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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
