Prove here!

a , b , c a, b, c are positive integers such that:
5 a + 5 b + 2 a b = 92 5a+5b+2ab= 92 ;
5 b + 5 c + 2 b c = 136 5b+5c+2bc= 136 ;
5 c + 5 a + 2 a c = 244 5c+5a+2ac= 244 .
Then ( 7 a + 8 b + 9 c ) (7a+8b+9c) = ?


The answer is 172.

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1 solution

Sachin Sharma
Oct 29, 2015

The three equations may be written as

( 2 a + 5 ) ( 2 b + 5 ) = 209 = 11 19 (2a+5)(2b+5) = 209 = 11\cdot19

( 2 b + 5 ) ( 2 c + 5 ) = 297 = 11 27 (2b+5)(2c+5) = 297 = 11\cdot27

( 2 c + 5 ) ( 2 a + 5 ) = 513 = 19 27 (2c+5)(2a+5) = 513= 19\cdot27

which gives a = 7 , b = 3 , c = 11 a=7, b=3, c=11

and finally, 7 a + 8 b + 9 c = 172 7a+8b+9c = \boxed{172}

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