in a 2 digit number sum of two digit is 7 of the difference of two digit number and number obtained by reversing the degit is 9 find the numbers
0, 7
1, 6
2, 5
3, 4
Which of those combinations meets the second criterion?
Let m be the ten digit, and n be the ones. Then:
m+n=7
(10m+n)-(10n+m)=9
Then:
10m-10n+n-m=9
Solve for m and n
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To solve this problem, let's break it down into steps:
Step 1: Define the two-digit number
Let's say the two-digit number is represented as '10a + b', where 'a' represents the tens digit and 'b' represents the ones digit.
Step 2: Define the reverse of the number
The reverse of '10a + b' can be represented as '10b + a'.
Step 3: Set up the equations
According to the problem statement, we have two conditions:
1. The sum of the two digits is 7: a + b = 7
2. The difference between the original number and its reverse is 9: (10a + b) - (10b + a) = 9
Step 4: Simplify the equations
Expanding the second equation, we get: 10a + b - 10b - a = 9
Simplifying further, we have: 9a - 9b = 9
Dividing both sides by 9, we get: a - b = 1
Step 5: Solve the equations
Now we have a system of equations:
a + b = 7 (Equation 1)
a - b = 1 (Equation 2)
Solving Equation 2 for 'a', we get: a = b + 1
Substituting this value into Equation 1, we have: (b + 1) + b = 7
Simplifying, we get: 2b + 1 = 7
Subtracting 1 from both sides, we have: 2b = 6
Dividing both sides by 2, we get: b = 3
Now that we have the value of 'b', we can substitute it back into Equation 2 to find 'a':
a - 3 = 1
Adding 3 to both sides, we get: a = 4
Therefore, the two-digit number is 43.
So, the numbers are 4 and 3.