﻿ Find the remainder of \dfrac{5^{22}+7}{8}

#### Question

Find the remainder of \dfrac{5^{22}+7}{8}

Collected in the board: Number Theory

Steven Zheng posted 4 months ago

Let's denote r as remainder and evaluate the remainder of 8 divide 5^n

n =1, 5/8, r =5

n=2, 25/8, r=1

n=3, 125/8, r=5

n=4, 625/8, r =1

...

It is observated when n is odd, the reminder of 8 divide 5^n is 5 and when is even, the remainder of 8 divide 5^n is 1.

Therefor

The remainder of \dfrac{5^{22}}{8} is 1. The remainder of \dfrac{5^{22}+7}{8} is 0

Steven Zheng posted 4 months ago

Using the identity

x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)

Using the power rule for exponent, the original expression is transformed to

\dfrac{5^{22}+7}{8} = \dfrac{25^{11}-1+8}{8}

Apply the binomial identity

25^{11}-1

= (25-1)(25^{10}+25^9+\dots+25+1)

=24\cdot (25^{10}+25^9+\dots+25+1)

Let k = 25^{10}+25^9+\dots+25+1

Obviously k is an integer

Then,

\dfrac{5^{22}+7}{8}

=\dfrac{24k+8}{8}

=3k+1

which leaves no remainder.

Steven Zheng posted 4 months ago

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