Question 946177
let a = the 10's digit
let b = the units
then
10a + b = the number
and
10b + a = the number with digits reversed
:
A two digit number is 54 more than the number formed by interchanging the digits
10a + b = 10b + a + 54
10a - a = 10b - b + 54
9a = 9b + 54
simplify, divide by 9
a = b + 6
:
 the difference between the product and the sum of the digits is 15.
ab - (a+b) = 15
replace a with (b+6)
b(b+6) - ((b+6)+b) = 15
b^2 + 6b - (2b + 6) = 15
combine like terms
b^2 + 6b - 2b - 6 - 15 = 0
b^2 + 4b - 21 = 0
Factors to
(b + 7)(b - 3) = 0
The positive solution is what we want here
b = 3
Find a
a = 3 + 6
a = 9
:
93 is the number.
:
:
Check this in the statement:
"A two digit number is 54 more than the number formed by interchanging the digits "
93 = 39 + 54
:
You can check it in the other statement
" the difference between the product and the sum of the digits is 15.