What is the original number?

Question

The units digit of a two-digit number is two more than four times the tens digit. When the digits are reversed, the new number is thirteen more than three times the original number. What is the original number?

Answer 1

U = 4T+2

The original number is 10T + U

The number formed by reversing the digits is 10U + T

10U + T = 3(10T + U) + 13 <--- the new number formed by reversing the digits is 13 more than 3 times the

original number

10U + T = 30T + 3U + 13

7U = 29T + 13

7(4T+2) = 29T + 13

28T + 14 = 29T + 13

1=T

U = 2+4(1) = 2+4 = 6

The original number is 16

3(16)+13 = 48 + 13 = 61, so yes it checks.

The original number is 16

Note that you COULD have done the problem by trial and error rather than doing the algebra:

Suppose the tens digit is 1. THen the units digit must be 4(1)+2 = 4+2 = 6.

So the starting number is 16. Reversing the digits is 61, and 3(16)+ 13 = 48+13 = 61.

So the conditions in the problem are met, so the answer is 16.

Proceeding anyway... of the tens digit is 2. then the units digit must be 4(2)+2 = 10

which cannot be. The condition(s) fail for starting tens digit of 3 or more.

THe winner is 16 by default/process of elimination