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

To find: a + b + c

If we try Hit and try method by considering that 9^2 > 75. that means we need to find combination lower than 9

8^2 doesn't satisfy

7^2 = 49 ==> 75 - 49 ==> 26 = 25 + 1 = (5)^2 + (1)^2

so, a + b + c = 7 + 5 + 1 = 13

[spoiler]{E}[/spoiler]

I like Brent's way better than the one I'm about to give, but here's another approach.

We know that 25 + 25 + 25 = 75. 25 = 5², so we can guess that ONE of our unique integers is 5. Now we just need to find two that sum to 50. 7² is close, and hey, 1² + 7² is the difference!

Feels a little goofy, but this is how you solve these sort of problems ...

kaudes11114 wrote: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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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