my thousandth digit is 3 times my hundredth digit. The sum of the tenth digit and unit digit is 1 less than my thousandth digit. My hundreth digit is 2 less than my tenth digit. What four digit I am?

my thousandth digit is 3 times my hundredth digit

---> only possible cases: 31xx , 62xx, and 93xx

My hundredth digit is 2 less than my tenth digit.
---> only possible cases: 313x , 624x, and 935x

The sum of the tenth digit and unit digit is 1 less than my thousandth digit
---> possible cases: 6241, 9353

6241 or 9353

hint: consider the cases where the thousands digit is three times the hundreds digit... so

Case 1: the thousands digit is a 3 while the hundreds digit is a 1
Case 2: the thousands digit is a 6 while the hundreds digit is a 2
Case 3: the thousands digit is a 9 while the undreds digit is a 3
Then work on the Tens digit in each of these three cases : )

That is a lovely solution for a Thursday morning Reiny!

I really like it : )

Thank you Ms π

To find the four-digit number that satisfies these conditions, we will use algebraic reasoning.

Let's assume the number is represented as ABCD, where A is the thousandth digit, B is the hundredth digit, C is the tenth digit, and D is the unit digit.

Based on the given information:
1. The thousandth digit is 3 times the hundredth digit: A = 3B.
2. The sum of the tenth digit and unit digit is 1 less than the thousandth digit: C + D = A - 1.
3. The hundredth digit is 2 less than the tenth digit: B = C - 2.

Substituting equation (3) into equation (1), we get:
B = C - 2.

Now we have three equations with three unknowns, A, B, and C. We can solve them simultaneously:
1. A = 3B.
2. C + D = A - 1.
3. B = C - 2.

Substituting equation (1) into equation (2), we have:
C + D = 3B - 1.

Now substitute equation (3) into the above equation:
C + D = 3(C - 2) - 1,
C + D = 3C - 6 - 1,
C + D = 3C - 7.

Rearrange the equation:
2C - D = 7.

Now, we need to find the possible values for C and D that satisfy equation (4).

Since we are dealing with a four-digit number, the thousandth digit (A) cannot be zero.

Let's try different values for C and D:
When C = 1, D = 9,
When C = 2, D = 5,
When C = 3, D = 1,
When C = 4, D = 7,
When C = 5, D = 3,
When C = 6, D = 9,
When C = 7, D = 5,
When C = 8, D = 1,
When C = 9, D = 7.

Now, let's substitute these values of C and D into equation (3) to find B:
For C = 1, D = 9, B = 1 - 2 = -1 (which is not possible as a digit),
For C = 2, D = 5, B = 2 - 2 = 0 (which is not possible as a digit),
For C = 3, D = 1, B = 3 - 2 = 1,
For C = 4, D = 7, B = 4 - 2 = 2,
For C = 5, D = 3, B = 5 - 2 = 3,
For C = 6, D = 9, B = 6 - 2 = 4,
For C = 7, D = 5, B = 7 - 2 = 5,
For C = 8, D = 1, B = 8 - 2 = 6,
For C = 9, D = 7, B = 9 - 2 = 7.

As we can see, only when C = 3, D = 1, and B = 1 satisfies all the given conditions.

Therefore, the four-digit number is 3171.