SOLUTION: If a, b, and c are positive integers, find the sum a+b+c if {{{ (a^3)*b=1375 }}} , {{{ (b^3)*c=3993 }}} , and {{{ ac^3=135 }}} =

Algebra ->  Exponents -> SOLUTION: If a, b, and c are positive integers, find the sum a+b+c if {{{ (a^3)*b=1375 }}} , {{{ (b^3)*c=3993 }}} , and {{{ ac^3=135 }}} =       Log On


   



Question 1104241: If a, b, and c are positive integers, find the sum a+b+c if +%28a%5E3%29%2Ab=1375+ , +%28b%5E3%29%2Ac=3993+ , and +ac%5E3=135+ =
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
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If a, b, and c are positive integers, find the sum a+b+c if +%28a%5E3%29%2Ab=1375+ , +%28b%5E3%29%2Ac=3993+ , and +ac%5E3=135+
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a%5E3%2Ab = 1375,    (1)
b%5E3%2Ac = 3993,    (2)
a%2Ac%5E3 = 135      (3)    


implies   (after multipluing all three equations, both left and right sides)


a%5E4%2Ab%5E4%2Ac%5E4 = 741200625  ====>

%28a%2Ab%2Ac%29%5E4 = 741200625  ====> abc = root%284%2C741200625%29 = 165.


Again:  abc = 165 = 3*5*11.    (4)

3, 5 and 11 are prime integers.

Since b%5E3%2Ac = 3993  is not divided by 5,  neither "b" nor "c" are multiple of 5.


Hence, from (4), "a" is multiple of 5.


Having it, it is not difficult to conclude further that  b = 11  and  c = 3.


a + b + c = 5 + 11 + 3 = 19.


Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Probably the easiest place to start is with ac%5E3=135. The prime factorization of 135 is 135+=+3%2A3%2A3%2A5; that means a=5 an c=3.

Then a%5E3b+=+5%5E3%2Ab+=+125b+=+1375 so b is 11.

Then a+b+c = 5+11+3 = 19.