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integers

by Akansha » Thu May 26, 2011 9:26 am
75 can be written as the sum of the squares of 3 different positive integers.
What is sum of these integers?
a. 17
b. 16
c. 15
d. 14
e. 13

OA is E
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by Frankenstein » Thu May 26, 2011 9:37 am
Hi,
I went by trial and error and got 75 = 1^2+5^2+7^2.
So, sum of integers is 1+5+7=13.
Hence, answer E

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by amar66 » Thu May 26, 2011 10:42 am
Another Approach:

We can write 75=x^2+ y^2 + z^2
then 75= 25+50
25 is the perfect square of 5 so you have x.
50= 49 +1 ; 49 is the perfect square of 7 so you have y and 1 is square of 1 so z.
x+y+z=13
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by manpsingh87 » Thu May 26, 2011 10:56 am
Akansha wrote:75 can be written as the sum of the squares of 3 different positive integers.
What is sum of these integers?
a. 17
b. 16
c. 15
d. 14
e. 13

OA is E
lets analyze this question, lets concentrate on the last digit of 5..!! we can have last digit of sum of squares of three integers
case 1)if all the three integers have 5 as its last digit or
case 2) if the square of two of the numbers add up to give zero as last digit and the remaining number's square will give 5 as a last digit, and the whole sum will have 5 as its last digit (x0+x5=x5);

case 1) now case 1 is possible if all the integers are equal to 5; which is not possible as integers are distinct hence case 1 here is not applicable;

case 2) now here last digit of 1 of the number for sure will be 5; hence 1 of the number will be 5;
and also 25+x^2+y^2=75; x^2+y^2=50;
now x^2+y^2=50; it is possible only if x (or y)=7 and y(or x)=1;

hence possible sum is 7+1+5=13; E
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