The smallest number which, when divisible by 16, 18 and 21 leaves the remainder 3, 5 and 8 respectively is-

Options

A. 893
B. 992
C. 995
D. 1024

Answer

To find the smallest number which, when divisible by 16, 18, and 21 leaves the remainders 3, 5, and 8 respectively, we can follow these steps:

### Step 1: Understand the problem
We need to find a number \( N \) such that:
- \( N \equiv 3 \mod 16 \)
- \( N \equiv 5 \mod 18 \)
- \( N \equiv 8 \mod 21 \)

### Step 2: Rewrite the congruences
We can express these congruences in terms of a common number:
- From \( N \equiv 3 \mod 16 \), we can write \( N = 16k + 3 \) for some integer \( k \).
- From \( N \equiv 5 \mod 18 \), we can write \( N = 18m + 5 \) for some integer \( m \).
- From \( N \equiv 8 \mod 21 \), we can write \( N = 21n + 8 \) for some integer \( n \).

### Step 3: Find the differences
To simplify the calculations, we can find the differences between the divisors and their respective remainders:
- \( 16 - 3 = 13 \)
- \( 18 - 5 = 13 \)
- \( 21 - 8 = 13 \)

Since all differences are equal, we can conclude that the number we are looking for can be expressed in terms of the least common multiple (LCM) of the divisors minus this common difference.

### Step 4: Calculate the LCM of 16, 18, and 21
To find the LCM:
- The prime factorization of 16 is \( 2^4 \).
- The prime factorization of 18 is \( 2^1 \times 3^2 \).
- The prime factorization of 21 is \( 3^1 \times 7^1 \).

The LCM is found by taking the highest power of each prime:
- For \( 2 \): \( 2^4 \)
- For \( 3 \): \( 3^2 \)
- For \( 7 \): \( 7^1 \)

Thus, the LCM is:
\[ \text{LCM} = 2^4 \times 3^2 \times 7^1 = 16 \times 9 \times 7 \]

Calculating this step-by-step:
1. \( 16 \times 9 = 144 \)
2. \( 144 \times 7 = 1008 \)

So, the LCM of 16, 18, and 21 is 1008.

### Step 5: Subtract the common difference
Now we subtract the common difference (13) from the LCM:
\[ N = 1008 - 13 = 995 \]

### Conclusion
Thus, the smallest number which, when divisible by 16, 18, and 21 leaves the remainders 3, 5, and 8 respectively is **995**.