CAT 2025 Slot3 Quant: For a 4-digit number (greater than 1000), the sum of the digits in the thousands, hundreds, and tens places is 15

Step 1: Set up the equations
Let the 4-digit number be $N = 1000a + 100b + 10c + d$.
1. Thousands + Hundreds + Tens: $a + b + c = 15 \dots(i)$
2. Hundreds + Tens + Units: $b + c + d = 16 \dots(ii)$
3. Tens is 6 more than Units: $c = d + 6 \dots(iii)$

Step 2: Relate the digits
Subtract equation (ii) from (i):
$(a + b + c) - (b + c + d) = 15 - 16$
$a - d = -1 \implies \mathbf{a = d - 1}$

Step 3: Determine the range for $d$
All digits must be between $0$ and $9$. Also, $a \ge 1$ (for a 4-digit number).
- From $a = d - 1$: If $a \ge 1$, then $d \ge 2$.
- From $c = d + 6$: If $c \le 9$, then $d \le 3$.
So, $d$ can only be **2** or **3**.

Step 4: Calculate possible values
Case 1: $d = 3$
$a = 3 - 1 = 2$
$c = 3 + 6 = 9$
Substitute in (i): $2 + b + 9 = 15 \implies b = 4$.
Number = $\mathbf{2493}$.

Case 2: $d = 2$
$a = 2 - 1 = 1$
$c = 2 + 6 = 8$
Substitute in (i): $1 + b + 8 = 15 \implies b = 6$.
Number = $\mathbf{1682}$.

Step 5: Find the difference
Difference $= 2493 - 1682 = \mathbf{811}$.