Tower of 31 part 5 (For Nihar 2)

Number Theory Level 4

∑ n = 1 5 x 31 n \sum_{n=1}^5 x^{31n}

Nihar was working on a problem where he found following pattern for x 31 n x^{31n} where n = 1 , 2 , 3 , 4 , 5 n=1,2,3,4,5 . Find the last 3 digits of the summation above when x = 31 x=31 .


The answer is 455.

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1 solution

Alex Spagnoletti
Mar 24, 2016

∑ n = 1 5 x 31 n \displaystyle \sum_{n=1}^5 x^{31n} for n = 1 , 2 , 3 , 4 , 5 n= 1, 2, 3, 4, 5 is the same of 3 1 31 ( 1 + 3 1 31 ( 1 + 3 1 31 ( 1 + 3 1 31 ( 1 + 3 1 31 ) ) ) ) 31^{31}(1+31^{31}(1+31^{31}(1+31^{31}(1+31^{31})))) . Being 3 1 31 ≡ 431 ( m o d 1000 ) 31^{31} \equiv 431 \pmod {1000} , I can rewrite the expression as 431 ( 1 + 431 ( 1 + 431 ( 1 + 431 ( 1 + 431 ) ) ) ) 431(1+431(1+431(1+431(1+431)))) and I can easily calculate the products ( m o d 1000 ) \pmod {1000} . The result is 455 \boxed {455}

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