Number Theory

By division we find that 𝑛3+100=(𝑛+10)(𝑛2−10𝑛+100)−900n3+100=(n+10)(n2−10n+100)−900.

Therefore, if 𝑛+10n+10 divides 𝑛3+100n3+100, then it must also divide 900. Since we are looking for largest 𝑛n, 𝑛n is maximized whenever 𝑛+10n+10 is, and since the largest divisor of 900 is 900, we must have 𝑛+10=900⇒𝑛=890n+10=900⇒n=890

The largest 𝑛 is therefore 890

 

 

 

ok so the n3 and n2 is just \(n^3 or N^2\)

n takes the following values:

 

n ==2,  8,  10,  32  and  98

 

Therefore, the largest n==98

Check: [98^3 + 100] mod [98 + 1] ==0