What is the smallest number which when divided by 8, 10 or 12 gives the remainder 7 ?

Options

A. 127
B. 137
C. 267
D. 177

Answer

To find the smallest number which, when divided by 8, 10, or 12, gives a remainder of 7, we can follow these steps:

### Step 1: Understand the problem
We need to find a number \( x \) such that:
- \( x \mod 8 = 7 \)
- \( x \mod 10 = 7 \)
- \( x \mod 12 = 7 \)

This means that the number \( x \) is 7 less than a multiple of 8, 10, and 12.

### Step 2: Set up the equations
We can express \( x \) in terms of the least common multiple (LCM) of the divisors:
- \( x = 8k + 7 \)
- \( x = 10m + 7 \)
- \( x = 12n + 7 \)

Where \( k, m, n \) are integers.

### Step 3: Find the LCM of 8, 10, and 12
To find the LCM, we can use the prime factorization method:
- \( 8 = 2^3 \)
- \( 10 = 2^1 \times 5^1 \)
- \( 12 = 2^2 \times 3^1 \)

The LCM is found by taking the highest power of each prime:
- For \( 2 \): \( 2^3 \)
- For \( 3 \): \( 3^1 \)
- For \( 5 \): \( 5^1 \)

Thus, the LCM is:
\[ \text{LCM} = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120 \]

### Step 4: Find the smallest number
Since we want \( x \) to be 7 more than a multiple of the LCM, we can write:
\[ x = 120k + 7 \]
To find the smallest positive \( x \), we set \( k = 1 \):
\[ x = 120 \times 1 + 7 = 127 \]

### Step 5: Verify the solution
Now we need to check if 127 gives a remainder of 7 when divided by 8, 10, and 12:
- \( 127 \div 8 = 15 \) remainder \( 7 \) (since \( 8 \times 15 = 120 \))
- \( 127 \div 10 = 12 \) remainder \( 7 \) (since \( 10 \times 12 = 120 \))
- \( 127 \div 12 = 10 \) remainder \( 7 \) (since \( 12 \times 10 = 120 \))

Since all conditions are satisfied, the smallest number is:
\[ \boxed{127} \]