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Two different numbers when divided by the same divisor the remainder are 26 and 47 respectively. When the sum of these two numbers are divided by the same divisor the remainder is 15. Find the divisor.

Two different numbers when divided by the same divisor the remainder are 26 and 47 respectively. When the sum of these two numbers are divided by the same divisor the remainder is 15. Find the divisor. Correct Answer 58

Given:

When two different number is divided by the same divisor the remainder are 26 and 47 respectively. And when the sum of these two numbers is divided by the same divisor the remainder is 15

Concept Used:

Dividend = Divisor × Quotient + Remainder

Calculation:

Let the divisor be ‘d’ and the quotient be q1 for the dividend x, q2 for the dividend y and q3 for the dividend (x + y)

∴ By the given condition,

x = q1 × d + 26

y = q2 × d + 47

And (x + y) = q3 × d + 15

Since q1, q2, q3 are whole numbers and in all cases, the remainder are 26, 47 and 15 then it is obvious that the divisor d > 47

∴ (x – 26), (y – 47) and (x + y – 15) all will completely divisible by d      ….(1)

∴ {(x – 26) + (y – 47)} is also completely divisible by d

⇒ (x + y – 73) is also completely divisible by d      ….(2)

Now, from (1) and (2) we can write {(x + y – 15) – (x + y – 73)} = 58, must be divisible by ‘d’ or ‘d’ be a factor of 58

Now, factors of 58 are 1, 2, 29 and 58

Since d > 47, the value of ‘d’ must be 58

∴ The required divisor is 58

 

Divisor = 1st remainder + 2nd remainder – 3rd remainder

Here 1st remainder = 26

2nd remainder = 47

3rd remainder = 15

∴ Required divisor = 26 + 47 – 15

⇒ 58

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