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I'm working on a problem for school where I have to evaluate by hand $(6619255^{19} + 6619255^5) mod 10$.

I'm really lost and not really sure how to go about this. I've done the following:

$(6619255^{19}mod10 + 6619255^5mod10) mod 10$ $((6619255 mod 10)^{19} + (6619255 mod 10)^5) mod 10$

Is this right? And how do I move on from here. I'm so confused and disheartened...

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  • $\begingroup$ Hint: for any number $k$ ending in a $5$, we have $k \text{mod} 10 = 5$, and $k$ raised to any positive integer power $n$ also ends in a $5$. $\endgroup$
    – WaveX
    Nov 10, 2017 at 2:08

2 Answers 2

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Hint: $$(10a+b)^n \equiv b^n \mod{10}.$$

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Note that $a=6619255$ is a multiple of $5$, so $a^n$ will still be a multiple of $5$ for any $n>0$.

Also $a$ is odd, and so will be $a^n$.

Finally $a^{19}$ and $a^5$ are odd and divisible by $5$, so the sum $a^{19}+a^5$ is even and divisible by $5$, this is equivalent to say that it is divisible by $10$ (divisible by $2$ and $5$ simultaneously).

Thus $6619255^{19}+6619255^5\equiv 0\pmod{10}$

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