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Given that S=7a+b+7c and also S=6a+8b+6c, where 51< S < 149 and S is an integer, find the sum a+b+c. 

Guest Nov 12, 2017
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I have assumed that a,b,c are integers.....don't know if this is correct....!!!!

 

S=7a+b+7c 

S=6a+8b+6c       

 

Subtract the second equation from the first and rearrange

 

a + c  = 7b   

 

So  using the first equation.........

 

S  =  7 (a + c)  + b  ⇒  S  = 49b + b =   50b

 

But since     51 < S < 149    and S is an integer....  then     51 < 50b < 149  ⇒  b = 2

 

So    S  = 100

 

So  manipulating the first equation, we have

 

100 = 7a + 2 + 7c

 

98  = 7 (a + c)     →  a + c  =  14

 

So.....the sum of a + b + c  =   (a + c) + b =  14 + 2  =  16

 

 

 

 

cool cool cool

CPhill  Nov 12, 2017
edited by CPhill  Nov 12, 2017

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