Question 275504: if you take a certain two-digit number and reverse its digits to get another two-digit number, then add these two numbers together, their sum is 132. what is the original number?
Found 5 solutions by mananth, ikleyn, n2, josgarithmetic, greenestamps:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website! if you take a certain two-digit number and reverse its digits to get another two-digit number, then add these two numbers together, their sum is 132. what is the original number?
Let the digit in the units place be y
& in the tens place be x
So the number will be 10x+y
On reversing the digits
the number becomes 10y+x
The sum of the two = 132
10x+y + 10y+x= 132
11x+11y=132
x+y=12
To satisfy this equation only three pairs are possible
48, 57, 66
All the three numbers wheen reversed and added give you 132
Answer by ikleyn(53748)  (Show Source):
You can put this solution on YOUR website! .
if you take a certain two-digit number and reverse its digits to get another two-digit number,
then add these two numbers together, their sum is 132. what is the original number?
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The solution in the post by @mananth is incomplete: some other possible answers are missed.
I came to provide a complete accurate solution.
Let the digit in the units place be y
& in the tens place be x
So the number will be 10x+y
On reversing the digits
the number becomes 10y+x
The sum of the two = 132
10x+y + 10y+x= 132
11x+11y=132
x+y=12
From this equation, SEVEN different two-digit integer numbers are possible
93, 84, 75, 66, 57, 48, 39.
All the SEVEN numbers when reversed and added give you 132.
Solved completely and correctly.
Answer by n2(79)  (Show Source):
You can put this solution on YOUR website! .
if you take a certain two-digit number and reverse its digits to get another two-digit number,
then add these two numbers together, their sum is 132. what is the original number?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let the digit in the units place be y
& in the tens place be x
So the number will be 10x+y
On reversing the digits
the number becomes 10y+x
The sum of the two = 132
10x+y + 10y+x= 132
11x+11y=132
x+y=12
From this equation, SEVEN different two-digit integer numbers are possible
93, 84, 75, 66, 57, 48, 39.
All the SEVEN numbers when reversed and added give you 132.
Answer by josgarithmetic(39792) (Show Source):
Answer by greenestamps(13327)  (Show Source):
You can put this solution on YOUR website!
While a formal algebraic solution was probably wanted, note that you can also work this problem using logical reasoning and the basic process of adding two 2-digit numbers. In "coded" form, you have this addition, where A and B are the two digits of the original number and S is the sum of those two digits (S is not a digit, because the sum of the two 2-digit numbers is 3 digits):
A B
+ B A
------
S S
Now that sum has the sum "S" in the 10s column and also in the units column, so the value of that sum is 10S + 1S = 11S.
But the sum is 132, so 11S = 132, so S = 132/11 = 12.
So we know that the sum of the two digits A and B is 12.
However, we have no other information to use to find a unique solution to the problem. So the original number can be any 2-digit number in which the sum of the two digits is 12: 39, 48, 57, 66, 75, 84, or 93. Notice that the statement of the problem did not require the two digits of the original number to be different, so 66 is one of the possible answers.
ANSWERS: any of the numbers 39, 48, 57, 66, 75, 84, or 93
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