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| Question 205486:  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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