When a number is divided by 1,2,3,4,5, .... ,(n-1), n individually it leaves 0,1,2,3,4,.... ,(n-2), (n-1) respective remainders, then this number can be :
(i) `(n!)`
(ii) `(n!-1)`
(iii) [L.C.M. of 1,2,3, ..., n)-1]

Options

A. BOTH (i) AND (ii)
B. BOTH (ii) AND (iii)
C. ONLY (iii)
D. ONLY(ii)

Answer

To solve the problem, we need to determine which of the given options can represent a number that leaves specific remainders when divided by integers from 1 to n. The number must leave remainders of 0, 1, 2, ..., (n-1) when divided by 1, 2, 3, ..., n respectively.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need a number \( x \) such that:
- \( x \mod 1 = 0 \)
- \( x \mod 2 = 1 \)
- \( x \mod 3 = 2 \)
- ...
- \( x \mod n = n-1 \)

This means that \( x + 1 \) must be divisible by all numbers from 1 to n.

2. **Analyzing the Conditions**:
From the above conditions, we can deduce that:
\[ x + 1 \equiv 0 \mod k \quad \text{for } k = 1, 2, ..., n \]
This implies that \( x + 1 \) is a common multiple of all integers from 1 to n.

3. **Finding the Number**:
The least common multiple (LCM) of the numbers from 1 to n is denoted as \( \text{lcm}(1, 2, ..., n) \). Therefore, we can express \( x \) as:
\[ x = \text{lcm}(1, 2, ..., n) - 1 \]

4. **Evaluating the Options**:
Now we will evaluate the given options:
- **Option (i)**: \( n! \)
- \( n! \) is divisible by all integers from 1 to n, so \( n! \mod k = 0 \) for all \( k \). This does not satisfy our conditions.
- **Option (ii)**: \( n! - 1 \)
- \( n! - 1 \) will leave a remainder of \( n-1 \) when divided by \( n \), \( n-2 \) when divided by \( n-1 \), and so on, fulfilling the conditions.
- **Option (iii)**: \( \text{lcm}(1, 2, ..., n) - 1 \)
- This also satisfies the conditions as derived above.

5. **Conclusion**:
The correct options that satisfy the conditions are:
- (ii) \( n! - 1 \)
- (iii) \( \text{lcm}(1, 2, ..., n) - 1 \)

### Final Answer:
The number can be:
- (ii) \( n! - 1 \)
- (iii) \( \text{lcm}(1, 2, ..., n) - 1 \)