last week's MGRE Math Beast Challenge: correct!

Here's what MGRE has to say about last week's Math Beast Challenge problem:

Let’s start by translating this into algebra.

Abe has k ketchup packets, Beata has m mustard packets, Cruz has s soy sauce packets, and Dion has b barbecue sauce packets. We are trying to find the smallest possible value for k + m + s + b.

We know that 2k = 9m = 7s = 15b. In other words: 2k, 9m, 7s and 15b each equal the same integer number, but what is that number? It would have to be cleanly divisible by 2, 9, 7, and 15. In order to minimize the number of packets owned by the group as a whole, we would want to find the smallest such number. That would be least common multiple of 2, 9, 7, and 15.

First, we find the factors of each number:
2 = 2
9 = 3 * 3
7 = 7
15 = 3 * 5

Then we multiply only the necessary factors together:
2 * 3 * 3 * 7 * 5 = 630.

Note that we leave out a 3, compared to the list of all factors above. Remember that we only include the factors needed to build each of the starting numbers individually. With two 3’s and a 5, we could make either 9 or 15. That’s good enough for a least common multiple.

Now we know that 2k, 9m, 7s, and 15b each equals 630. Let’s find how many packets each person has.

2k = 630, so k = 315 9m = 630, so m = 70 7s = 630, so s = 90 15b = 630, so b = 42

To finish, just add together the individual number of packets (k + m + s + b) = 315 + 70 + 90 + 42 = 517.

It is worth analyzing how the other choices all represent a possible mistake, a quality that makes this question harder than it would be with different answer choices:

(A) is a trap. It’s just the number in the problem added together (2 + 9 + 7 + 15).

(B) CORRECT.

(C) is a trap for those who stop at 630 and don’t remember that it’s just a step on the way to finding the individual number of packets each person has.

(D) is the sum of the packets if starting with 1890 instead of 630 as the common multiple. 2k = 1890 (k = 945). 9m = 1890 (m = 210). 7s = 1890 (s = 270). 15b = 1890 (b = 126). Sum: 945 + 210 + 270 + 126 = 1551.

(E) is another possible multiple of 2, 9, 7 and 15. (2)(9)(7)(15) = 1890. But it’s not the least common multiple. It’s the trap for those who don’t omit a redundant 3, and who also forget to finish the solving. (The value of 2k = 9m = 7s = 15b is not the answer, k + m + s + b is the answer.)

The correct answer is B.

Woo-hoo! But I actually think my own method, with the compound ratio, is quicker.


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