What is the largest positive integer n for which n^3+100 is divisible by n+1?
+179 Because we can factorize n^3+1, we simply write n^3+100 as n^3+1+99.
Thus, because n+1 divides n^3+100, aka n^3+1+99, n+1 divides (n+1)(n^2-n+1)+99.
Knowing that (n+1) will always divide (n+1)(n^2-n+1), we simply need n+1 to divide 99.
The largest divisor of 99 is 99, meaning that n+1 will equal 99 to maximize n. With this, we have n=98.
The answer is 98.
+179 Because we can factorize n^3+1, we simply write n^3+100 as n^3+1+99.
Thus, because n+1 divides n^3+100, aka n^3+1+99, n+1 divides (n+1)(n^2-n+1)+99.
Knowing that (n+1) will always divide (n+1)(n^2-n+1), we simply need n+1 to divide 99.
The largest divisor of 99 is 99, meaning that n+1 will equal 99 to maximize n. With this, we have n=98.
The answer is 98.
