The least number of 3 digits when successively divided by 2, 5, 4, 3 gives respective remainders of 1,1,3,1 is :

Options

A. 372
B. 275
C. 273
D. 193

Answer

To find the least three-digit number that, when successively divided by 2, 5, 4, and 3 gives respective remainders of 1, 1, 3, and 1, we can follow these steps:

### Step 1: Set up the equations based on the remainders
We know that:
- When divided by 2, the remainder is 1.
- When divided by 5, the remainder is 1.
- When divided by 4, the remainder is 3.
- When divided by 3, the remainder is 1.

We can express these conditions mathematically:
1. \( x \equiv 1 \mod 2 \)
2. \( x \equiv 1 \mod 5 \)
3. \( x \equiv 3 \mod 4 \)
4. \( x \equiv 1 \mod 3 \)

### Step 2: Solve the equations
From the first two equations, we can combine them:
- Since \( x \equiv 1 \mod 2 \) and \( x \equiv 1 \mod 5 \), we can say that \( x \equiv 1 \mod 10 \) (the least common multiple of 2 and 5).

Now we have:
1. \( x \equiv 1 \mod 10 \)
2. \( x \equiv 3 \mod 4 \)
3. \( x \equiv 1 \mod 3 \)

### Step 3: Solve \( x \equiv 3 \mod 4 \)
Let’s express \( x \) in terms of the first equation:
- From \( x \equiv 1 \mod 10 \), we can write \( x = 10k + 1 \) for some integer \( k \).

Now substitute this into the equation \( x \equiv 3 \mod 4 \):
- \( 10k + 1 \equiv 3 \mod 4 \)
- \( 2k + 1 \equiv 3 \mod 4 \) (since \( 10 \mod 4 = 2 \))
- \( 2k \equiv 2 \mod 4 \)
- \( k \equiv 1 \mod 2 \)

This means \( k \) is odd. We can express \( k \) as:
- \( k = 2m + 1 \) for some integer \( m \).

### Step 4: Substitute back to find \( x \)
Substituting \( k \) back into the equation for \( x \):
- \( x = 10(2m + 1) + 1 = 20m + 10 + 1 = 20m + 11 \).

### Step 5: Solve \( x \equiv 1 \mod 3 \)
Now substitute \( x = 20m + 11 \) into \( x \equiv 1 \mod 3 \):
- \( 20m + 11 \equiv 1 \mod 3 \)
- \( 2m + 2 \equiv 1 \mod 3 \) (since \( 20 \mod 3 = 2 \) and \( 11 \mod 3 = 2 \))
- \( 2m \equiv -1 \equiv 2 \mod 3 \)
- \( m \equiv 1 \mod 3 \)

This means \( m = 3n + 1 \) for some integer \( n \).

### Step 6: Substitute back to find \( x \)
Substituting \( m \) back into the equation for \( x \):
- \( x = 20(3n + 1) + 11 = 60n + 20 + 11 = 60n + 31 \).

### Step 7: Find the least three-digit number
To find the least three-digit number, we set \( 60n + 31 \geq 100 \):
- \( 60n \geq 69 \)
- \( n \geq 1.15 \)

Thus, the smallest integer \( n \) is 2:
- \( x = 60(2) + 31 = 120 + 31 = 151 \).

### Step 8: Check if \( x = 151 \) satisfies all conditions
1. \( 151 \div 2 = 75 \) remainder \( 1 \) (satisfied)
2. \( 151 \div 5 = 30 \) remainder \( 1 \) (satisfied)
3. \( 151 \div 4 = 37 \) remainder \( 3 \) (satisfied)
4. \( 151 \div 3 = 50 \) remainder \( 1 \) (satisfied)

### Conclusion
The least three-digit number that satisfies all the conditions is **151**.