Track / Stage 5 / 246 of 400 #846 of 1964
Problem 846 AIME late Number theory Difficulty 5.5 Prove it
Prove that the number 30 239 + 239 30 30^{239} + 239^{30} 3 0 239 + 23 9 30 is composite.
This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.
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Official solution This number is divisible by 31.
## Solution
30 239 + 239 30 ≡ ( − 1 ) 239 + 1 = 0 ( m o d 31 ) 30^{239}+239^{30} \equiv(-1)^{239}+1=0(\bmod 31) 3 0 239 + 23 9 30 ≡ ( − 1 ) 239 + 1 = 0 ( mod 31 ) .
Let p p p be a prime number. Prove that ( a + b ) p ≡ a p + b p ( m o d p ) (a+b)^{p} \equiv a^{p}+b^{p}(\bmod p) ( a + b ) p ≡ a p + b p ( mod p ) for any integers a a a and b b b .
## Solution
( a + b ) p ≡ a + b ≡ a p + b p ( m o d p ) (a+b)^{p} \equiv a+b \equiv a^{p}+b^{p}(\bmod p) ( a + b ) p ≡ a + b ≡ a p + b p ( mod p )
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Source: NuminaMath-1.5 ,
licensed Apache-2.0 .
Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.