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Power and Digit Poser (Posted on 2015-12-23)

Find all possible pairs (M,N) of positive integers such that 10^{N^2} contains precisely
N*M - M digits.

No Solution Yet Submitted by K Sengupta

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Analytical solution

Comment 2 of 2 |

Starting with Charlie's equation,

m = (n^2 + 1) / (n - 1)

Complete the square to get

m= ((n^2-2n+1) + 2n)/(n-1)

= (n-1) + 2n/(n-1)

n-1 is relatively prime to n,

so the only way the right hand side is integral is if

(n-1) = 1 or (n-1) = 2

So the only solutions are n = 2 or 3. In both cases, m = 5.

Posted by Steve Herman on 2015-12-23 11:34:48

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