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Number Theory Level 3

Which of the following expression is divisible by 9 with ∀ n ∈ Z + \forall n \in \mathbb Z^+ ?

2 2 n ( 2 n + 1 − 1 ) − 1 2^{2n}(2^{n + 1} - 1) - 1 2 n ( 2 n + 1 − 1 ) − 1 2^{n}(2^{n + 1} - 1) - 1 2 2 n ( 2 2 n + 1 − 1 ) − 1 2^{2n}(2^{2n + 1} - 1) - 1 2 n ( 2 2 n + 1 − 1 ) − 1 2^{n}(2^{2n + 1} - 1) - 1

Thành Đạt Lê by Thành Đạt Lê , 17, Singapore

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1 solution

Otto Bretscher
Dec 9, 2018

You made it rather easy for us, since the first three expressions are not divisible by 9 9 for n = 1 n=1 ;)

The last expression can be written as 2 2 n ( 2 2 n + 1 − 1 ) − 1 = ( 4 n − 1 ) ( 2 × 4 n + 1 ) \boxed{2^{2n}(2^{2n+1}-1)-1}=(4^n-1)(2\times4^n+1) , and both factors are divisible by 3 3 since 4 ≡ 1 ( m o d 3 ) 4\equiv 1 \pmod{3} .

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Thanks! I will not be that nice sometime soon. (Actually, I am not sure when.) Have a good day!

Thành Đạt Lê - 2 years, 6 months ago

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