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

Answer 1

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.

Answer 2

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 :)

Answer 3

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.