Find the least number which divided by 12,18,36 and 45 leaves 8,14,32 and 41 as remainder respectively.

Options

A. 176
B. 180
C. 178
D. 186

Answer

To find the least number which, when divided by 12, 18, 36, and 45, leaves remainders of 8, 14, 32, and 41 respectively, we can follow these steps:

### Step 1: Set Up the Equations
We can express the problem in terms of congruences:
- Let \( N \) be the least number we are looking for.
- From the problem, we have the following equations:
\[ N \equiv 8 \mod{12} \]
\[ N \equiv 14 \mod{18} \]
\[ N \equiv 32 \mod{36} \]
\[ N \equiv 41 \mod{45} \]

### Step 2: Adjust the Remainders
To simplify our calculations, we can adjust the remainders by subtracting the respective remainders from \( N \):
- Let’s define \( M = N - 8 \). Then we can rewrite the equations as:
\[ M \equiv 0 \mod{12} \]
\[ M \equiv 6 \mod{18} \]
\[ M \equiv 24 \mod{36} \]
\[ M \equiv 33 \mod{45} \]

### Step 3: Find the LCM of the Divisors
Next, we need to find the least common multiple (LCM) of the divisors (12, 18, 36, 45):
- The prime factorization of each number is:
- \( 12 = 2^2 \times 3^1 \)
- \( 18 = 2^1 \times 3^2 \)
- \( 36 = 2^2 \times 3^2 \)
- \( 45 = 3^2 \times 5^1 \)

- The LCM is found by taking the highest power of each prime:
- \( LCM = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 = 180 \)

### Step 4: Solve for \( M \)
Now we have:
- \( M \equiv 0 \mod{12} \)
- \( M \equiv 6 \mod{18} \)
- \( M \equiv 24 \mod{36} \)
- \( M \equiv 33 \mod{45} \)

We can start with \( M = 180k \) for some integer \( k \) and check which value satisfies the other congruences.

### Step 5: Check Values of \( M \)
We can check \( M = 180 \) (when \( k = 1 \)):
- \( 180 \mod{12} = 0 \) (satisfied)
- \( 180 \mod{18} = 0 \) (not satisfied, we need 6)
- \( 180 \mod{36} = 0 \) (not satisfied, we need 24)
- \( 180 \mod{45} = 0 \) (not satisfied, we need 33)

Next, let’s check \( M = 180 - 4 = 176 \):
- \( 176 \mod{12} = 8 \) (satisfied)
- \( 176 \mod{18} = 14 \) (satisfied)
- \( 176 \mod{36} = 32 \) (satisfied)
- \( 176 \mod{45} = 41 \) (satisfied)

### Step 6: Calculate \( N \)
Finally, we can find \( N \):
\[ N = M + 8 = 176 + 8 = 184 \]

### Conclusion
The least number which, when divided by 12, 18, 36, and 45, leaves remainders of 8, 14, 32, and 41 respectively is:
\[ \boxed{184} \]