Question 205487:  Find three consecutive odd numbers such that 1/7 of the first plus 1/3 of the second plus 1/5  of the third is 63. 
 Answer by checkley77(12844)      (Show Source): 
You can  put this solution on YOUR website! Let x, x+2, x+4 be the 3 odd numbers. 
x/7+(x+2)/3+(x+4)/5=63 
[(15x+35(x+2)+21(x+4)]/105=63 
[15x+35x+70+21x+84]/105=63 
[71x+154)/105=63 
71x+154=63*105 
71x+154=6,615 
71x=6,615-154 
71x=6,461 
x=6,461/71 
x=91 the first number 
91+2=93 the second number 
91+4=95 the third number 
Proof: 
91/7+93/3+95/5=63 
13+31+19=63 
63=63 
 
 
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