Question 902053
let a = the 100's digit
let b = the 10's digit
let c = the units
then
100a + 10b + c = the value of the 3 digit number
:
write an equation for each statement
:
the sum of the digits of a three digit number is 13.
a + b + c = 13
:
when 11 is subtracted from the sum of the hundreds digit and tens digit, the answer is equal to the units digit.
(a+b) - 11 = c
we can rearrange it
a + b - c = 11
:
 additionally, when the digits are reversed, the new value is 495 less than the original number.
100c + 10b + a = 100a + 10b + c - 495
combine like terms
100c - c + 10b - 10b + a - 100a = -495
99c - 99a = -495
simplify divide by 99
c - a = -5
:
Use elimination with the 1st two equations
a + b + c = 13
a + b - c = 11
---------------Subtraction eliminates a and b, find c
2c = 1
c = 1
replace c with 1 in the last equation
1 - a = -5
-a = -5 - 1
-a = -6
a = 6
Find b
6 + b + 1 = 13
b = 13 - 7
b = 6
:
661 is the original number
:
:
See if the works
subtract
661
166
----
495