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If p and q are positive integers...

Expert replies
by BTGmoderatorLU » Tue Oct 24, 2017 1:37 pm
If p and q are positive integers, what is the remainder when 9^2p × 5^(p+q) + 11^q × 6^pq is divided by 10?

A)0
B)1
C)3
D)4
E)5

The OA is B.

Can any expert explain this PS question please? I don't have it clear. Thanks.
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Source: — Problem Solving |

by Jay@ManhattanReview » Wed Oct 25, 2017 5:47 am
LUANDATO wrote:If p and q are positive integers, what is the remainder when 9^2p × 5^(p+q) + 11^q × 6^pq is divided by 10?

A)0
B)1
C)3
D)4
E)5

The OA is B.

Can any expert explain this PS question please? I don't have it clear. Thanks.
The remainder when a number is divided by 10 is its unit digit. For example, the remainder when 234 divided by 10 is 4.

We have 9^2p × 5^(p+q) + 11^q × 6^pq.

9^2p × 5^(p+q) + 11^q × 6^pq = 3^4p × 5^(p+q) + 11^q × 6^pq

Let's understand the deduction of the unit digit of exponents.

1. unit digit of the exponent of 3:
It follows a cycle of 4. Unit digit of 3^1 --> 3; 3^2 --> 9; 3^3 --> 7; 3^4 --> 1.

Thus, unit digit of 3^4p --> 3.

2. Unit digit of the exponent of 5:
Unit digit of the exponent of 5 is always 5.

3. Unit digit of the exponent of 6:
Unit digit of the exponent of 6 is always 6.

4. Unit digit of the exponent of 11:
Unit digit of the exponent of 11 is same as the unit digit of the exponent of 1, which is always 1.

Thus, unit digit of 3^4p × 5^(p+q) + 11^q × 6^pq --> 3 x 5 + 1 x 6 --> 5 + 6 --> 1.

The correct answer: B

Hope this helps!

-Jay

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BTGmoderatorLU wrote: ↑
Tue Oct 24, 2017 1:37 pm
If p and q are positive integers, what is the remainder when 9^2p × 5^(p+q) + 11^q × 6^pq is divided by 10?

A)0
B)1
C)3
D)4
E)5

The OA is B.

Can any expert explain this PS question please? I don't have it clear. Thanks.
When a number is divided by 10, we need to know or identify the units digit of the number since that will be the remainder when the number is divided by 10. Since the given expression has 4 terms, we need to know or identify the units digit of each term.

For 9^(2p), we see that 2p is even and 9 raised to an even power always has 1 as the units digit.

As for the other three terms, 5^(p+q) always has a units digit of 5 since 5 raised to any power has 5 as the units digit, 11^q always has a units digit of 1 since 11 raised to any power has 1 as the units digit, and 6^(pq) always has a units digit of 6 since 6 raised to any power has 6 as the units digit.

Therefore, the “units digit” of the expression is 1 - 5 + 1 - 6 = 2 - 11 = -9. Of course, a units digit can’t be negative; however, we can fix it by adding 10 to it to obtain -9 + 10 = 1 as the true units digit.

Answer: B

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