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Manhattan Question Set # 12

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Manhattan Question Set # 12

by richachampion » Wed Oct 12, 2016 7:39 am
D is the set of all the multiples of 3 between 20 and 100. E is the set of all the factors of 400. Set D and Set E have how many numbers in common?

(A)0
(B)1
(C)3
(D)5
(E)12

OA: A
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by [email protected] » Wed Oct 12, 2016 9:46 am
Hi richachampion,

This question has an interesting Number Property 'shortcut' built into it. If you know the 'rule of 3', then you know that the number 400 is NOT evenly divisible by 3 (since 4+0+0 = 4, which is not divisible by 3). Since 400 is not divisible by 3, NONE of its factors will be a multiple of 3. Thus, none of the numbers in Set D (which are all multiples of 3) are also in Set E.

Final Answer: A

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by richachampion » Wed Oct 12, 2016 10:05 am
[email protected] wrote:Hi richachampion,

This question has an interesting Number Property 'shortcut' built into it. If you know the 'rule of 3', then you know that the number 400 is NOT evenly divisible by 3 (since 4+0+0 = 4, which is not divisible by 3). Since 400 is not divisible by 3, NONE of its factors will be a multiple of 3. Thus, none of the numbers in Set D (which are all multiples of 3) are also in Set E.

Final Answer: A

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by Matt@VeritasPrep » Fri Oct 14, 2016 1:27 am
We might want to get at a more general solution here.

Start by factoring 400 into primes:

400 = 2 * 2 * 2 * 2 * 5 * 5

Now consider the set of multiples of 3 between 20 and 100, which would be:

3*7, 3*8, ..., 3*33

Given any sets of numbers, you'd want to look for numbers in the second set whose prime factorization is ENTIRELY contained in the prime factorization of the first number (here, 400).

Since 400 doesn't contain a 3, we know right away that none of 3*7, 3*8, ..., 3*33 will appear in the prime factorization of 400. But if the question were a bit more complex, we'd need an approach like this to solve it reasonably. (And with this one, it certainly makes the inspiration for the solution clearer!)
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