Automorphic

Number Theory Level 3

Find the value of
∑ n = 1 99 987 6 n (mod 100) \sum _{n=1}^{99}9876^{n} \text{(mod 100)} hints


The answer is 24.

Jeff Giff by Jeff Giff , 13, China

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1 solution

Jeff Giff
Jun 4, 2020

Since 76 is an automorphic number ,
7 6 1 = 7 6 2 = 7 6 3 = … = 7 6 99 = 987 6 1 = 987 6 2 = … = 987 6 99 = 76 ( m o d 100 ) . 76^1=76^2=76^3=…=76^{99}=9876^1=9876^2=…=9876^{99}=76(mod 100).
Therefore, the sum is equal to 76 + 76 + 76 … = 76 × 99 = 7524 ( m o d 100 ) 76+76+76…=76\times99=7524(mod100) .
So the answer is 7524 mod 100=24.


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Nice problem!

Mahdi Raza - 1 year ago

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Thanks! BTW, I was inspired by yours 😊

Jeff Giff - 1 year ago

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Thanks, This is one level further though, Nice!! And that 98 in front totally deceived the problem as if it was hard, but we just need to know the fact that the last two digits of the product are determined by last two digits of the number itself

Mahdi Raza - 1 year ago

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@Mahdi Raza – You’re welcome :)

Jeff Giff - 1 year ago

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