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: Subtract the two sum-conditions first to kill three digits at once, then use the gap clue to lock the answer with almost no algebra.

Step 1: Subtract the conditions. $(b+c+d) - (a+b+c) = 16 - 15$ gives $d - a = 1$, i.e. the units digit is exactly one more than the thousands digit.

Step 2: Use the tens clue. $c = d + 6$. Since $c \le 9$, we need $d \le 3$; and $d = a+1$ with $a \ge 1$ means $d \ge 2$. So $d \in \{2, 3\}$ — only two cases.

Step 3: Case $d = 2$. Then $a = 1$, $c = 8$, and from $a+b+c = 15$: $b = 15 - 1 - 8 = 6$. Number $= 1682$.

Step 4: Case $d = 3$. Then $a = 2$, $c = 9$, and $b = 15 - 2 - 9 = 4$. Number $= 2493$.

Step 5: Difference. Largest $= 2493$, smallest $= 1682$. \[ 2493 - 1682 = 811. \] Answer: option (c), $811$.