A natural number ,when divided by 9,10,12 or 15 leaves a rremainder of 3 in each case .What is the smallest of all such numbers ?

Options

A. 183
B. 153
C. 63
D. 123

Answer

To solve the problem step by step, we need to find the smallest natural number that leaves a remainder of 3 when divided by 9, 10, 12, or 15.

### Step 1: Understand the problem
We need a number \( N \) such that:
- \( N \mod 9 = 3 \)
- \( N \mod 10 = 3 \)
- \( N \mod 12 = 3 \)
- \( N \mod 15 = 3 \)

### Step 2: Rewrite the conditions
We can rewrite the conditions as:
- \( N - 3 \) is divisible by 9
- \( N - 3 \) is divisible by 10
- \( N - 3 \) is divisible by 12
- \( N - 3 \) is divisible by 15

Let \( M = N - 3 \). Then, we need to find \( M \) such that:
- \( M \mod 9 = 0 \)
- \( M \mod 10 = 0 \)
- \( M \mod 12 = 0 \)
- \( M \mod 15 = 0 \)

### Step 3: Find the Least Common Multiple (LCM)
Now, we need to find the least common multiple (LCM) of the numbers 9, 10, 12, and 15.

- The prime factorization of each number is:
- \( 9 = 3^2 \)
- \( 10 = 2 \times 5 \)
- \( 12 = 2^2 \times 3 \)
- \( 15 = 3 \times 5 \)

To find the LCM, we take the highest power of each prime:
- For \( 2 \): \( 2^2 \) (from 12)
- For \( 3 \): \( 3^2 \) (from 9)
- For \( 5 \): \( 5^1 \) (from 10 and 15)

Thus, the LCM is:
\[ \text{LCM} = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 \]

Calculating this:
\[ 4 \times 9 = 36 \]
\[ 36 \times 5 = 180 \]

So, \( M = 180 \).

### Step 4: Find \( N \)
Since \( M = N - 3 \), we can find \( N \):
\[ N = M + 3 = 180 + 3 = 183 \]

### Conclusion
The smallest natural number that leaves a remainder of 3 when divided by 9, 10, 12, or 15 is **183**.