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What is the largest positive integer n for which n^3+100 is divisible by n+1?

 Jan 14, 2022
 #1
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By division we find that π‘›3+100=(𝑛+10)(𝑛2βˆ’10𝑛+100)βˆ’900n3+100=(n+10)(n2βˆ’10n+100)βˆ’900.

Therefore, if π‘›+10n+10 divides π‘›3+100n3+100, then it must also divide 900. Since we are looking for largest π‘›n, π‘›n is maximized whenever π‘›+10n+10 is, and since the largest divisor of 900 is 900, we must have π‘›+10=900⇒𝑛=890n+10=900β‡’n=890

The largest π‘› is therefore 890

 

 

 

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

 Jan 14, 2022
 #2
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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

 Jan 14, 2022

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