Question

For a 4-digit number (greater than 1000), sum of the digits in the thousands, hundreds, and tens places is 15. Sum of the digits in the hundreds, tens, and units places is 16. Also, the digit in the tens place is 6 more than the digit in the units place. The difference between the largest and smallest possible value of the number is

• A. \(40\)
• B. \(78\)
• C. \(811\)
• D. \(735\)

Hint:

When dealing with digit problems, always convert the verbal conditions into equations using place-value notation, and apply the digit constraints \(0 \leq \text{digit} \leq 9\) to narrow down possible values quickly.

Answer 1

The Correct Option is C

Approach: Three conditions tie four digits together—express every digit in terms of one variable (the tens digit), then let the digit bounds \(0\)–\(9\) squeeze out the only possibilities.

Step 1: Name the digits. Number \(= abcd\), where \(a\) is the thousands digit, \(b\) the hundreds digit, \(c\) the tens digit, and \(d\) the units digit, with \(a\ge1\). Given:
\(a+b+c=15\)
\(b+c+d=16\)
\(c=d+6\)

Step 2: Reduce to one variable \(c\). From \(c=d+6\), we get \(d=c-6\). Substitute into \(b+c+d=16\):
\(b+c+(c-6)=16\Rightarrow b=22-2c.\)
Now substitute \(b\) into \(a+b+c=15\):
\(a+(22-2c)+c=15\Rightarrow a=c-7.\)

Step 3: Apply digit limits. Now everything depends on \(c\):
\(a=c-7\ge1\Rightarrow c\ge8.\)
Since \(c\le9\), the only possibilities are \(c=8\) or \(c=9\).

Intuition: Subtracting \(a+b+c=15\) from \(b+c+d=16\) gives \(d-a=1\), which provides a quick check for the two cases below.

Step 4: Build both numbers.
\(c=8:\ a=1,\ b=6,\ d=2\Rightarrow1682.\)
\(c=9:\ a=2,\ b=4,\ d=3\Rightarrow2493.\)
Both satisfy all three conditions. For example, for \(2493\): \(2+4+9=15,\ 4+9+3=16,\ 9=3+6.\) ✓

Step 5: Required difference.
\(2493-1682=\boxed{811}\), matching option (c).

Answer 2

Let the 4-digit number be \( \overline{abcd} \), where: \[ a = \text{thousands digit}, \quad b = \text{hundreds digit}, \quad c = \text{tens digit}, \quad d = \text{units digit} \] with \( a \neq 0 \). 
Step 1: Write equations from the given conditions. From the question: \[ a + b + c = 15 \quad \cdots (1) \] \[ b + c + d = 16 \quad \cdots (2) \] Also, \[ c = d + 6 \quad \Rightarrow \quad d = c - 6 \quad \cdots (3) \] 
Step 2: Substitute \( d = c - 6 \) in equation (2). From (2): \[ b + c + d = 16 \Rightarrow b + c + (c - 6) = 16 \Rightarrow b + 2c - 6 = 16 \Rightarrow b + 2c = 22 \Rightarrow b = 22 - 2c \quad \cdots (4) \] 
Step 3: Use equation (1) to express \( a \) in terms of \( c \). From (1): \[ a + b + c = 15 \Rightarrow a + (22 - 2c) + c = 15 \Rightarrow a + 22 - c = 15 \Rightarrow a = 15 - 22 + c = c - 7 \quad \cdots (5) \] 
Step 4: Use digit constraints. Digits must satisfy \( 0 \leq a,b,c,d \leq 9 \) and \( a \geq 1 \). From (5): \( a = c - 7 \geq 1 \Rightarrow c \geq 8 \). Also \( a \leq 9 \Rightarrow c - 7 \leq 9 \Rightarrow c \leq 16 \). Since \( c \) is a digit, \( c \in \{8,9\} \). 
\underline{Case 1:} \( c = 8 \) \[ a = c - 7 = 1, \quad b = 22 - 2c = 22 - 16 = 6, \quad d = c - 6 = 2 \] Number: \( \overline{abcd} = 1682 \). Check: \[ 1 + 6 + 8 = 15, \quad 6 + 8 + 2 = 16, \quad 8 = 2 + 6 \; \text{(OK)} \] 
Case 2: \( c = 9 \) \[ a = c - 7 = 2, \quad b = 22 - 2c = 22 - 18 = 4, \quad d = c - 6 = 3 \] Number: \( \overline{abcd} = 2493 \). Check: \[ 2 + 4 + 9 = 15, \quad 4 + 9 + 3 = 16, \quad 9 = 3 + 6 \; \text{(OK)} \] 
Step 5: Find the required difference. Largest possible number \( = 2493 \) 
Smallest possible number \( = 1682 \) \[ \text{Difference} = 2493 - 1682 = 811 \] So, the required difference is \( 811 \).