https://file.aiquickdraw.com/mathbot/file/1728505982262dfd62025.jpg

The problem in the image asks to find the smallest integer \( n \) such that:

\[
n \equiv 0 \pmod{2}, \quad n \equiv 0 \pmod{3}, \quad n \equiv 0 \pmod{5}, \dots, \quad n \equiv 0 \pmod{23}
\]

This means that \( n \) must be divisible by the consecutive primes starting from 2 up to 23. Essentially, we are asked to find the least common multiple (LCM) of these primes.

The primes involved are: 2, 3, 5, 7, 11, 13, 17, 19, and 23.

Thus, the smallest possible value of \( n \) will be the least common multiple of these primes:

\[
n = \text{LCM}(2, 3, 5, 7, 11, 13, 17, 19, 23)
\]

Since these are distinct prime numbers, the LCM is simply the product of all the primes:

\[
n = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23
\]

Let me compute this for you:

\[
n = 223092870
\]

Therefore, the smallest possible value of \( n \) that satisfies all the given conditions is \( 223092870 \).

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Would you like further details or clarifications on this solution?

Here are 5 related questions you might find helpful:
1. What is the significance of the least common multiple in problems like this?
2. How do you compute the LCM for a set of numbers in general?
3. Can this type of problem be solved using other methods like the Chinese Remainder Theorem?
4. What happens if non-prime numbers are included in the conditions?
5. How would the solution change if one of the primes were excluded?

**Tip:** When solving problems involving divisibility by several primes, finding the LCM is a straightforward and efficient method to get the smallest solution.

Let n be an integer such that n ≡ 0 (mod 2), n ≡ 0 (mod 3), n ≡ 0 (mod 5), ..., n ≡ 0 (mod 23) (modulo consecutive primes). What is the smallest possible value of n satisfying the conditions above?

Learn how to find the smallest integer n such that it is divisible by all consecutive primes from 2 to 23. This involves using concepts of modular arithmetic and finding the least common multiple (LCM). Suitable for Grades 10-12.

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