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Value of x?

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Value of x?

by PauloAH » Sun Oct 05, 2014 7:57 am
What is the value of positive integer x?

1) The sum of all unique factors of x is 31
2) x = y2, where y is an integer





I understand that one might do it by guessing numbers, but is there a less time consuming approach?
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by [email protected] » Sun Oct 05, 2014 12:01 pm
Hi PauloAH,

In this DS prompt, we're told that X is a positive integer. We're asked for the value of X. This question is perfect for TESTing VALUES.

You ask a good question - is there a "fast" way to find some values to TEST. There IS, but you have to think about how the question is designed. Notice in Fact 2 that X = Y^2 and Y is an integer. This means that X is a PERFECT SQUARE. That's an interesting "limitation" and it might help you to deal with Fact 1 a bit faster.

Fact 1: The sum of the unique factors of X = 31.

Can you find any perfect squares that have unique factors that sum to 31?

16 = 1 & 16, 2 & 8, 4 ---> These factors sum to 31, so X COULD = 16

25 = 1 & 25, 5 --> These factors sum to 31, so X COULD = 25
Fact 1 is iNSUFFICIENT

Fact 2: X = Y^2 and Y is an integer.

The TESTs that we used in Fact 1 ALSO fit Fact 2....

Y = 4, X = 16
Y = 5, X = 25
Fact 2 is INSUFFICIENT.

Combined, we have 2 different answers that fit both facts: 16 and 25.
Combined, INSUFFICIENT.

Final Answer: E

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by Matt@VeritasPrep » Sun Oct 12, 2014 10:49 pm
Here's a nice trick that would work on this question, or any question involving the sum of the factors of a number; I wanted to share it because I think it's (i) cool and (ii) new to most students.

If we know the prime factorization of an integer, we can find the sum of the factors as follows:

1:: Find the highest power of each prime factor
2:: Find the sum of ALL the powers of that prime factor, up to the highest power present
3:: Multiply these sums together

For instance, if I want the sum of the factors of 360, I follow those steps like so::

1:: 360 = 2³ * 3² * 5
2:: The sums of the powers are (2� + 2¹ + 2² + 2³) and (3� + 3¹ + 3²) and (5� + 5¹).
3:: Multiplying those sums gives (2� + 2¹ + 2² + 2³) * (3� + 3¹ + 3²) * (5� + 5¹) = 15 * 13 * 6 = 1170.

Too cool!

Now let's look at our problem.

S1 tells us the sum is 31. 31 is prime, so it can only be represented as the product 1 * 31. This means that we can only have ONE prime factor, as we can't be multiplying two sums greater than 1.

Now let's see what sums of the first few powers of certain primes give us 31.

(2� + 2¹ + 2² + 2³ + 2�) = 31, so this is a possibility
(3� + 3¹ + 3² + 3³) = 40, so no power of 3 will work
(5� + 5¹ + 5²) = 31, so this is a possibility too

Hence we could have 2� or 5²; INSUFFICIENT.

S2 obviously doesn't do much on its own.

Taking the two together, we still have 16 and 25 as possibilities; INSUFFICIENT.
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