What is the units digit of $13^4\cdot17^2\cdot29^3$? 
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.
 A: 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 $1.$
A: $13^1$ 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 ($13^4$)($17^2$)($29^3$)=$1$
A: 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}.
A: 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.
A: 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 
$$\rm mod\ 10\!:\ \ 13^4\, 17^2\, 29^3 \equiv\, \color{#C00}3^4\, \color{#0A0}7^2\, 9^3\equiv {\it \color{#C00}i}^{\,4} (\color{#0A0}{-{\it i}})^2 (-1)^3 \equiv\, 1 (-1) (-1)\,\equiv\, 1$$
