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What is the units digit of $13^4\cdot17^2\cdot29^3$?

I saw this on a GMAT practice test and was wondering how to approach it without using a calculator. Thanks.

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5 Answers 5

up vote 7 down vote accepted

If you compute modulo $10,$ then you'll get $$13^4 17^2 29^3 \equiv 3^4 7^2 (-1)^3\equiv -81\cdot49\equiv (-1)^2\equiv 1 (\mathrm{mod}~10).$$ Thus the last digit is $1.$

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3  
This is exactly the right technique, but we can make things slightly easier for mental math by taking $13^4 17^2 29^3=(3^2)^2(-3)^2(-1)^3=(9^2)(9)(-1)=(-1)^2(-1)(-1)=1\pmod{10}$. –  Aaron Aug 5 '12 at 4:12

$13^1$ will give $3$ at the units place

$13^2$ will give $9$ at the units place, we get $9$ by taking the units digit of the result $3^2=9$

$13^3$ will give $7$ at the units place, we get $7$ by taking the units digit of the result $3^3=27$

$13^4$ gives $1$ at the units place, we get $1$ by taking the units digit of the result $3^4=81$

$17^2$ gives $9$ at the units place, we get $9$ by taking the units digit of the result $7^2=49$

$29^3$ gives $9$ at the units place, we get $9$ by taking the units digit of the result $9^3=729$

units digit of ($13^4$)($17^2$)($29^3$)=(units digit of $13^4$)(units digit of $17^2$)(units digit of $29^3$)

units digit of ($13^4$)($17^2$)($29^3$)=units digit of $(1)(9)(9)$

units digit of ($13^4$)($17^2$)($29^3$)=units digit of $81$

i.e. units digit of ($13^4$)($17^2$)($29^3$)=$1$

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explanation is lengthy but this can be done quickly in mind. –  Rajesh K Singh Aug 5 '12 at 5:39

My approach was to look at the unit digits (ie work mod 10) as follows:

$3 \times 7 =21$ and $9 \times 9 = 81$

This enables me to reduce to $3^2 \times 9 = 9 \times 9 = 81$ (I have taken out $13^2 17^2 29^2$)

And the answer is 1.

I think searching for simple products which are congruent to 1 is a useful practical technique for such questions.

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The last digit $(10x+t)^n$ (where n is positive integers, t odd digit so that (t,10)=1)has a period(d) which must divide $\phi(10)=4$ as $a^\phi(m)≡1(mod\ m)$ for(a,m)=1(this is known as Euler's Totient theorem).

The period of $(10x+1)^n$ is 1{1},

the period of $(10x+3)^n$ is 4 {3,9,7,1},

the period of $(10x+7)^n$ is 4 {7,9,3,1} and

the period of $(10x+7)^n$ is 2 {9,1}.

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Hint $\rm\ mod\ 10\!:\ 3^2\equiv -1,\,$ so $3$ is a sqrt of $\,-1$. As in $\Bbb C,\,$ put $\, {\it\color{#C00} i} \equiv \color{#C00}3\equiv \sqrt{-1},\, $ so $\color{#0A0}{\rm\: 7\equiv -3\equiv -{\it i}},\:$ so

$$\rm mod\ 10\!:\ \ 13^4\, 17^2\, 29^3 \equiv\, \color{#C00}3^4\, \color{#0A0}7^2\, 9^3\equiv {\it \color{#C00}i}^{\,4} (\color{#0A0}{-{\it i}})^2 (-1)^3 \equiv\, 1 (-1) (-1)\,\equiv\, 1$$

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