Question

When an idiot get 999 rejections he says this. ​

Answer 1

Three consecutive integers

Let the first integer be x, second be (x+1) & third be (x+2)

x+x+1+x+2= 51

3x+3 = 51

3x= 51–3

3x = 48

x= 48/3

x = 16

3x = 48

x = 48/3

x = 16

x +1 = 16+1 =17

x+2 = 16+2 = 18

Hence the integers are 16,17 & 18

Answer 2

Aɴsᴡᴇʀ࿐

Let Z be the number

Z = 13x + 11 where x is the quotient when Z is divided by 13

Z = 17y + 9 where y is the quotient when Z is divided by 17

13x + 11 = 17y + 9

13x + 2 = 17y since x and y are quotients they should be whole numbers . Since y has to be a whole number the left hand side should be multiple of 17

The least possible value of x satisfying the condition is 9 and y will be 7

The answer is 13*9 + 11 = 128 or it is 17*7 + 9 = 128

This is the least number possible. There will be multiple answers and will increase in multiples 17*13 = 221 like 349 , 570, etc.

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