I am trying to get a head start on practicing for UIL mathematics, and I am stumped on this random practice problem. I have no clue how to solve it. The question is:

What is the sum of the digits in the tens place and the units place of 7^65

I don't just want the answer, I have that, I want to be able to solve other questions like this. I appreciate the help.

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TopNewestRelevant article: finding the last few digits of a power

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You can notice that the unit digit change properly:

\(\color{red}\dots 7, \color{blue}\dots 9, \color{green}\dots 3, \color{yellow}\dots 1,\color{red}\dots 7, \color{blue}\dots 9, \color{green}\dots 3, \color{yellow}\dots 1,\color{red}\dots 7, \color{blue}\dots 9, \color{green}\dots 3, \color{yellow}\dots 1,\color{black}\dots\)

Since \[65\equiv 1 \ \mod 4\], the unit digit of the \(7^{65}\) will be

7.You can notice that the tens digit change properly:

\(\color{pink}\dots0., \color{green}\dots4., \ \dots4., \color{pink}\dots0., \ \dots0., \color{green}\dots4., \ \dots4., \color{pink}\dots0., \ \dots0., \color{green}\dots4., \ \dots4., \color{pink}\dots0., \ \dots0.,\color{black}\dots\)

So the first is 0, then 4400 repeat. So we subtract 1 from 65 (since the first, alone 0), and we know that

\[64\equiv0 \ \mod4\]

so the tens digit of \(7^{65}\) will be 0 (the last one from the repeating sequence).

So the sum is 7.

Do you need proof for the notices, or it is enough?

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Thank you so much! This helped a lot.

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