Question 1049169

A two-digit number is such that the sum of its digit is 1/8 of the number.
When the digits of the number are reversed and the number is subtracted from the original number, the result obtained is 45. Find the original number please.
<pre>Let the tens and units digits of the original number be T, and U, respectively
Then number is: 10T + U, and when reversed, it becomes: 10U + T
Then: {{{T + U = (1/8)(10T + U)}}}
{{{T + U = (10T + U)/8}}}
10T + U = 8(T + U) ------ Cross-multiplying
10T + U = 8T + 8U
10T - 8T = 8U - U
2T = 7U ------- eq (i)

Also, 10T + U - (10U + T) = 45
10T + U - 10U - T = 45 =====> 9T - 9U = 45 ======> 9(T - U) = 9(5) ======> T - U = 5 ======> T = 5 + U ------ eq (ii)
2(5 + U)  = 7U ------ Substituting 5 + U for T in eq (i)
10 + 2U = 7U
10 = 7U - 2U
10 = 5U
U, or units digit of original number = {{{10/5}}} = 2

T = 5 + 2 ------- Substituting 2 for U in eq (ii)
T, or tens digit of original number = 7

Original number: {{{highlight_green(72)}}}