Problem Solving

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

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

(2^16)*(3^16)*(7^16) = 42^16

42^16 / 10 = (40+2)^16/10
if we expand all the terms, all will have 40^x with the exception of the last term
since 40^x/10 leaves 0 as a remainder, we need to worry about the last term, 2^16

2^16 / 10
2^1 / 10 - remainder =2
2^2 / 10 - remainder =4
2^3 / 10 - remainder =8
2^4 / 10 - remainder =6

2^5 / 10 - remainder =2
2^6 / 10 - remainder =4
2^7 / 10 - remainder =8
2^8 / 10 - remainder =6

notice that there is a pattern that repeats!

if 2^8 / 10 leaves a remainder of 6; 2^16 will leave a remainder of 6 as well

ans = d

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.

I think it's much simpler than that.

Since the exponents are the same, we could write this as

42¹�

Since the question is essentially asking for the units digit, we can ignore the tens digit, and treat this as

"What is the remainder when 2¹� is divided by 10?"

2¹� = (2�)� = 16�. Anything that ends in 6 must end in 6 when raised to a positive power, so the number must end in 6, and we're done!

Matt@VeritasPrep wrote:I think it's much simpler than that.

Since the exponents are the same, we could write this as 42¹�

Haha. I did that one at 2AM and I was thinking like, "I dunno if this is a realistic question. Takes too much work."

Guess I had the too much work part right!

Marty Murray wrote:

Matt@VeritasPrep wrote:I think it's much simpler than that.

Since the exponents are the same, we could write this as 42¹�

Haha. I did that one at 2AM and I was thinking like, "I dunno if this is a realistic question. Takes too much work."

Guess I had the too much work part right!

Even I solved the problem just like u marty.
but I did not feel it has much work. It is straight word that we need to find the units digit of the product.
But it is true that Matt reduced the amount of work needed to 1/3rd(no need to solve for 3 and 7).

Smart work indeed................