A 2-digit number is twice the difference of its digits. What is the number?
I suspect a typo. The difference cannot be more than 9, so the number must be at most 18.
It is even, so the only candidates are
10 12 14 16 18
None of those works.
It is impossible. 12, 14, or 16 may work, but I think you should give up on this equation...
To find the answer, we need to apply some logic and algebraic reasoning.
Let's assume the 2-digit number is represented as "AB," where A represents the tens digit and B represents the units digit.
According to the given information, the number is twice the difference of its digits. So, we can write the equation as:
AB = 2 * (A - B)
Expanding the right side, we have:
10A + B = 2A - 2B
Combining like terms, we get:
10A - 2A = 2B + B
Simplifying further, we have:
8A = 3B
To find the 2-digit number, we need to find values for A and B that satisfy the equation.
Since A and B must be digits, there are a limited number of possibilities.
Starting with A = 1, we can substitute values for B (from 0 to 9) to see if they satisfy the equation.
For example, when A = 1 and B = 0, the equation becomes:
8(1) = 3(0)
This is not true since 8 is not equal to 0.
Using this method, we find that there is no pair of A and B digits that satisfy the equation.
Therefore, there is no 2-digit number that is twice the difference between its digits.