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How can I find the hundreds’ place digit of the following number:

$$1993^3 – 913^3 – 1083^3$$

I have not tried this one since I don't know how to begin. I can tell the units digit of this but how to approach for hundred's place. I am looking for approaches on how to proceed with these types of questions.

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You can compute modulo 1000: $-7^3+87^3-83^3 \mod 1000$. Then take the leading digit. –  WimC Feb 24 '12 at 10:10
    
These numbers aren't that big, are they? You could compute the exact result in a couple of minutes (it's 5,885,006,373). –  PseudoNeo Feb 24 '12 at 16:19
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2 Answers

up vote 3 down vote accepted

They key here is to use modular arithmetic. What you want to find is the last three digits of $1993^3-913^3-1083^3$, and to do so you can take advantage of the fact that if $x-y$ is a multiple of $1000$ (in which case we say $x\equiv y\mod 1000$), then $x$ and $y$ have the same last three digits. In your case, we have $$\begin{eqnarray}1993^3-913^3-1083^3&\equiv&(-7)^3-(-87)^3-83^3\mod 1000\\ &\equiv&-7^3+87^3-83^3\mod 1000\\ &\equiv&-343+658503-571787\mod 1000\\ &\equiv&-343+503-787\mod 1000\\ &\equiv&-627\mod 1000\\ &\equiv& 373\mod 1000\end{eqnarray}$$ so the last 3 digits are $373$. This approach works for finding other places (replace $10^3$ with $10^n$) and in other bases as well (replace $10^n$ with $b^n$).

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Thanks so much.Your answer is correct. But i am not able to get this line "if x−y is a multiple of 1000 (in which case we say x≡ymod1000)" –  vikiiii Feb 24 '12 at 10:14
    
@anna Have you seen modular arithmetic before? –  Alex Becker Feb 24 '12 at 10:21
    
i know mod function. Actually tell me what are x and y? –  vikiiii Feb 24 '12 at 10:24
    
@anna I was just making the general statement that for any integers $x$ and $y$, if $x-y$ is a multiple of $1000$ then $x$ and $y$ have the same last three digits. –  Alex Becker Feb 24 '12 at 10:27
    
OK..alex thank you..got it..:) –  vikiiii Feb 24 '12 at 10:29
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One can do this with mental arithmetic if one first employs the binomial theorem, viz.

$\begin{eqnarray} (100 a + 10 b + c)^3\ &\equiv& &\ \ \ c^3& + 30 c (bc &+& 10 (ac +\ b^2))\pmod{1000},\quad \rm therefore \\ 1993^3\equiv 993^3\ &\equiv& &\ \ \ 27& +\ 90 (27 &+& 10 (27 + 81)) \\ 913^3\ &\equiv& &\ \ \ 27& +\ 90 (\ 3 &+& 10 (27 +\ 1\ )) \\ 1083^3\equiv 083^3 \ &\equiv& &\ \ \ 27& +\ 90 (24 &+& 10 (\ 0\ + 64)) \\ 1993^3-913^3-1083^3\ &\equiv& &-27& +\ 90 (\ 0 &+& 10(81-65))\equiv -27 - 100\cdot 16 \equiv -627 \equiv 373 \end{eqnarray}$

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