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PQ and QP represent two-digit numbers

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PQ and QP represent two-digit numbers

by stevecultt » Fri Jun 09, 2017 4:17 am
PQ and QP represent two-digit numbers having P and Q as their digits. RSR is a three-digit
number having the digits R and S. What is the value of P + Q + R + S?

(1) PQ + QP = RSR.
(2) P, Q, R, and S are distinct non-zero digits.

OA A

Need help how the answer is A.
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by GMATGuruNY » Fri Jun 09, 2017 4:35 am
stevecultt wrote:Need help how the answer is A.
To satisfy Statement 1, P and Q must sum to 11, with the result that RSR = 121:
29+92 = 121
38+83 = 121
47+74 = 121
56+65 = 121.
In every case, P+Q = 11 and R+S = 3.
Thus, P+Q+R+S = 11+3 = 14.
SUFFICIENT.
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by Jay@ManhattanReview » Fri Jun 09, 2017 4:46 am
stevecultt wrote:PQ and QP represent two-digit numbers having P and Q as their digits. RSR is a three-digit
number having the digits R and S. What is the value of P + Q + R + S?

(1) PQ + QP = RSR.
(2) P, Q, R, and S are distinct non-zero digits.

OA A

Need help how the answer is A.
Statement 1:

Since the two-digit numbers, PQ and QP add to form a three-digit number RSR, the value
of the digit in the hundreds place must be '1.'

=ƒ> R = 1

Thus, P and Q should be such digits which add up to give 1 as the unit digit (so that the unit digit of R = 1 is obtained in the sum).

Thus, possible values of P and Q are:
  • P = 2, Q = 9 ƒ=> PQ + QP = 29 + 92 = 121 ƒ=> P + Q + R + S = 2 + 9 + 1 + 2 = 14
    P = 3, Q = 8 ƒ=> PQ + QP = 38 + 83 = 121 ƒ=> P + Q + R + S = 3 + 8 + 1 + 2 = 14
    P = 4, Q = 7 ƒ=> PQ + QP = 47 + 74 = 121 ƒ=> P + Q + R + S = 4 + 7 + 1 + 2 = 14
    P = 5, Q = 6 ƒ=> PQ + QP = 56 + 65 = 121 ƒ=> P + Q + R + S = 5 + 6 + 1 + 2 = 14
Thus, the unique answer is 14. - Sufficient

Statement 2:

We know that P, Q, R, and S are distinct non-zero digits.

However, their values cannot be determined. - Insufficient

The correct answer: A

Hope this helps!

Relevant book: Manhattan Review GMAT Data Sufficiency Guide

-Jay
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by [email protected] » Fri Jun 09, 2017 9:51 am
Hi stevecultt,

While this question might look complex, it's actually based on some basic arithmetic rules. If you don't recognize the "theory" behind this question, then you could still solve it with a bit of 'brute force' arithmetic and a little logic.

We're told that QP and PQ are two 2-digit numbers that have the same digits (just in reverse-order) and that RSR is a 3-digit number. We're asked for the value of P+Q+R+S.

1) PQ + QP = RSR.

With Fact 1, notice that the sum of the two 2-digit numbers is a 3-digit number. In this situation, the 3-digit number MUST begin with a 1 (there's no other possibility since the highest sum of two 2-digit numbers is 99+99 = 198). Thus, R = 1....

PQ + QP = 1S1

By extension, Q+P must "end" in a 1 AND PQ+QP must be large enough to create a 3-digit sum. From here, you can brute-force the options and see what happens....
P=2, Q=9... 29+92 = 121.... so S=2 and the answer to the question is 2+9+1+2 = 14
P=3, Q=8... 38+83 = 121..... so S=2 and the answer to the question is 3+8+1+2 = 14

Interesting that the resulting sum stayed exactly the SAME. There are only a few options left, but if you map them out, you'll end up with the exact same answer every time... the answer to the question is ALWAYS 14.
Fact 1 is SUFFICIENT

2) P, Q, R, and S are distinct non-zero digits.

With this Fact, we know that the 4 digits are all DIFFERENT non-zero integers, but there are multiple possible answers to the given question.
Fact 2 is INSUFFICIENT

Final Answer: A

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by Matt@VeritasPrep » Thu Jun 22, 2017 10:59 pm
PQ = 10P + Q
QP = 10Q + P

so

A slightly more thorough breakdown of S1::

PQ + QP = 10P + Q + 10Q + P => 11P + 11Q => 11 * (P + Q)

RSR = 100R + 10S + R = 101R + 10S

Since 11*(P+Q) is at most 11*(9+9), we know that it's less than 200.

From that 101R + 10S must also be less than 200, so R = 1. (By convention in these cryptarithms, the leading digit can't be 0.) That gives us

11*(P + Q) = 101 + 10S

or

11*(P+Q) = 1S1

The only threedigit multiple of 11 that begins and ends in 1 is 121, so S1 is sufficient.
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