the sum of the digits of a two digit number is 9. if the digits are reversed, the new number if 63 grater than the original number. what is the original number?
The number is AB
A+B=9
10B+A=63+10A+B
Those two equations will solve it.
7,2 ---> 2,7
5,4 ----> 4,5
1,8 ----->8,1
9,0 -----> can't be reversed because it doesn't become a two digit number.
3,6 ----->6,3
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all these two digit numbers add up to 9, and if reversed, you should subtract
72- 27= does it equal greater than 63, no
54-45= ..........................., no
81-18 = yes it does ===63
so the original number is 18
To solve this problem, we need to break it down into smaller steps. Let's start by assigning variables to the unknowns.
Let the original two-digit number be represented by the variables '10a + b', where 'a' and 'b' represent the tens and units digits, respectively.
From the given information, we can establish two equations:
Equation 1: a + b = 9 (sum of the digits is 9)
Equation 2: 10b + a = 10a + b + 63 (reversed number is 63 greater)
To solve these equations, we can use substitution or elimination method. Let's use substitution.
From Equation 1, we can write 'a' in terms of 'b':
a = 9 - b
Substituting this value of 'a' into Equation 2, we have:
10b + (9 - b) = 10(9 - b) + b + 63
Simplifying the equation:
10b + 9 - b = 90 - 10b + b + 63
Combining like terms:
9b + 9 = 90 + 63
Further simplifying the equation:
9b + 9 = 153
Now, we can solve for 'b':
9b = 153 - 9
9b = 144
b = 16
Substituting the value of 'b' back into Equation 1:
a + 16 = 9
a = 9 - 16
a = -7
Since 'a' represents the tens digit, it cannot be negative. Thus, there is no valid two-digit number that satisfies the given conditions.
Therefore, there is no solution to this problem.