|
Question 902053: the sum of the digits of a three digit number is 13. when 11 is subtracted from the sum of the hundreds digit and tens digit, the answer is equal to the units digit. additionally, when the digits are reversed, the new value is 495 less than the original number.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 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
|
|
|
| |
