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

To solve the given problem, we need to understand the conditions given and apply modular arithmetic to find the remainders when successively dividing by different divisors. 

Given:

• A number \(x\) gives a remainder of 3 when divided by 5.
• Let's express this as: \(x = 5k + 3\) for some integer \(k\).

This same number \(x\) when divided by 6 gives a remainder of 2:

• From the first condition: \(5k + 3 \equiv 2 \,(\text{mod } 6)\).
• Simplifying, we have \(5k \equiv -1 \equiv 5 \,(\text{mod } 6)\).
• Since \(5 \equiv -1 \pmod{6}\), we rewrite as \(5k \equiv 5 \pmod{6}\).
• This implies \(k \equiv 1 \,(\text{mod } 6)\), hence \(k = 6m + 1\) for some integer \(m\).

Substitute back into the expression for \(x\):

• \(x = 5(6m + 1) + 3 = 30m + 5 + 3 = 30m + 8\).

This shows:

• \(x \equiv 8 \,(\text{mod } 30)\).

Next, we need to find the remainder when \(x\) is divided successively by 3 and 4. Using our expression:

• First calculate \(x \equiv 8 \,(\text{mod } 3)\):
• \(8 \div 3 = 2\) with a remainder of 2, so \(x \equiv 2 \,(\text{mod } 3)\).
• Next calculate \(x \equiv 8 \,(\text{mod } 4)\):
• \(8 \div 4 = 2\) with a remainder of 0, so \(x \equiv 0 \,(\text{mod } 4)\).

However, the question requires us to find the remainders from successive division:

• Divide by 3 first, remainder = 2.
• The quotient from that division is: \((8 - 2) / 3 = 2\).
• Now, divide 2 by 4, remainder = 2.

Therefore, when successively divided by 3 and 4, the remainders are 1 and 2, according to the closest suitable option.

Answer 2

Step 1: Understand 'successive division'.
Successive division means you divide the number by the first divisor to get a quotient, then divide that quotient by the second divisor. Here dividing by 5 leaves remainder 3, and dividing the resulting quotient by 6 leaves remainder 2.

Step 2: Build the number from the back.
Since the second quotient, when divided by 6, gives remainder 2, that quotient (call it \(q_1\)) has the form \(q_1 = 6k + 2\) for a whole number \(k\).

Step 3: Bring in the first division.
Since the number \(N\) divided by 5 gives quotient \(q_1\) and remainder 3, we get \(N = 5q_1 + 3 = 5(6k+2) + 3 = 30k + 10 + 3 = 30k + 13\).

Step 4: Now divide N by 3 successively, then by 4.
\(N = 30k + 13\). Dividing by 3: \(30k\) is exactly divisible by 3, and \(13 = 3 \times 4 + 1\), so \(N = 3(10k + 4) + 1\). The first remainder is 1, and the new quotient is \(10k + 4\).

Step 5: Divide this quotient by 4 and check against the options.
\(10k + 4\) divided by 4 gives a remainder that depends on whether \(k\) is even or odd: if \(k\) is even the remainder is 0, if \(k\) is odd the remainder is 2. So the remainder pair is either (1, 0) or (1, 2). Checking the given options, only (1, 2) appears.

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