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# An elegant way to solve this problem:

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CEO
Joined: 17 Nov 2007
Posts: 3574
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
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Kudos [?]: 2244 [1] , given: 359

Re: An elegant way to solve this problem: [#permalink]  05 Jul 2009, 11:05
1
KUDOS
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I would solve it with playing with numbers.

1) integers cannot by grater than 8, because 9^2 = 81 > 75
2) write out all squares : 1, 4, 9, 16, 25, 36, 49
3) Do you see the answer? No?
4) 75 is a big integer, so let's one of integers is 7. 75-49=26 - for 2 squares. it is possible for 1+25 (1 and 5)
5) we could begin with 6. 75 - 36 = 39. 25+16=41, 25+9=34, others options are less than 39.
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Manager
Status: Stanford GSB
Joined: 02 Jun 2008
Posts: 101
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Kudos [?]: 107 [0], given: 4

Re: An elegant way to solve this problem: [#permalink]  07 Jul 2009, 23:02
I kudo to Walker for his clarity of thinking.
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Joined: 07 Jul 2009
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Re: An elegant way to solve this problem: [#permalink]  08 Jul 2009, 00:34
you can also notice that at least one number has to be odd.
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Joined: 30 May 2009
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Kudos [?]: 1 [0], given: 0

Re: An elegant way to solve this problem: [#permalink]  08 Jul 2009, 01:58
The number 75 can be written as the sum of the squares of 3 different positive integers. What is the sum of these integers:

75 is odd , and 75 must be made from squares of 3 different positive integers , the integers must be

Observations : 1.0 integer^2 = int X int so if int is odd then int^2 = odd , even , int^2 = even
2.0 Since 75 is odd then three integers must be { e,e,o} or {o,o,o} hence their sum must be odd

so then answer choices B and D are out ( B, D the sum is even)

3.0 We know that the largest of the 3 must be less than 9 ( 9^2 = 81)
integers = { 1,2,3,4,5,6,7,8} integer squared (unit digit) = { 1,4,9,6,5,6,9,4}

there are three odd units( 1,5,9) and 2 even units (2,4)

Look at the unit digit of 75 , unit digit = 5 , so when we add the squares we must get 5 as the unit

From 2.0 check odd,odd,odd -----> 1+9+5 = 5; 5+5+5=5 , so it satisfies the unit place value , so the numbers could be 1,3,5 : 5,5,5 or 1,7,5 only 1,7,5 will satisfy the given condition (1,3,5 is too small and 5,5,5 is out because the integers must be different)

so sum = 1+5+7 =13
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Joined: 18 Jun 2009
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Location: Tbilisi, Georgia
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Kudos [?]: 21 [0], given: 5

Re: An elegant way to solve this problem: [#permalink]  08 Jul 2009, 09:00
walker wrote:
I would solve it with playing with numbers.

1) integers cannot by grater than 8, because 9^2 = 81 > 75
2) write out all squares : 1, 4, 9, 16, 25, 36, 49
3) Do you see the answer? No?
4) 75 is a big integer, so let's one of integers is 7. 75-49=26 - for 2 squares. it is possible for 1+25 (1 and 5)
5) we could begin with 6. 75 - 36 = 39. 25+16=41, 25+9=34, others options are less than 39.

did exactly the same way
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Re: An elegant way to solve this problem:   [#permalink] 08 Jul 2009, 09:00
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