Question

Find the least possible 3 digit number which when successively divided by 2, 5, 4 and 3 leaves the remainders 1, 1, 3 and 1 respectively.

• A. 273
• B. 197
• C. 193
• D. 527

Hint:

For successive division problems, always work backward from the last division. Choose the smallest non-negative final quotient (usually 0 or 1) that will eventually lead to a number in the desired range (e.g., 3-digit number).

Answer 1

The Correct Option is C

Step 1: Start from the smallest possible value after the last division.
The last division is by 3, leaving remainder 1. The smallest quotient that keeps the final number a 3-digit number turns out to be 1, so this final quotient is taken as 1.

Step 2: Rebuild the number one division at a time, from the last step back to the first.
Before the division by 3: \( 3 \times 1 + 1 = 4 \). Before the division by 4: \( 4 \times 4 + 3 = 19 \). Before the division by 5: \( 5 \times 19 + 1 = 96 \). Before the division by 2, which gives the original number: \( 2 \times 96 + 1 = 193 \).

Step 3: Check the result satisfies the 3-digit requirement and confirm by dividing forward.
193 is a 3-digit number, and dividing it forward by 2, 5, 4, 3 in order gives remainders 1, 1, 3, 1, matching every condition given. \[ \boxed{193} \]