The least number, which when divided by 5, 10, 12 and 15 leaves a remainder of 2, in each case but when divided by 7 leaves no remainder is

Options

A. 189
B. 182
C. 175
D. 91

Answer

To find the least number that, when divided by 5, 10, 12, and 15 leaves a remainder of 2, and when divided by 7 leaves no remainder, we can follow these steps:

### Step 1: Find the LCM of 5, 10, 12, and 15

To find the least common multiple (LCM), we can use the prime factorization method.

- **Prime factorization:**
- 5 = 5
- 10 = 2 × 5
- 12 = 2² × 3
- 15 = 3 × 5

Now, we take the highest power of each prime number:
- The highest power of 2 is 2² (from 12)
- The highest power of 3 is 3¹ (from 12 and 15)
- The highest power of 5 is 5¹ (from 5, 10, and 15)

Thus, the LCM is calculated as:
\[ \text{LCM} = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60 \]

### Step 2: Set up the equation for the required number

Let the required number be \( x \). According to the problem, \( x \) leaves a remainder of 2 when divided by 5, 10, 12, and 15. Therefore, we can express this as:
\[ x = 60k + 2 \]
where \( k \) is a non-negative integer.

### Step 3: Ensure the number is divisible by 7

We also need \( x \) to be divisible by 7. Thus, we can set up the equation:
\[ 60k + 2 \equiv 0 \ (\text{mod} \ 7) \]

### Step 4: Simplify the equation

First, we need to find \( 60 \mod 7 \):
\[ 60 \div 7 = 8 \quad \text{(remainder 4)} \]
So,
\[ 60 \equiv 4 \ (\text{mod} \ 7) \]

Now substituting this back into our equation:
\[ 4k + 2 \equiv 0 \ (\text{mod} \ 7) \]
This simplifies to:
\[ 4k \equiv -2 \ (\text{mod} \ 7) \]
or equivalently,
\[ 4k \equiv 5 \ (\text{mod} \ 7) \]

### Step 5: Solve for \( k \)

To solve for \( k \), we can multiply both sides by the modular inverse of 4 modulo 7. The inverse of 4 modulo 7 is 2 (since \( 4 \times 2 = 8 \equiv 1 \ (\text{mod} \ 7) \)):
\[ k \equiv 2 \times 5 \ (\text{mod} \ 7) \]
\[ k \equiv 10 \ (\text{mod} \ 7) \]
\[ k \equiv 3 \ (\text{mod} \ 7) \]

Thus, \( k \) can be expressed as:
\[ k = 7m + 3 \]
for some integer \( m \).

### Step 6: Substitute back to find \( x \)

Now substituting \( k \) back into our equation for \( x \):
\[ x = 60(7m + 3) + 2 \]
\[ x = 420m + 180 + 2 \]
\[ x = 420m + 182 \]

### Step 7: Find the least number

To find the least number, we set \( m = 0 \):
\[ x = 182 \]

### Conclusion

Thus, the least number that meets the conditions is **182**.

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