Find that minimum value of x

Positive integer x x is such that the remainder of x 2 x^{2} when divided by 9 is 7. If x > 23 x>23 , what is the minimal value of x x ?


The answer is 31.

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1 solution

Jared Low
Nov 15, 2018

For positive integers x x , x 2 x^2 leaves a remainder of 7 when divided by 9 if and only if x 4 , 5 ( m o d 9 ) x \equiv 4, 5 \pmod{9} . The sequence of such numbers is 4 , 5 , 22 , 23 , 31 , 32 , 4,5,22,23,31,32, \ldots , the minimal value of which that exceeds 23 being 31 \boxed{31}

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