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GMAT Prep - Factors

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GMAT Prep - Factors

by tonebeeze » Thu Jun 23, 2011 12:43 pm
What is the most efficient method to solve this problem? Please jot down any logical deductions or number theory applications that you utilize. thanks!
The number 75 can be written as the sum of squares of three different positive integers. What is the sum of these three integers?

a. 17
b. 16
c. 15
d. 14
e. 13

OA = e
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by Ian Stewart » Thu Jun 23, 2011 12:51 pm
I don't think there is any quicker way than testing values here, but fortunately 75 is small enough that we can arrive at an answer within two minutes. I'd start by looking at the squares closest to 75, since that will narrow down our possibilities most quickly. We know a^2 + b^2 + c^2 = 75. First we might 'guess' that a=8. In that case b^2 + c^2 must be 11, and we can quickly see that no integer combination of b and c is possible. Next we could 'guess' a = 7. In that case, b^2 + c^2 would be 26, and now we can see that 5^2 + 1^2 = 26, so our numbers can be 7, 5 and 1.
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by Ozlemg » Thu Jun 23, 2011 12:54 pm
my 2 cents:

what is being asked : x^2+y^2+z^2=75, what is x+y+z?

I wrote down the integer numbers :
1,2,3,4,5,6,7,8,9 (because 9 square is 81, it will be excluded!)

Then, I start to take the square of 8 which is 64, subtract it from 75 which is 11, then I try to figure out what other numbers' squares add up to 11? Only 9, 2 leaeves alone...So I start all the process starting from 7 and I got the answer :

7*7 = 49, 75- 49 : 26, 5*5 + 1*1 is 26

so answer is 7+5+1 : 13

hope this helps!
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by Ashley@VeritasPrep » Thu Jun 23, 2011 1:12 pm
I'd only add one thing, with the disclaimer that it barely saves time on this problem, if indeed it saves any at all... I'm just saying this because it's good practice to get in the habit of thinking about number properties as often as possible ;)

In this problem, we're summing three squares to get an odd number (75). This means there are only two possibilities: either all three of those addends are odd, or two are even and one is odd. Since all integers have the same parity (i.e. even-or-odd-ness) as their squares, we can also conclude that the breakdown among the three integers themselves is the same -- either all odd, or two even and one odd -- and in any case, those integers themselves will also have an odd sum. Of course, that only knocks out two of the answer choices given here, and as I said, I don't think it's really a time-saver on this problem -- just good practice thinking about it. But you can imagine a problem where it might be more helpful -- for instance, if GMAC had wanted to test your number properties facility and so had given some much larger odd number and provided answer choices such that only one was odd:

The number 461 can be written as the sum of squares of three different positive integers. What is the sum of these integers?

a. 30
b. 34
c. 37
d. 40
e. 44

In that case I wouldn't go looking for the integers. I'd just use the number properties trick.
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