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?
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Write
n = q ⋅ d + 5 n = q ′ ⋅ 2 d + 4 5
subtract the equations and get
( q − 2 q ′ ) d = 4 0
So the divisor d is a divisor of 4 0 . Since 4 5 is a remainder of the division by 2 d we have necessarily 4 5 < 2 d . This inequality is true for only one divisor d of 4 0 , namely d = 4 0 .