Question

On dividing 77 into three parts such that sum of the first and the second multiplied by 3 the sum of ii and iii diminished by 3 and the sum of first and third increased by 3 may all be equal we get

Answer 1

Let the numbers be a, b & c

So

3 (a+b) = b+c - 3 = a + c +3

Taking

b + c - 3 = a + c + 3

b -6 = a ....(1)

Now taking

3(a+b) = b+ c - 3

3a + 3b = b + c-3

Substituting the value of a = b-6

3(b-6) + 3b =b + c -3

3b - 18 +3b = b + c -3

6b -18 = b + c -3

5b - c = 15 .....(2)

Also we have

a + b + c = 77

b - 6 + b + c = 77

2b + c = 83 .......(3)

Adding (2) and (3)

7b = 98

b = 14

Now a = b- 6 = 14-6 = 8

And

2b + c = 83

2(14) + c = 83

28 + c = 83

c = 55

The three parts are

8, 14 & 55