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dferm: Master | Next Rank: 500 Posts; Posts: 446; Joined: Thu Jul 26, 2007 1:07 pm; Thanked: 6 times

by dferm » Tue Mar 18, 2008 5:44 pm

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

Please Explain....

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Source: Beat The GMAT — Problem Solving | Tue Mar 18, 2008 5:44 pm

siddarthd2919: Junior | Next Rank: 30 Posts; Posts: 16; Joined: Mon Feb 11, 2008 3:08 amFollowed by: 1 member; Location: chennai; Thanked: 6 times

by siddarthd2919 » Tue Mar 18, 2008 7:11 pm

the number 75 is the sum of squares of three numbers...... i solved it in trail and eror method....... 7+5+1=13 (i.e 49+25+1=75)

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Tue Mar 18, 2008 8:06 pm

siddarthd2919 wrote:the number 75 is the sum of squares of three numbers...... i solved it in trail and eror method....... 7+5+1=13 (i.e 49+25+1=75)

Trial and error is almost certainly the quickest way to answer this question.

Write out perfect squares under 75:

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

Find 3 different numbers on the list that sum to 75, add up their roots.

"Different" is a key word, otherwise 25 + 25 + 25 would also have been a valid solution.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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musicdaemon: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Wed Feb 13, 2008 10:34 am; Location: India; Thanked: 3 times

by musicdaemon » Tue Mar 18, 2008 8:29 pm

Given
x^2+y^2+z^2=75………………………… (1)
Suppose,
x+y+z = k…………………………………(2)
Now, we know from the standard formula

x^2+y^2+z^2 +2(xy+yz+zx) = (x+y+z)^2

Putting the values from (1) & (2)

=>75+2(xy+yz+zx) = k^2
=> xy+yz+zx = (k^2 -75)/2 ………..(3)

Since, x,y and z are all integers, therefore, (k^2 -75)/2 must be an integer,

So, (k^2 -75) must be an even integer
For (k^2 -75) to be even, k^2 must be odd, ( since, odd – odd = even)
For k^2 to be odd, k must be odd

Thus only k = 13, 15 and 17 can be the sum of integers,

Putting the value in (3), xy+yz+zx = 47,75,107

But from the inequality, xy+yz+zx < x^2+y^2+z^2

Only value of xy+yz+zx satisfying this is 47

Thus, k= x+y+z = 13…………………………………………………….Ans

Hope it is clear to you.

Let the Game begin

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Chick: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Thu Aug 14, 2008 4:17 am

by Chick » Tue Sep 29, 2009 9:22 pm

I really don't see how this question can be answered within 2 mins even by using trial and error. Can someone explain how it can be solved this way within a reasonable amount of time?

Thanks!

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jolivera: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Thu Mar 11, 2010 9:39 am; Thanked: 2 times

by jolivera » Thu Mar 11, 2010 9:42 am

Easier if you just say 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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Bek: Senior | Next Rank: 100 Posts; Posts: 68; Joined: Fri Dec 03, 2010 10:29 pmFollowed by: 1 member; Thanked: 5 times

by Bek » Fri Jun 10, 2011 7:26 pm

I don't know if it's correct, anyway:

75=5x5x3
5+5+3=13

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cans: Legendary Member; Posts: 1309; Joined: Mon Apr 04, 2011 5:34 amFollowed by: 123 members; Location: India; Thanked: 310 times; GMAT Score:750

by cans » Fri Jun 10, 2011 7:43 pm

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

75=a^2 + b^2 + c^2
75=25 + 49 + 1 = 5^2 + 7^2 + 1^2
a+B+c = 5+7+1 = 13

If my post helped you- let me know by pushing the thanks button

Contact me about long distance tutoring!
[email protected]

Cans!!

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shingik: Junior | Next Rank: 30 Posts; Posts: 28; Joined: Thu May 19, 2011 8:47 am

by shingik » Thu Jul 21, 2011 10:34 pm

Here is how I worked it. I broke down 75 into its prime factors ie
75 = 5 x 5 x 3
I saw that the prime factors made up one perfect square I could see immediately ie
5 x 5 = 25 ------ So I figured 25 was one of the perfect squares
I subtracted 25 from 75 and was left with 50
This meant x-squared + y-squared MUST equal 50
ALSO I could see that both x-squared and y-squared MUST be odd (odd/even properties)
This enabled me to elminate for consideration any even perfect squares under 50
SO I was left with ONE, NINE and FORTY-NINE as the only possibilities.
The rest is a peice of cake

*IF the answer is not obvious by this time then you could do this
(1 + 9 + 49)-50 = 9 (this means the odd man out is nine and 1 and 49 remain as your perfect squares!) Consider taking the GRE if you have to do this last step !!!!!!(ha ha that was mean)

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