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I am currently working on some algebra and I am studying the modulus chapter of my book. One question was finding the last digit of:

$$37^{100}$$

They give a hint about how the solution should be about calculating a remainder...

Is there any easy solution to this?

I know that there is a solution of just finding the pattern of the last digit of $ 37^x$. But I don't understand how that has to do with calculating a remainder...

Thank you!

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1  
Last digit, you mean? –  Thomas Andrews Feb 6 '13 at 15:26
    
Yes, that is right. –  Lukas Arvidsson Feb 6 '13 at 15:27
1  
Hint: What do you get if you look at some number mod $10$? –  Tobias Kildetoft Feb 6 '13 at 15:28
    
Use Euler's totient theorem –  tetori Feb 6 '13 at 15:29

5 Answers 5

up vote 14 down vote accepted

The last digit of $37^{100}$ is the same as the last digit of $7^{100}$ or of $492038497^{100}$.

When you multiply two positive integers, the last digit in the product depends on those two integers only through their last digits. That will become obvious if you look at the way they told you to multiply numbers by hand in third or fourth grade or whenever it was: $$ \begin{array}{rrrrrrrrrr} & & & 4 & 2 & 7 \\ & & \times & 3 & 1 & 9 \\ \hline & & \bullet & \bullet & \bullet & 3 \\ & & \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet & \bullet \\ \hline \bullet & \bullet & \bullet & \bullet & \bullet & 3 \end{array} $$ Where did that last $3$ come from??

Now multiply $7$, and discard all but the last digit: \begin{align} 7 \times 7 & & & = \cdots\ 9 \\ 7 \times 7 \times 7 & = (\cdots\ 9\times 7) & &= \cdots \bullet \\ 7 \times 7 \times 7 \times 7 & = (\cdots \bullet\times 7) & & = \cdots \bullet \\ 7 \times 7 \times 7 \times 7 \times 7 & = (\cdots \bullet\times 7) & & = \cdots\bullet \\ \end{align} Just throw away all but the last digit at each step.

Now notice something: There are only $9$ digits that could possibly appear as the last digit (you can't get $0$ when you're multiplying non-zero digits). That means you can't keep getting new digits there that you haven't seen before.

All this will lead you to see a pattern that occurs as you continue the process. And you'll see why that happens. And that will lead you quickly to the answer.

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Thank you for an excellent answer! –  Lukas Arvidsson Feb 6 '13 at 20:29
    
@LukasArvidsson: welcome ;) –  Chris's sis Feb 6 '13 at 20:30

Since $37^2=1369$, then $$37^{100}=(37^2)^{50}=(1369)^{50}$$ and the last digit of $(1369)^{n}$ oscillates between $9$ if $n$ odd and $1$ if $n$ even.

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Thank you for taking the time to answer my question, it was very helpful! –  Lukas Arvidsson Feb 6 '13 at 20:29

This is a question about calculating modulo 10. So the last digit of $37^{100}$ is the same as the last digit of $7^{100}$; which is the same as the last digit of $49^{50}$, and so the same as the last digit of $9^{50}$. Perhaps you can carry on from there.

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Thank you for that explanation, it makes sense now ;) –  Lukas Arvidsson Feb 6 '13 at 20:30

HINT

You want to calculate the remainder after dividing by $10$. It's also good to know that $37^2$ has remainder $9$ after dividing by $10$, which may be written as $37^2 \equiv -1 \pmod{10}$.

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One finds the modulus of $37^{100}$ in decimal, by noting that in decimal, anything raised to the fourth power must end in $0$, $1$, $5$, or $6$. When various powers of 5 are added, then the number must be adjent to a multiple of $5*5^n$, where $5^n$ divides the index.

So, we look for solutions, ending in any of these three digits: $000$, $001$, $625$, or $376$, depending on what the common divisor with 10 is. Here it's 1, so it ends in $001$.

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