Find the least number which when divided by 12, 24, 36 and 40 leaves a remainder 1, but when divided by 7 leaves no remainder.

Options

A. 361
B. 1080
C. 721
D. 371

Answer

To find the least number which, when divided by 12, 24, 36, and 40 leaves a remainder of 1, and when divided by 7 leaves no remainder, we can follow these steps:

### Step 1: Find the LCM of 12, 24, 36, and 40
To find the least common multiple (LCM), we can use the prime factorization method.

- **Prime factorization:**
- 12 = 2² × 3¹
- 24 = 2³ × 3¹
- 36 = 2² × 3²
- 40 = 2³ × 5¹

- **LCM calculation:**
- Take the highest power of each prime factor:
- For 2: max(2², 2³, 2², 2³) = 2³
- For 3: max(3¹, 3¹, 3², 3⁰) = 3²
- For 5: max(5⁰, 5⁰, 5⁰, 5¹) = 5¹

Thus, the LCM = 2³ × 3² × 5¹ = 8 × 9 × 5 = 360.

### Step 2: Formulate the required number
The required number can be expressed in the form:
\[ N = 360k + 1 \]
where \( k \) is a natural number.

### Step 3: Find \( k \) such that \( N \) is divisible by 7
We need \( N \) to be divisible by 7:
\[ 360k + 1 \equiv 0 \ (\text{mod} \ 7) \]
This simplifies to:
\[ 360k \equiv -1 \ (\text{mod} \ 7) \]

### Step 4: Calculate \( 360 \mod 7 \)
Now, we calculate \( 360 \mod 7 \):
- \( 360 \div 7 = 51 \) remainder \( 3 \)
So, \( 360 \equiv 3 \ (\text{mod} \ 7) \).

### Step 5: Substitute and solve for \( k \)
Now substituting back:
\[ 3k \equiv -1 \ (\text{mod} \ 7) \]
This can be rewritten as:
\[ 3k \equiv 6 \ (\text{mod} \ 7) \]
To solve for \( k \), we can find the multiplicative inverse of 3 modulo 7.

### Step 6: Find the inverse of 3 modulo 7
Testing values:
- \( 3 \times 5 = 15 \equiv 1 \ (\text{mod} \ 7) \)

So, the inverse of 3 is 5. Now, multiply both sides of the equation \( 3k \equiv 6 \) by 5:
\[ k \equiv 30 \ (\text{mod} \ 7) \]
Calculating \( 30 \mod 7 \):
- \( 30 \div 7 = 4 \) remainder \( 2 \)
So, \( k \equiv 2 \ (\text{mod} \ 7) \).

### Step 7: Find the smallest \( k \)
The smallest positive integer \( k \) that satisfies this is \( k = 2 \).

### Step 8: Calculate the required number
Now substitute \( k \) back into the equation for \( N \):
\[ N = 360 \times 2 + 1 = 720 + 1 = 721 \]

### Final Answer
The least number is **721**.

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