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Is there a way to answer the following preferably without a calculator

What digit appears in unit place when $2^{320}$ is multiplied out ? a)$0$ b)$2$ c)$4$ d)$6$ e)$8$ ---- Ans(d)

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    You should try looking at the units digit of successive powers of 2 - 1,2,4,8,6,2,4,8,6,2,4,8,6 - notice a pattern?2012-07-31
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    Have you tried looking at the last digits of $2^1, 2^2, 2^3, 2^4, 2^5, 2^6$, etc.? Can you _prove_ that the pattern $2,4,8,6$ in the unit place digit must repeat over and over again? If so, what do you think unit place digit of $2^{320}$ is?2012-07-31

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Quickly look at the last digit of $2^n$, for $n=1$, $2$, $3$, and so on. So we keep multiplying by $2$. For determining the last digit of $2^{n+1}$, only the last digit of $2^n$ matters.

We get $2$, $4$, $8$, $6$, $2$ and the pattern starts all over again. The pattern is periodic with period $4$. So at $n$ a multiple of $4$, we get a $6$. But $320$ is a multiple of $4$.

Remark: If we wanted the last digit of $2^{999}$, note that $996$ is a multiple of $4$. So at $996$ we get a $6$. Now count forward: $2$, $4$, $8$: the answer is $8$. Or else $1000$ is a multiple of $4$. Go backwards one step from $6$: the last digit is $8$.

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    Nice method but $320$ is also a multiple of 2 so how do we choose weather its n=2 or n=4 ?2012-07-31
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    I did not look at $320$, at least not for a while. First I found that the last digits were periodic with period $4$. So the $4$ came first. It happens to divide $320$, which makes life easy. But look at the example in the remark. there our exponent is $999$, we look for a multiple of $4$ near it but below. So $2^{996}$ has last digit $6$. Now we get to $2^{999}$ by moving forward by $3$: from $6$ we get to $2$ then $4$ then $8$.2012-07-31
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    Yeah that also helps..2012-07-31
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$\rm mod\ 5\!:\, \color{#0A0}{2^4}\equiv \color{#C00}1\Rightarrow2^{320}\equiv (\color{#0A0}{2^4})^{80}\equiv \color{#C00}1^{80}\!\equiv \color{#C00}1.\,$ The only choice $\:\equiv \color{#C00}1\!\pmod 5\:$ is $6,\: $ in d).

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    Not really good at solving congruences but I got till $x = 320 \mod4$ and I get $x = 80$. But dont know how $x=80$ helps.2012-07-31
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    @Mis If congruences aren't familiar, then divisibility works: $\rm\:x\!-\!1\:|\:x^n\!-\!1,\:$ so $\rm\:5\:|\:2^4\!-\!1\:|\:(2^4)^{80}\!-\!1.\:$ Alternatively, using the Binomial Theorem $\rm\:2^{320}\! = (2^4)^{80}\! = (1 + 5 \cdot 3)^{80}\! = 1\!+\!5k\ \ $2012-07-31
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As the ending digits for powers of two are $${2}^{0} \mapsto 1$$ $${2}^{1} \mapsto 2$$ $${2}^{2} \mapsto 4$$ $${2}^{3} \mapsto8$$ $${2}^{4} \mapsto6$$ $${2}^{5} \mapsto2$$ You only have to do ${2}^{320 \mod(4)}$ to get the ending digit.

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    Not really good at solving congruences but I got till $x = 320 \mod4$ and I get $x = 80$. But dont know how $x=80$ helps.2012-07-31
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    To solve congruences you need the remainder, not the quotient. To find $x \pmod 4$ you express $x=4q+r$, then $x \equiv r \pmod 4$2012-07-31
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    @RossMillikan I can see how you got $x=4q+r$ from $x \equiv r ( \mod 4)$ however could you show how these equations might be used to solve the problem2012-07-31
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    @MistyD: The periodicity says you only need the remainder on dividing 320 by 4 to get the last digit. $2^{(4q)}$ will end in $6$ (0 is a startup transient), $2^{(4q+1)}$ will end in 2, etc. So once we know $320 \equiv 0 \pmod 4$, we know that $2^{320}$ ends in $6$2012-07-31
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Since $2^{320}$ is a square $2^{320}=(2^{150})^2$ and squares last digit is always $0,1,4,9,6,5$, two of them, $2$ and $8$, are ruled out. $0$ as well, since you need a $10$ for that. Write it as $(2^{75})^4$, and check that $4$th powers end with $0,1,6,1,6,5$ to get the remaining as result.