What positive, two-digit number, when doubled and added to two, gives a sum that is the original number with its digits reversed?
25
To solve this problem, let's break it down step by step.
We are looking for a positive, two-digit number. Let's represent the tens digit as 'x' and the units digit as 'y'. So, the two-digit number can be written as 10x + y.
According to the problem, when this number is doubled and added to 2, it gives a sum that is equal to the original number with its digits reversed.
Thus, we can form the equation as follows:
2(10x + y) + 2 = 10y + x
Let's solve this equation:
Expanding:
20x + 2y + 2 = 10y + x
Combining like terms:
20x + 2y - x - 10y = -2
Simplifying:
19x - 8y = -2
Now, we want to find a positive, two-digit number. Since both x and y represent the digits, they can be any whole numbers from 0 to 9 inclusive.
To make the problem more approachable, we can start by trying different values for y and then solve for x.
Let's start with y = 0:
19x - 8(0) = -2
19x = -2
x = -2/19 (which is not a whole number)
Since x is not a whole number, let's try the next value of y.
Let's try y = 1:
19x - 8(1) = -2
19x - 8 = -2
19x = 6
x = 6/19 (which is not a whole number)
Now, let's try y = 2:
19x - 8(2) = -2
19x - 16 = -2
19x = 14
x = 14/19 (which is not a whole number)
Continuing this process, we will find that there are no whole number solutions for x and y that satisfy the given conditions.
Therefore, there is no positive, two-digit number that, when doubled and added to two, gives a sum that is the original number with its digits reversed.