The number 75 can be written as the sum of the squares of 3 different positive integers. What is the sum of these 3 integers?

A. 17
B. 16
C. 15
D. 14
E. 13
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75 can be expressed as addition of squares of same integer as follows

\(75 = 5^2 + 5^2 + 5^2\)

Leaving one square term as it is; try to search for other options

\(75 = 5^2 + 50\)

Closest is 1 & 49

\(75 = 5^2 + 1 + 49\)

\(75 = 5^2 + 1^2 + 7^2\)

Answer = 5+1+7 = 13 = E

Hi, there. I'm happy to give my 2 cents on this.

First of all, this is a relatively difficult question for GMAT math. You would only see a question of this difficulty if you were already answering almost all of the math questions correctly.

I can't really suggest much other than trial-and-error. Since 25 is a factor of 75, I figured it might make sense if 25 were one of the three squares. That means the other two squares would have to add up to 50. Well, conveniently, 7^2 = 49, so 7^2 + 1^2 = 50, and 7^2 + 1^2 + 5^2 = 75. The three numbers are 1, 5, and 7, and those have a sum of 13. Answer = E.

Here, I was really just following my intuition for numbers, which is really just a notch above pure guess-and-check. I don't know the source of this question, but the real GMAT tends to give questions that admit of either a methodical approach or an elegant solution, whereas this more or less requires some guess & check, some poking around in the dark. That's just not the style of questions that the GMAC dishes out.

Please let me know if you have any questions on what I've said here.

Mike
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Very elegant solution. I got this question in my GMAT Prep 1. I also figured that it would require brute force. But Bunuel's point about 75 being an odd number and therefore being a sum of either odd-odd-odd or odd-even-even adds an elegant touch to the brute force approach.

is there any way to solve this one faster? I find myself just trying to add the different numbers and spending 3 mins on this questions, just due to the calculations. any tricks on this one? Thanks.

Yeah I misread this question and thought it was asking for factors so I had 5 5 3 which also add up to 13

Oh God.
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Bunuel's bruteforce using odd and even looks promising! I'd also like to add that you could eliminate some answer options using the same approach.
Since 75 is an odd number. The squares must be of the form -
\(Odd^2 + Odd^2 + Odd^2\) OR
\(Even^2 + Even^2 + Odd^2\)

Alternatively -
\(Odd + Odd + Even\) OR \(Even + Even + Odd\). Thus, the sum of these numbers must be ODD. Eliminate answer option B and C.
Breaking 13, 15, and 17 using further bruteforce. We get \(7^2 + 1^2 + 5^2\). 7 + 1 + 5 = 13. Answer Option A

Bunuel - Please correct me I am wrong, but if this is a general question that appears on the exam, we can conclude -
Sum of squares equal to a number?
* If the number is odd --> Correct answer option will be Odd.
* If the number is even --> Correct answer option will be Even
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1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
7^2 = 49
8^2 = 64

9^2 = 81
10^2 = 100



Start from 8^2 = 64

75 - 64 = 11 ; we need 2 other squares to add upto 11 - No possibility

Then 7^2 = 49

75 - 49 = 26 ; we need 2 other squares to add upto 26 - Possibile squares are 25 , 1

So, the required numbers are 7 , 5 and 1

Hence the numbers are 1,5,7

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If the sum of 3 different perfect squares is 75, each must be less than 75. So, we want to start by writing out the perfect squares that are less than 75:

1, 4, 9, 16, 25, 36, 49, 64

In this problem, there is no shortcut to determining which 3 squares add up to 75; however, it is strategic to start with the largest one, 64, and move down the list if it doesn’t work:

64 + 11 = 75

There is no way 11 can be written as a sum of two squares. So, we move down to 49:

49 + 26 = 75

We see that 26 can be written as the sum of 25 and 1; that is:

49 + 25 + 1 = 75

We have found the three perfect squares that sum to 75. In other words:

7^2 + 5^2 + 1^2 = 75

The sum of 7, 5, and 1 is 7 + 5 + 1 = 13.

Answer: E
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Similar question to practice: https://gmatclub.com/forum/the-number-9 ... 88946.html
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