The number 75 can be written as the sum of the squares of 3 different

We're looking for 3 DIFFERENT squares that add to 75

Here are the only squares we need to consider: 1, 4, 9, 16, 25, 36, 49, 64
Can you find 3 that add to 75?
After some fiddling, we may notice that 1 + 25 + 49
In other words, 1² + 5² + 7² = 75
We want the SUM of 1 + 5 + 7, which is 13

Answer: E

Cheers,
Brent

I think this question has been discussed earlier also.
squares of natural numbers, which are below 75,are 1,4,9,16,25,36,49,64
1+25+49=75 is the only option
so numbers are 1,5,7
sum =13

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

Thanks for a very clear explanation Bunuel.

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.

thanks for the solution... was hoping for another solution other than just brute calculation. bunel you rock tho.

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.

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

Hi All,

This question can be solved with a bit of 'brute force' arithmetic, but the speed with which you solve it will likely depend on how quickly you can write everything down and how you do the work.

We're asked to find the three positive integers, whose squares add up to 75. To start, we should write down the list of possible squares:

1
4
9
16
25
36
49
64

We're limited to these 8 numbers. To work efficiently, we should work from 'greatest to least'..

If we use 64, then the other two numbers have to add up to 11....but there's NO WAY to make that happen (so 64 is NOT one of the numbers we need).

Next, let's try 49... then the other two numbers have to add up to 26....25 and 1 'fit', so the three numbers are 49, 25 and 1 (7, 5 and 1 --> that sum = 13).

Final Answer:
GMAT assassins aren't born, they're made,
Rich

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

I got this question on a practice test I took via GMAC.

Oddly enough, if you prime factorize 75 (15->5 and 3 and 5), you will get 5+5+3= 13. Is that a coincidence or is there actually something behind it?

I know that the question is not asking for this, but I found it a bit strange. Any insight on this would be helpful! Thank you

Hi MTLGMAT514,

Since you can't actually get to MOST integers by adding up three squares of integers, this result is just a coincidence.

GMAT assassins aren't born, they're made,
Rich

newish here - how come this is labeled sub 600 if it's tough? I found this on a official practice gmat

Hi maxraxstax,

As far as the 'knowledge' and the 'skills' needed to answer this question are concerned, this is not a particularly difficult question (at its core, you're just adding some 1-digit and 2-digit numbers together - and that's low level Arithmetic). As such, the sub-600 'label' is appropriate.

GMAT assassins aren't born, they're made,
Rich

­Thanks Bunel for the approach

Another Method:

75 = 3*25 = 3 * 5^2
If all numbers could have been same, then sum would have been 5*3 = 15. Given this, the sum cannot be more than 15, since for any number more than 5, the other number must be less than 5.
This eliminates A,B,C.

Given the square of values add to odd number, all the values must add to odd. So only 13 works