A number theory problem by Anubhav Jain

let the number be 'x' and the divisor be 'd' .

By division algorithm ,

x = dm + 24 .......... (1)

2x = dn + 11 ........... (2)

Multiplying (1) by 2 ,

2x = 2dm + 48 ........ (3)

Equating (2) , (3) ,

dn + 11 = 2dm + 48

dn - 2dm = 48 - 11

d( n - 2m) = 37

Since 37 is a prime number and d,n,m are positive integers , only 2 cases come into consideration -

1) d = 1 ; n - 2m = 37

But if d = 1 , then d | x .

So remainder must be 0 always!

So , d = 1 is not possible.

2) So we are left with the only case , d = 37 , which is possible.

Hence , the required divisor is 37 :)

Let the original number be 'a'

Let the divisor be 'd'

Let the quotient of the division of a by d be 'x'

Therefore, we can write the relation as a by d = x and the remainder is 24.

i.e., a = dx + 24

When twice the original number is divided by d, 2a is divided by d.

We know that a = dx + 24. Therefore, 2a = 2dx + 48

The problem states that 2dx + 48 by d leaves a remainder of 11.

2dx is perfectly divisible by d and will therefore, not leave a remainder.

The remainder of 11 was obtained by dividing 48 by d.

When 48 is divided by 37, the remainder that one will obtain is 11.

Hence, the divisor is 37.