Let m be the resultant integer. Then n/(100-n) = m.
Cross multiply and rearrange the terms: mn - 100m + n = 0
Subtract 100 from each side and factor: (m+1)*(n-100) = -100
So now I can conclude that there is an integer solution for each of the signed factorizationss of -100. 100=2^2+5^2, so the number of factorizations is 2*(2+1)*(2+1)=18.
The 18 values of n can be derived by taking any signed factor of -100 and adding 100 to that factor.
Edited on September 9, 2024, 1:26 pm
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