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Can you figure out the digit in the tens place of the solution to $4^{2015} \cdot 9^{2016}$ without using a calculator?

By tens place I mean, for example if you have number $2451$, the number in tens place here is $5$.

I know the answer, but I don't know how to get it, so if you got any ideas, please share. :)

PiGuy314
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Mykybo
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3 Answers3

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$$ 4^{2015}\cdot9^{2016} = 9\cdot36^{2015} $$

So look at $9\cdot36^n$ for $n=1,2,3,\ldots$: \begin{align} 9\cdot36^1 & = 324 \\ & = \cdots24 \\[6pt] 9\cdot36^2 & = \cdots64 \\ 9\cdot36^3 & = \cdots04 \\ 9\cdot36^4 & = \cdots44 \\ 9\cdot36^5 & = \cdots84 \\ 9\cdot36^6 & = \cdots24 \longleftarrow\text{Now we're back to where we were when }n=1, \\ & \phantom{=\cdots24 \longleftarrow\text{a}}\text{so it starts over again.} \end{align} After five steps we return to where we started; thus at $n=1,6,11,16,21,26,\ldots$ we get $\cdots24$.

Of course, in order for this to make sense, you have to realize that when you multiply two numbers, the last two digits of the answer are determined by the last two digits of each of the numbers you're multiplying, and are not affected by any earlier digits. That is clear if you think about the algorithm for multiplication that you learned in elementary school, and you can also show it via some simple algebra.

So we get $24$ whenever the exponent is congruent mod $5$ to $1$, i.e. the last digit is $1$ or $6$. So $9\cdot36^{2016}=24$ and $9\cdot{2015}$ is one step before that, thus $84$.

Michael Hardy
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    I think it should be 6 * 335, not 355. Other than that it makes sense, but I am wondering whether there isn't some simpler solution. Since it should be done without calculator, you would have to count all the 7 items. (up to 9*36^7).. But I guess this is the simplest it can get. Thanks. – Mykybo Apr 10 '15 at 22:40
  • @Mykybo : Correct. I've fixed it. – Michael Hardy Apr 10 '15 at 22:44
  • $9\times36^{7} \equiv 64 \pmod{100}$, whereas $9\times36^{6} \equiv 24 \pmod{100}$. Thus you have actually to divide by $5$, and you'll get $9\times36^{402\times5+1}\equiv24 \pmod{100}$, then remains $36^4\equiv 16$ (because $2015=5\times 403$), and finally $24\times16\equiv 84$. – Jean-Claude Arbaut Apr 10 '15 at 22:48
  • @Jean-ClaudeArbaut : You're right: somehow I let $\cdots16$ get into the sequence and that threw off the rest of it. I hope I've got all the arithmetic right now. ${}\qquad{}$ – Michael Hardy Apr 10 '15 at 22:54
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You have

$$4^{11}=4194304\equiv 4 \pmod{100}$$

And since $2015=183\times 11+2$, and $183=16\times 11+7$, you have

$$4^{2015}=4^{11\times183+2}=4^2\times(4^{11})^{183}\equiv 4^2\times 4^{183} \pmod{100}$$

$$=4^2\times4^{11\times16+7}=4^9\times(4^{11})^{16}\equiv 4^9\times4^{16}=4^{25}=4^3\times(4^{11})^2\equiv4^5 \pmod{100}$$ $$=1024\equiv 24\pmod{100}$$

Likewise,

$$9^{10}=3486784401\equiv1\pmod{100}$$

Thus

$$9^{2016}=9^6\times (9^{10})^{201}\equiv 9^6=531441\equiv 41\pmod{100}$$

Therefore

$$4^{2015}\times9^{2016}\equiv 24\times41=984\equiv 84\pmod{100}$$

And the digit in tens place is $8$.

Jean-Claude Arbaut
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Hint, consider computing the value $4^{2015}9^{2016}\mod 100$.

See the technique here: calculating $a^b \!\mod c$

Community
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TravisJ
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