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Number Theory; Difficulty: Hard

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by [email protected] » Tue Jul 26, 2016 9:37 am
Hi Alphonsaj,

When dealing with Roman Numeral questions, it's often easiest to do the opposite of what the question asks (so that you can eliminate answers). Here, we're asked for what MUST be true... so let's try to prove what is NOT always true. We can TEST VALUES rather easily.

IF....
X = 1, the only factor is 1
(1)(13) = 13, which has two factors (1, 13)
So, X COULD be 1.
Roman Numeral 1 is NOT always true.
Eliminate Answers A and D

IF...
X = 2, there are two factors (1, 2)
(2)(13) = 26, which has four factors (1, 2, 13, 26)
So, X COULD be 2.
Roman Numeral 3 is NOT always true.
Eliminate Answers B and E

There's only one answer remaining...

Final Answer: C

GMAT assassins aren't born, they're made,
Rich
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by ceilidh.erickson » Fri Jul 29, 2016 5:51 am
I would recommend Rich's approach: test numbers and try to disprove the Roman Numeral statements.

However, there is a larger theory here:
Whenever you multiply a given number by a prime (such as 13), you always double the number of distinct factors... unless that original number already contained that particular prime as a factor.

If x = 6, x has factors of 1, 2, 3, and 6. If we multiply it by 13, then 13x will still contain all of those factors, but it will also contain 13 times each of those factors: (1)(13) = 13, (2)(13) = 26, (3)(13) = 39, and (6)(13) = 78. So we've doubled the number of factors.

The only exception is if x already contained a factor of 13. For example, if x = 26, it will have factors of 1, 2, 13, and 26. 13x will have factors of 1, 2, 13, 26, 169, and 338. (1)(13) and (2)(13) were already included in our original count, and we don't double-count factors.

Therefore, if multiplying x by a prime doubles the number of factors, x must not have contained that prime as a factor.
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by Matt@VeritasPrep » Thu Aug 04, 2016 9:07 pm
On test day, just try a few easy numbers.

Suppose x = 1. x has one factor. Multiplied by 13, we get 13, which has two factors. So "x is not a multiple of 13" is possible, and "x is a multiple of 13" must be false. Eliminate A and D.

Since we found an odd solution, let's try an even one. Suppose x = 2. x has two factors. 2*13 = 26, which has four factors, so we've again doubled the number of factors. We know III is false: eliminate B and E.

C is the only answer left, so we're good to go!
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