conclusion: Sometimes true.

(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.
Solution:
Let the two numbers be a and b.
Not divisible by 6 means they do not satisfy 6∣a or 6∣b.
But their sum can still be divisible by 6.
Example: 2 and 4 → both not divisible by 6.
But, 2 + 4 = 6 → divisible by 6.
Conclusion: Sometimes true.

(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.
Solution:
Let the multiple of 6 be 6a, the multiple of 9 be 9b.
Sum: 6a + 9b = 3(2a + 3b)→ clearly divisible by 3.
Example:
6 + 9 = 15 → divisible by 3.
12 + 18 = 30 → divisible by 3.
Conclusion: Always true.

(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
Solution:
Let multiple of 6 be 6a, multiple of 3 be 3b.
Sum: 6a + 3b = 3(2a + b).
For it to be divisible by 9, 2a + b must be divisible by 3.
Example:
6 (6 × 1) + 3 (3 × 1) = 9 →divisible by 9
6 + 6 = 12 → not divisible by 9
Conclusion: Sometimes true.

4. Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.
Solution:
L.C.M of 3 and 4 = 12.
All such numbers are given by the expression = 12a + 2.
Examples:
(i) 12 × 1 + 2 = 12 + 2 = 14.
(ii) 12 × 2 + 2 = 24 + 2 = 26.
(iii) 12 × 3 + 2 = 36 + 2 = 38.

5. “I hold some pebbles, not too many, When I group them in 3’s, one stays with me. Try pairing them up — it simply won’t do, A stubborn odd pebble remains in my view. Group them by 5, yet one’s still around,
But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?”

Solution:
Grouped in 3’s leaves 1.
Pairing (2’s) leaves 1.
Grouped by 5 leaves 1.
Grouped by 7 is perfect.
Number ≤ 100.

L.C.M of 2, 3, and 5 = 30.
In all those cases, when we group them, 1 pebble remains.
So, the actual number of pebbles must be = 30 + 1 = 31, but 31 is not divisible by 7.

The next multiple of 30 is 2 × 30 = 60.
So, 60 + 1 = 61, but this is also not divisible by 7.

Similarly, the next number is 90 + 1 = 91.
And 91 is divisible by 7.
Hence, the number of pebbles I hold = 91.

6. Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, “If you add any three such numbers, the sum will always be a multiple of 6.” Is Tathagat’s claim true?
Solution:
A number that leaves remainder of 2 when divided by 6 can be written as 6k + 2.
Three such numbers are: (6a + 2), (6b + 2), (6c + 2).
(6a + 2) + (6b + 2) + (6c + 2) = 6(a + b + c) + 6 = 6(a + b + c + 1).
This sum is divisible by 6.
So yes, Tathagat’s claim is always true.
Example:
Take 20, 26, 32 → sum = 78 → divisible by 6.
Take 2, 8, 14 → sum = 24 → divisible by 6.

7. When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually.
(i) 4779 + 661 (ii) 4779 – 661
Solution:
(i) 4779 + 661
= Remainder 5 + Remainder 3
= Remainder 8
8 divided by 7 → remainder 1.

(ii) 4779 – 661
= Remainder 5 – Remainder 3
= Remainder 2

8. Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?
Solution:
A number that leaves a remainder of 2 when divided by 3 is = 3x + 2
A number that leaves a remainder of 3 when divided by 4 is = 4x + 3
A number that leaves a remainder of 4 when divided by 5 is = 5x + 4
L.C.M of 3, 4, and 5 = 60
All the numbers are the same, so 4x + 3 = 3x + 2
4x – 3x = 2 – 3
x = -1
Each remainder is 1 less than the divisor.
Hence, the number is 1 less than the L.C.M = (60 – 1) = 59.

So, 59 is the smallest number that satisfies all the given conditions.