A least number which when divide by 2, 3 and 5 successively, the remainder ob- tained are 1, 2 and 3 respectively. If the same number is divided by 7 , the re- mainder obtained is -

Options

A. 2
B. 3
C. 4
D. 5

Answer

To find the least number that, when divided by 2, 3, and 5 successively gives remainders of 1, 2, and 3 respectively, we can follow these steps:

### Step 1: Set Up the Problem
Let the number we are looking for be \( x \). According to the problem:
- When \( x \) is divided by 2, the remainder is 1.
- When \( x \) is divided by 3, the remainder is 2.
- When \( x \) is divided by 5, the remainder is 3.

We can express these conditions mathematically:
1. \( x \equiv 1 \mod 2 \)
2. \( x \equiv 2 \mod 3 \)
3. \( x \equiv 3 \mod 5 \)

### Step 2: Rewrite the Conditions
From the above congruences, we can rewrite \( x \) in terms of the moduli:
- From \( x \equiv 1 \mod 2 \), we can say \( x = 2k + 1 \) for some integer \( k \).
- From \( x \equiv 2 \mod 3 \), we can say \( x = 3m + 2 \) for some integer \( m \).
- From \( x \equiv 3 \mod 5 \), we can say \( x = 5n + 3 \) for some integer \( n \).

### Step 3: Solve the Congruences
We can start solving these congruences step by step.

1. **Start with the last two congruences**:
- Set \( x = 5n + 3 \) and substitute into the second congruence.
- So, \( 5n + 3 \equiv 2 \mod 3 \).
- Simplifying gives \( 2n + 0 \equiv 2 \mod 3 \) (since \( 5 \equiv 2 \mod 3 \)).
- This simplifies to \( 2n \equiv 2 \mod 3 \), which gives \( n \equiv 1 \mod 3 \).
- Therefore, \( n = 3p + 1 \) for some integer \( p \).

2. **Substituting back**:
- Substitute \( n \) back into \( x \):
\[ x = 5(3p + 1) + 3 = 15p + 5 + 3 = 15p + 8 \]
- Thus, \( x \equiv 8 \mod 15 \).

3. **Now use the first congruence**:
- Substitute \( x = 15p + 8 \) into \( x \equiv 1 \mod 2 \):
\[ 15p + 8 \equiv 1 \mod 2 \]
- Since \( 15 \equiv 1 \mod 2 \) and \( 8 \equiv 0 \mod 2 \):
\[ p \equiv 1 \mod 2 \]
- Therefore, \( p = 2q + 1 \) for some integer \( q \).

4. **Final substitution**:
- Substitute \( p \) back into \( x \):
\[ x = 15(2q + 1) + 8 = 30q + 15 + 8 = 30q + 23 \]
- Thus, \( x \equiv 23 \mod 30 \).

### Step 4: Find the Least Value
The least value of \( x \) that satisfies all conditions is when \( q = 0 \):
\[ x = 23 \]

### Step 5: Find the Remainder When Divided by 7
Now, we need to find the remainder when \( x = 23 \) is divided by 7:
\[ 23 \div 7 = 3 \quad \text{(quotient)} \quad \text{with a remainder of } 23 - 21 = 2. \]
Thus, the remainder is \( 2 \).

### Final Answer
The remainder when the least number is divided by 7 is **2**.