Question

A number when successively divided by 5 and 6 gives remainders 3 and 2 respectively. What will be the remainders if the number is successively divided by 3 and 4?

• A. 2, 3
• B. 2, 1
• C. 1, 2
• D. 3, 4

Hint:

Rebuild the general form of the number from the two given remainders, then divide that same expression successively by 3 and 4.

Answer 1

The Correct Option is C

Step 1: Find one actual number that fits the original condition.
We need a number that leaves remainder 3 when divided by 5, and when you divide that quotient by 6, you get remainder 2. Try quotient $q_1 = 2$ (the smallest value fitting $q_1 = 6k+2$ with $k=0$): then $N = 5(2) + 3 = 13$. Check: $13 \div 5 = 2$ remainder $3$, correct, and $2 \div 6 = 0$ remainder $2$, correct. So $N = 13$ works.

Step 2: Successively divide 13 by 3 and then 4.
$13 \div 3 = 4$ remainder $1$. Now take the quotient 4 and divide by 4: $4 \div 4 = 1$ remainder $0$. So this particular case gives remainders $(1, 0)$, which is not among the listed options.

Step 3: Try the next value of the number (k=1) to check for a different remainder pattern.
For $k=1$: $N = 30(1) + 13 = 43$. Check: $43 \div 5 = 8$ remainder $3$, and $8 \div 6 = 1$ remainder $2$, correct. Now $43 \div 3 = 14$ remainder $1$, and $14 \div 4 = 3$ remainder $2$. So this case gives remainders $(1, 2)$.

Step 4: Match with the given options.
Since the question expects one fixed pair of remainders from the choices, and $(1, 2)$ is the pair that appears among them, that is the intended answer.

Final Answer:
The remainders are 1 and 2. \[ \boxed{1, 2} \]