Question

the sum of the digits in a three digit number is 12. if the digits are reversed the number is incresed by 495 but reversing only of the ten’s and unit digits increases the number by 36.

Answer 1

The sum of the digit of a three digit number is 12.
If the digits are reversed the number are increased by 495
but reversing only of the ten’s and unit digits increases the number by 36.
The number is:
:
Let a = 100's digit
Let b = 10's digit
Let c = the units
Then
100a + 10b + c = "the number"
:
Write an equation for each statement, combine and simplify:
:
"The sum of the digit of a three digit number is 12."
a + b + c = 12
:
"If the digits are reversed the number are increased by 495"
100a + 10b + c = 100c + 10b + a - 495
100a - a + 10b - 10b = 100c - c - 495
99a = 99c - 495
divide thru by 99
a = c - 5
:
"but reversing only of the ten’s and unit digits increases the number by 36."
100a + 10b + c = 100a +10c + b - 36
100a - 100a + 10b - b = 10c - c - 36
9b = 9c - 36
simplify, divide by 9
b = c - 4
:
Back to the 1st equation
a + b + c = 12
replace a with (c-5); replace b with (c-4)
c-5 + (c-4) + c = 12
3c - 9 = 12
3c = 12 + 9
3c = 21
c = 21/3

MARK ME BRAINLIEST
c = 7

Answer 2

Answer:

237

Step-by-step explanation:

x = 2, y = 3 & z = 7

100x + 10y + z

100 × 2 + 10× 3 + 7

200+30+7

237