Question

when the digits of a two digit number are reversed the new number is 54 less than the original number. if the sum of its digits is 8, what is the original number

Answer 1

Let the first and second digits of this two-digit number be f and d.

We are told that f+d = 8.

The original number is 10f+d.

The value of 10d+f is 54 less than 10f+d: 10d+f = 10f+d-54.

Let's eliminate f: If f+d=8, then f=8-d. Subst. 8-d for f in the equation
10d+f = 10f+d-54:

10d + 8 - d = 10(8-d) - 54
= 80 - 10d - 54

Grouping all the terms in d on the left side results in 19d.

Grouping all the constant terms together on the right side results in
80-54-8, or 18.

Thus, we have 19d = 18

Unfortunately, this cannot be correct, since both f and d must be positive integers.

I've based my argument on the fact that a number such as 27 equals 2 times 10 plus 1. Thus, if the two unknown digits form the 2-digit number fd, the actual number is 10f+d.

Answer 2

17

71 - 17 = 54

1 + 7 = 8

17 is your answer

hope this helps