If P, Q, and R are distinct positive digits and the product of the two

Question

If P, Q, and R are distinct positive digits and the product of the two-digit integers PQ and PR is 221, what is the sum of the digits P, Q, and R?

A. 5
B. 11
C. 13
D. 21
E. 23

BrentGMATPrepNow · Answer

221 = (13)(17)

Since the two-digit integers PQ and PR both have the same tens digit (P), we can see that one possible solution is: 
P = 1
Q = 3
R = 7

So, P + Q + R = 1 + 3 + 7 = 11

Answer: B

julialynn240 · Answer

In this case, the units digit and the tens digit of a product is the units and tens digit of the product of the units digits being multiplied.

For the product of the two-digit integers PR and PQ to yield 221, R and Q must be 3 and 7. Since as stated in the previous comment, PR and PQ both have the same tens digit P, the only possible value for P is 1.

17*13 = 221

1 + 7 + 3 = 11

Answer B

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IanStewart · Answer

That does turn out to be true in this particular case, but in general, if you want to predict the tens digit of a product, you need to take account of both the tens and the units digits of the numbers you're multiplying together. For example, 121 = 11*11, and we can't guess that 11*11 will end in '21' by considering units digits alone.

But in this particular question, we don't even need to consider tens digits. Our units digits are the distinct digits Q and R, and we're getting a product ending in '1'. If Q and R are different, they can only be 3 and 7, since those are the only distinct one-digit numbers with a product ending in 1. Since P is clearly 1 here (if P were 2 or greater, both two-digit numbers would be greater than 20, and their product would be greater than 400), the answer must be 1+3+7 = 11.

Or you can notice that 221 = 225 - 4 = 15^2 - 2^2 = (15+2)(15-2) = 17*13, and arrive at the values of P, Q and R that way (in some order -- we can't tell which of Q or R is 3 and which is 7, but that's not important here).

Or I suppose once you notice P = 1, you can then tell that the sum must be somewhere between 1+3+5 and 1+7+9, since the other two digits must be odd and distinct, and that means only two answer choices B and C are even plausible, and then it's not too time-consuming to test the various possibilities. But that seems the slowest approach here.

MagnusSamIBA · Answer

We need to find distinct digits P, Q, and R such that the product of the two-digit integers PQ and PR equals 221.

First, notice that 221 = 13 × 17, which means PQ could be 13 and PR could be 17 (or vice versa). The common digit P in both numbers must be 1, while Q is 3 and R is 7.

Thus, the sum of the digits P, Q, and R is 1 + 3 + 7 = 11.

Answer: B (11)

If P, Q, and R are distinct positive digits and the product of the two

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If P, Q, and R are distinct positive digits and the product of the two

Prep Toolkit

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