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Arithmetic Problem

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Arithmetic Problem

by tamaracao » Tue Mar 11, 2014 2:29 am
The number 75 can be written as the sum of the squares of 3 diff positive integers. What is the sum of these 3 integers?
a) 17
b) 16
c) 15
d) 14
e) 13

This question was on the GMATPrep exam. I was wondering if it was a typo and should be 65 instead of 75 because I can't get any numbers to work for 75.
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by theCodeToGMAT » Tue Mar 11, 2014 2:51 am
75 = 7^2 + 5^2 + 1^2

So, 7+5+1 = 13
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by GMATGuruNY » Tue Mar 11, 2014 5:44 am
tamaracao wrote:The number 75 can be written as the sum of the squares of 3 diff positive integers. What is the sum of these 3 integers?
a) 17
b) 16
c) 15
d) 14
e) 13

This question was on the GMATPrep exam. I was wondering if it was a typo and should be 65 instead of 75 because I can't get any numbers to work for 75.
75 = the sum of three PERFECT SQUARES.
List the perfect squares less than 75.
Ask yourself the following:
Which perfect square will the average test-taker forget to consider?
Answer:
1² = 1.
Don't be an average test-taker.
Be sure to include 1² = 1 in your list:
1² = 1.
2² = 4.
3² = 9.
4² = 16.
5² = 25.
6² = 36.
7² = 49.
8² = 64.

The sum of the 3 values in red is 75:
1²+ 5² + 7² = 1 + 25 + 49 = 75.

Thus, the sum of the 3 integers = 1+5+7 = 13.

The correct answer is E.
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by [email protected] » Tue Mar 11, 2014 10:26 pm
Hi tamaracao,

A few questions on the GMAT Quant section are going to come down to 'limited options' - there usually not a fancy way to solve these types of questions, there's just "brute force" - pound on this question until you find the answer.

Here, we're told that the sum of the squares of 3 positive integers = 75, so the options are severely limited....

Since 9^2 = 81, we know that all 3 of the integers must be between 1 and 8.

From there, it's just a matter of "working down"....

If one of the numbers was 8^2, then you'd have 64 and the other two squares would have to add up to 11. You won't find this in the possibilities. As Mitch pointed out, it helps to write them down.

Next, try 7^2 = 49, the other two squares have to add up to 26. THAT'S pretty easy...5^2 + 1^2.

Now you've got the 3 integers and can sum them up.

Final Answer: E

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by Abhishek009 » Wed Mar 12, 2014 10:32 am
tamaracao wrote:The number 75 can be written as the sum of the squares of 3 diff positive integers. What is the sum of these 3 integers?

a^2 + b^2 + c^2 = 75

Seems difficult actually its not at all difficult.

See the numbers ( a , b and c ) are lesser than 10 , since 10^2 = 100

Now progress back wards -

10^2 = 100

9^2 = 81

8^2 = 64

7^2 = 49

6^2 = 36

5^2 = 25

4^2 = 16

3^2 = 9

2^2 = 4

1^1 = 1


The last digit of the number to be formed is 5 keep in mind so we have to select three numbers where the sum constitutes a number whose last digit is 5.

7^2 + 5^2 + 1^1 is only possible...
Abhishek
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