A number when divided by 10 leaves a remiander 9, when divided by 9 leaves a remainder 8, when divided by 8 leaves a remainder 7 ............ When divided by 2 leaves a remainder 1. Determine the number.

Options

A. 31
B. 1029
C. 2519
D. 1679

Answer

To solve the problem step by step, we need to find a number \( x \) that satisfies the following conditions:

1. \( x \mod 10 = 9 \)
2. \( x \mod 9 = 8 \)
3. \( x \mod 8 = 7 \)
4. \( x \mod 7 = 6 \)
5. \( x \mod 6 = 5 \)
6. \( x \mod 5 = 4 \)
7. \( x \mod 4 = 3 \)
8. \( x \mod 3 = 2 \)
9. \( x \mod 2 = 1 \)

From the above conditions, we can observe a pattern. Each condition can be rewritten as:

- \( x = 10k + 9 \) for some integer \( k \)
- \( x = 9m + 8 \) for some integer \( m \)
- \( x = 8n + 7 \) for some integer \( n \)
- ...
- \( x = 2p + 1 \) for some integer \( p \)

This means that the number \( x \) is always 1 less than the divisor. Therefore, we can express \( x \) as:

\[ x = n - 1 \]

Where \( n \) is the least common multiple (LCM) of the divisors (2 through 10).

### Step 1: Calculate the LCM of the numbers from 2 to 10

The prime factorization of each number is:

- \( 2 = 2 \)
- \( 3 = 3 \)
- \( 4 = 2^2 \)
- \( 5 = 5 \)
- \( 6 = 2 \times 3 \)
- \( 7 = 7 \)
- \( 8 = 2^3 \)
- \( 9 = 3^2 \)
- \( 10 = 2 \times 5 \)

To find the LCM, we take the highest power of each prime that appears:

- Highest power of \( 2 \) is \( 2^3 \) (from 8)
- Highest power of \( 3 \) is \( 3^2 \) (from 9)
- Highest power of \( 5 \) is \( 5^1 \) (from 5 or 10)
- Highest power of \( 7 \) is \( 7^1 \) (from 7)

Thus, the LCM is:

\[ \text{LCM} = 2^3 \times 3^2 \times 5^1 \times 7^1 \]

Calculating this:

\[ = 8 \times 9 \times 5 \times 7 \]

Calculating step-by-step:

1. \( 8 \times 9 = 72 \)
2. \( 72 \times 5 = 360 \)
3. \( 360 \times 7 = 2520 \)

So, \( \text{LCM}(2, 3, 4, 5, 6, 7, 8, 9, 10) = 2520 \).

### Step 2: Determine the number \( x \)

Since \( x = \text{LCM} - 1 \):

\[ x = 2520 - 1 = 2519 \]

### Conclusion

The number that satisfies all the conditions is \( \boxed{2519} \).