What is the least possible number which when divided by 2, 3,4,5,6 it leaves the remainders 1,2,3,4,5 respectively ?

Answer

To find the least possible number which, when divided by 2, 3, 4, 5, and 6, leaves the remainders 1, 2, 3, 4, and 5 respectively, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( N \) such that:
- \( N \mod 2 = 1 \)
- \( N \mod 3 = 2 \)
- \( N \mod 4 = 3 \)
- \( N \mod 5 = 4 \)
- \( N \mod 6 = 5 \)

### Step 2: Rewrite the Conditions
From the conditions above, we can rewrite them as:
- \( N + 1 \) is divisible by 2
- \( N + 1 \) is divisible by 3
- \( N + 1 \) is divisible by 4
- \( N + 1 \) is divisible by 5
- \( N + 1 \) is divisible by 6

This means that \( N + 1 \) must be a common multiple of these numbers.

### Step 3: Find the Least Common Multiple (LCM)
We need to find the least common multiple (LCM) of the numbers 2, 3, 4, 5, and 6.

- The prime factorization of each number is:
- \( 2 = 2^1 \)
- \( 3 = 3^1 \)
- \( 4 = 2^2 \)
- \( 5 = 5^1 \)
- \( 6 = 2^1 \times 3^1 \)

To find the LCM, we take the highest power of each prime:
- For 2, the highest power is \( 2^2 \) (from 4).
- For 3, the highest power is \( 3^1 \) (from 3 or 6).
- For 5, the highest power is \( 5^1 \) (from 5).

Thus, the LCM is:
\[ \text{LCM} = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60 \]

### Step 4: Calculate \( N \)
Since \( N + 1 \) must be equal to the LCM, we have:
\[ N + 1 = 60 \]
Thus, we can find \( N \):
\[ N = 60 - 1 = 59 \]

### Conclusion
The least possible number \( N \) which satisfies all the conditions is:
\[ \boxed{59} \]