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 Post subject: Arithmetic Diagnostic Test - Factors ProblemPosted: Wed Aug 04, 2010 2:57 pm

Joined: Thu Jul 03, 2008 2:13 pm
Posts: 117
Spoiler alert - if you haven't yet taken the Arithmetic Diagnostic, you may want to wait until you have since this question is on it! I had a student ask for an alternate explanation, so I figured I'd share while I was writing it.

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Question:

How many different positive integers are factors of 420?

(A) 6

(B) 12

(C) 16

(D) 21

(E) 24

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Solution:

As with any factor-based problems, it is a good idea to begin by breaking 420 into its prime factors: 2*2*3*5*7. From that set, in order to find the total number of factors of 420, we can multiply combinations of those prime factors – as well as 1, a factor of all positive integers – to determine the entire set of factors. A quick glance at the answer choices in this problem is helpful, as well; 420 will have quite a few factors, but because the numbers are spread out, we only need to find 22 factors in order to conclude that the answer is E, and similarly if we reach 19 or 20 and can’t determine which combination we may have omitted, we can be fairly certain that the number is 21 (definitely larger than 16 but not quite 24). The answer choices should put you at ease that you can afford to be slightly less than perfect in calculating the each factor and still be able to answer correctly.

Calculating these factors methodically is helpful in order to stay organized and not omit factors, so begin with the “obvious” factors, 1 and 420 (running total: 2). Then, take all of the unique prime factors: 2, 3, 5, 7 (running total: 6). Next, find the combinations of multiplying two prime factors together: 2*2, 2*3, 2*5, 2*7; 3*5, 3*7; 5*7 (running total: 13). Then, do the same for three prime factors: 2*2*3, 2*2*5, 2*2*7; 2*3*5, 2*3*7; 2*5*7; 3*5*7 (running total: 20). Note that, because choice D is 21, we only need to find two more combinations to pick E. If we try combinations of four primes, we’ll get there: 2*2*3*5, 2*2*5*7, 2*3*5*7, 2*2*5*7, 2*2*3*7 (running total: 24).

NOTE: You could use combinations rules instead of listing out the possibilities, but as this question comes in the Arithmetic diagnostic - before we cover combinatorics - I didn't want to confuse in the solution. And, honestly, you'd have to perform so many combinations problems that it's probably not worth it.

There is a shortcut that works for this problem, as well, playing on rules for combinatorics: Take the prime factors of the number and express them as exponents (22, 31, 51, 71). Then, add one to each exponent and multiply the exponents together: 3*2*2*2. That will tell you the number of unique factors of the original number (24).

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