Expert: Samuel Clement - 10/23/2006

Question
hi there
My question is-The tens digit of a two digit number is 3 more than the units digit.the number is 8 more than six times the sum of the digits.find the number???
Can anyone help me out and explain this one to me as im lost-Ive been trying to form an equation with the formula (10t+u)? thanks

Answer
Let d1 be the ones digit and d2 be the tens digit.

Then you are told that d2 = d1 + 3.

This tells us that the number can be expressed as

         10(d2) + d1 = 10(d1 + 3) + d1
                     = 10(d1) + 30 + d1
                     = 11(d1) + 30.

We are also told that the number is 8 more than 6 times the sum of the digits.

       So 11(d1) + 30 = 6(d2 + d1) + 8. Right?

But this is the same as

          11(d1) + 30 = 6(d1 + 3 + d1) + 8
                      = 6(2(d1) + 3) + 8
                      = 12(d1) + 18 + 8
                      = 12(d1) + 26
                      

Now subtract 11(d1) and 26 from both sides of the equation to get

                  4 = d1 or rather d1 = 4.

This says that the one's digit is 4. If this is the case then the tens digit is 4+3 = 7. This would make the number 74.

Lets check our results in the original word problem.

Well, 7 does equal 4 + 3 so that part is good.

But is 74 equal to 8 more than six times the sum of the digits?

   6(4 + 7) + 8 = 6(11) + 8 = 66 + 8 = 74 which is what we want.

So the number we are after is 74.

Does this help?

Sam