infiniti007 wrote:What is the remainder when (2^16)*(3^16)*(7^16) is divided by 10?
A.) 0
B.) 2
C.) 4
D.) 6
E.) 8
When a positive integer is divided by 10, the remainder is equal to the value of the units digit. So what we really need here is simply the units digit of (2^16)*(3^16)*(7^16).
The units digit of a product is only affected by the units digits of the numbers being multiplied.
For example, the units digit of 1029 x 269 is the same at the units digit of 9 x 9. In both cases the units digit is 1.
So here all that will matter is the units digits of (2^16), (3^16), and (7^16).
When numbers are raised to consecutive powers, the units digits of those numbers fall into repeating patterns.
Let's look at 2 first.
2^1 = 2 (units digit of 2)
2^2 = 4 (units digit of 4)
2^3 = 8 (units digit of 8)
2^4 = 16 (units digit of 6)
Now since 2 x 16 = 32, with a units digit of 2, the pattern starts over again.
2^5 = 32 (units digit of 2)
2^6 = 64 (units digit of 4)
In fact the other digits don't even matter.
2^7 = _ _ 8 (units digit of 8)
2^8 = _ _ 6 (units digit of 6)
So as we raise 2 to consecutive powers the units digits follow the repeating pattern 2, 4, 8, 6.
So 2^16 will have a units digit of 6.
Now we just need to do the same with 3 and 7.
3^1 = 3 (units digit of 3)
3^2 = 9 (units digit of 9)
3^3 = 27 (units digit of 7)
3^4 = 81 (units digit of 1)
Since 1 x 3 = 3 we can tell that the pattern will now start over.
So the pattern for 3 is 3, 9, 7, 1, and 3^16 has a units digit of 1.
7^1 = 7
7^2 = 49
7^3 = _ _ 3 (I am not even going to bother figuring it out. 7 x 9 = 63. So I know the units digit of 7 x 49 is 3)
7^4 = _ _ _ 1 (I just need to know that the units digit of 7 x 3 is 1)
Since 1 x 7 = 7, I know that the pattern repeats again now.
So the pattern is 7, 9, 3, 1, and 7^16 has a units digit of 1.
Whoever wrote this question made it pretty easy for us now. We just multiply the units digits of (2^16), (3^16), and (7^16).
6 * 1 * 1 = 6
So the remainder is 6 and the answer is
D.
