SOLUTION: 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.

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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