LCM New Year

Consider all possible natural numbers a , b , c , x , y , a, b, c, x, y, and z z such that
1) L C M ( a , b , c ) = L C M ( x , y , z ) = 2015 , LCM(a,b,c) = LCM(x,y,z) = 2015,
2) both a + b + c a+b+c and x + y + z x+y+z are perfect squares,
3) a , b , c a, b, c are distinct,
4) x , y , z x, y, z are distinct.

Find the maximum value of a b c x y z . a - b - c - x - y - z.

94 91 93 92

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2 solutions

Srikanth Kummara
Dec 6, 2014

Here to have maximum value a should be maximum possible factor...that is greatest divisor....it is 155.....from here on rearranging we get two sets... 155;13;1 and 31;13;5.....hence solved

Patrick Corn
Dec 3, 2014

I don't think this is correctly stated. The only triples that work are permutations of ( 155 , 13 , 1 ) , ( 31 , 13 , 5 ) , (155,13,1), (31,13,5), and ( 155 , 65 , 5 ) (155,65,5) . It's not possible to find two triples with six distinct natural numbers as the problem asks for. But if you take the first two with a = 155 a = 155 , you get 92 92 for an answer.

Thanks. I have removed the condition that they are distinct.

Calvin Lin Staff - 6 years, 6 months ago

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I don't think that works; I assumed that a , b , c a,b,c were distinct and x , y , z x,y,z were distinct. Otherwise you could use e.g. ( a , b , c ) = ( 2015 , 5 , 5 ) (a,b,c) = (2015,5,5) to do even better

Patrick Corn - 6 years, 6 months ago

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Thanks. I've added that in.

Calvin Lin Staff - 6 years, 6 months ago

Th anks for the correction sir. Im beginner so ive got so many mistake here

Rimba Erlangga - 6 years, 6 months ago

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