Expert: Lynn Houston - 8/1/2009

Question
A two digit number is three less than seven times the sum of its digits. If the digits are reversed the new number is 18 less than the original number.Find the original number.

Answer
Sorry for the delay, I'm having computer issues...


Let x = the first digit (10's) and y = the second digit (1's), therefore 10x+y = the number you're looking for.

So, if the number is 3 less then 7 times the sum of the digits, then 10x+y+3 = 7(x+y)

When the number is reversed (10y+x), it is 18 less then the original number, so
10y+x +18 = 10x+y.  Now, you can solve one of the equations for x or y and then substitute the value into the other equation to solve for one of the variables.  

10x+y+3 = 7(x+y)
10x+y+3 = 7x+7y
3x = 6y -3
x = 2y -1 (substitute this value of x into the second eq)

10y+x +18 = 10x+y
10y+(2y-1) +18 = 10(2y-1)+y
10y+2y-1+18 = 20y -10 +y
12y+17 = 21y-10
27 = 9y
y = 3   (substitute y into either eq)

10y+x +18 = 10x+y
30+x+18 = 10x +3
48+x = 10x+3
45 = 9x
x = 5, so you're original number is 10x+y or 53.  You can check this by doing what the word problems tells you to do

53 is 3 less then 7(5+3) and 35 is 18 less then 53.