my thousands digit is 2 times my hundreds digit. the sum of my tens digi t and my ones digit is 2 less than my thousands digit. my hundreds digit is 1 less than my tens digit

To find the number that satisfies these conditions, we can break it down digit by digit. Let's call the thousands digit "A," the hundreds digit "B," the tens digit "C," and the ones digit "D."

We know that A = 2B (the thousands digit is two times the hundreds digit).

We also know that C + D = A - 2 (the sum of the tens digit and ones digit is two less than the thousands digit).

Furthermore, we know that B = C - 1 (the hundreds digit is one less than the tens digit).

Now, let's plug in the value of B in terms of C into the first equation:
A = 2(C - 1).

Now, let's substitute the value of A in terms of B into the second equation:
C + D = 2(C - 1) - 2.

Simplifying this equation, we get:
C + D = 2C - 4.

Finally, let's substitute the value of A in terms of B into the third equation:
C - 1 = C - 1.

This equation implies that C can have any value since it cancels out.

Therefore, the tens digit (C) can be any number, and we need more information to solve the problem.