Find the least number which when divided by 12, leaves a remainder 7, when divided by 15, leaves a remainder 10 and when divided by 16, leaves a remainder 11

Options

A. 115
B. 235
C. 247
D. 475

Answer

To find the least number that meets the given conditions, we can follow these steps:

### Step 1: Set Up the Congruences
We need to find a number \( x \) such that:
- \( x \equiv 7 \mod{12} \)
- \( x \equiv 10 \mod{15} \)
- \( x \equiv 11 \mod{16} \)

### Step 2: Rewrite the Congruences
From the congruences, we can express \( x \) in terms of the remainders:
- From \( x \equiv 7 \mod{12} \), we can write:
\[ x = 12k + 7 \] for some integer \( k \).

- From \( x \equiv 10 \mod{15} \), we can write:
\[ x = 15m + 10 \] for some integer \( m \]. - From \( x \equiv 11 \mod{16} \), we can write:
\[ x = 16n + 11 \] for some integer \( n \]. ### Step 3: Find the Differences Next, we calculate the differences between the divisors and the remainders: - For the first congruence: \( 12 - 7 = 5 \)
- For the second congruence: \( 15 - 10 = 5 \)
- For the third congruence: \( 16 - 11 = 5 \)

### Step 4: Find the LCM of the Divisors
Now, we need to find the least common multiple (LCM) of the divisors 12, 15, and 16.

- **Factorization**:
- \( 12 = 2^2 \times 3^1 \)
- \( 15 = 3^1 \times 5^1 \)
- \( 16 = 2^4 \)

- **LCM Calculation**:
- The LCM takes the highest power of each prime factor:
- For \( 2 \): \( 2^4 \) (from 16)
- For \( 3 \): \( 3^1 \) (from both 12 and 15)
- For \( 5 \): \( 5^1 \) (from 15)

Thus, the LCM is:
\[ \text{LCM} = 2^4 \times 3^1 \times 5^1 = 16 \times 3 \times 5 = 240 \]

### Step 5: Subtract the Common Difference
Now, we subtract the common difference (which is 5) from the LCM:
\[ x = 240 - 5 = 235 \]

### Final Answer
Thus, the least number that satisfies all the conditions is:
\[ \boxed{235} \]