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: Reversing the chain of successive divisions gives one combined expression for the number: \( N = 2\big(5\big(4(3q+1)+3\big)+1\big)+1 \), where \( q \) is the final quotient after all four divisions.

Step 2: Expand this step by step: \( 4(3q+1)+3 = 12q+7 \), then \( 5(12q+7)+1 = 60q+36 \), then \( 2(60q+36)+1 = 120q+73 \). So \( N = 120q + 73 \).

Step 3: Try the smallest non-negative values of \( q \): for \( q=0 \), \( N=73 \), only a 2-digit number, too small. For \( q=1 \), \( N = 120+73 = 193 \), a valid 3-digit number.
\[ \boxed{193} \]