A number when divided by a divisor leaves a remainder of 5 and when divided by twice the divisor leaves a remainder of 45. Find the divisor?

Let d be the divisor, Q1 and Q2 be the quotients of 1^st and 2^nd cases respectively.

According to the Euclid's Division Lemma,

If we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition a = b q + r where 0 ≤ r ≤ b. and q is the quotient and r is remainder.

In the first case, according to the given condition

Number = d × Q1 + 5 …… equation 1

Similarly, in the 2nd case, according to the given condition

Number = 2d × Q2 + 45 …… equation 2

Equating 1st and 2nd equations

d × Q1 + 5 = 2d × Q2 + 45

⇒ d × Q1 – 2d × Q2 = 45 – 5

⇒ d (Q1 – 2 Q2) = 40

⇒

Hence, divisor may be possible 40 or factor of 40 { e.g 2, 5, 8, 10, 20 } .

Now, in the first case

0 ≤ 5 < d and

In 2^nd case, 0 ≤ 45 < 2d

Or 0 ≤ 22.5 < d

It’s possible only when we take d = 40

Hence, divisor = 40.