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Question 269156: When the digits of a two-digit number is 9 more than the original number, and the sum of the digits of the original number is 11. What is the original number?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! I think this problem should read:
;
"When the digits of a two-digit number are reversed, the number is 9 more than the original number, and the sum of the digits of the original number is 11.
What is the original number?"
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Let x = the 10's digit
Let y = the units
the
10x + y = "the original two digit number
:
"When the digits of a two-digit number are reversed, the number is 9 more than the original number,"
10y + x = 10x + y + 9
10y - y = 10x - x + 9
9y = 9x + 9
Simplify divide by 9
y = x + 1
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"the sum of the digits of the original number is 11.
x + y = 11
Replace y with (x+1)
x + x + 1 = 11
2x = 11-1
x = 5
then
y = 5 + 1 = 6
:
56 is the original number
;
:
Check solution in the statement:
"the digits of a two-digit number are reversed, the number is 9 more than the original number,"
65 = 56 + 9
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