Reducto reducto reducto

Find the smallest positive integer N N , such that the fractions

1 N + 21 , 2 N + 22 , , 301 N + 321 \frac{1}{N+21}, \frac{2} {N+22}, \ldots, \frac{301} { N + 321}

are all in lowest terms.

Details and assumptions

The fraction a b \frac{a}{b} is in lowest terms if a a and b b are coprime.


The answer is 287.

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1 solution

Calvin Lin Staff
May 13, 2014

For k k from 1 1 to 301 301 , we must have gcd ( k , N + k + 20 ) = 1 \gcd (k, N+k + 20 ) = 1 . This is equivalent to gcd ( k , N + 20 ) = 1 \gcd(k, N + 20) = 1 .

Clearly, the values of N N from 1 1 to 281 281 will not work. Neither will any of the numbers N = 282 N= 282 to 286 286 , since n + 20 n + 20 will be composite, hence have a smaller factor. However, for N = 287 N= 287 , we have N + 20 = 307. N+20 = 307. It is a prime, so gcd ( k , N + 20 ) = 1 \gcd(k, N+20) = 1 for all k < 307. k<307. Thus, the smallest positive integer is 287.

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