A number us made up of two digits. The sum of the two digit is eleven. If the digits are interchanged the original number is increased
by 45.
Let the original number be 10a + b, where a and b are the two digits.
According to the given conditions:
a + b = 11
When the digits are interchanged, the new number is formed as 10b + a.
The original number is increased by 45, which means:
10a + b + 45 = 10b + a
From equation 1, we know that a = 11 - b.
Substitute this in the second equation:
10(11 - b) + b + 45 = 10b + (11 - b)
110 - 10b + b + 45 = 10b + 11 - b
155 - 9b = 11
9b = 144
b = 16
Substitute b back into equation 1:
a + 16 = 11
a = -5
However, since a and b are both digits, the solution a = -5 is not valid.
Therefore, the given conditions do not yield a valid solution.
checking
Substitute this in the second equation:
10(11 - b) + b + 45 = 10b + (11 - b)
110 - 10b + b + 45 = 10b + 11 - b
155 - 9b = 11 ......., +9b
so
18 b = 144
b = 8
Substitute b back into equation 1:
a + 8 = 11
a = 3
NOW check back in original
so we are talking about 38
reversed that is 83
83 - 38 = 45 YES :)
Apologies for the mistake. Thank you for bringing it to my attention.
The correct solution is as follows:
Given that a + b = 11, and 10a + b + 45 = 10b + a.
From a + b = 11, if we set b = 8, then a = 3.
So, the original number is 38, and upon reversing the digits, we get 83. Subtracting 38 from 83 does indeed yield 45, satisfying the given conditions.
Thank you for double-checking.