What is the least positive integer n such that n! is divisible by 1000?
AAAaannndd the bot gets it wrong yet again!
but 5! = 120, which is not divisible by 1000
To get 3 factors of 5, you will need to go up to 15, since 5 is a factor of 5,10,15.
15! = 1307674368000
You are correct, my previous response was incorrect. I apologize for the error.
To find the least positive integer $n$ such that $n!$ is divisible by 1000, we need to find the smallest value of $n$ for which we have at least three factors of 5 in the prime factorization of $n!$. As mentioned, 1000 is equal to $2^3 \cdot 5^3$.
To count the number of factors of 5, we can use the formula for the exponent of a prime in the prime factorization of $n!$, which is given by the equation $\left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \ldots$, where $\left\lfloor x \right\rfloor$ denotes the greatest integer less than or equal to $x$.
For 5, we have $\left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{5^2} \right\rfloor + \left\lfloor \frac{n}{5^3} \right\rfloor \geq 3$.
Plugging in values of $n$ starting from 15, we can check which one satisfies this condition:
For $n = 15$, we have $\left\lfloor \frac{15}{5} \right\rfloor + \left\lfloor \frac{15}{5^2} \right\rfloor + \left\lfloor \frac{15}{5^3} \right\rfloor = 3 + 0 + 0 = 3$.
Therefore, the least positive integer $n$ such that $n!$ is divisible by 1000 is $\boxed{15}$.
To find the least positive integer n such that n! is divisible by 1000, we need to determine the prime factorization of 1000.
The prime factorization of 1000 is:
1000 = 2^3 × 5^3
To have n! divisible by 1000, it needs to have at least three factors of 2 and three factors of 5.
Since n! is the product of all positive integers from 1 to n, we can count the number of factors of 2 and 5 in the multiples of those numbers.
For factors of 2:
- The multiples of 2 contribute one factor of 2 each.
- The multiples of 4 contribute an additional factor of 2 each.
- The multiples of 8 contribute yet another factor of 2 each.
Similarly, for factors of 5:
- The multiples of 5 contribute one factor of 5 each.
- The multiples of 25 contribute an additional factor of 5 each.
Based on this information, we need to find the least n such that there are at least three factors of 2 and three factors of 5 in the prime factorization of n!.
We can make a table to calculate the factors of 2 and 5:
```
n | Factors of 2 | Factors of 5
----|--------------|--------------
1 | 0 | 0
2 | 1 | 0
3 | 1 | 0
4 | 2 | 0
5 | 2 | 1
6 | 3 | 1
7 | 3 | 1
8 | 4 | 1
9 | 4 | 1
10 | 5 | 2
11 | 5 | 2
12 | 6 | 2
13 | 6 | 2
14 | 7 | 2
15 | 7 | 3
16 | 8 | 3
17 | 8 | 3
18 | 9 | 3
19 | 9 | 3
20 | 10 | 4
```
From the table, we can see that the least positive integer n such that n! is divisible by 1000 is 10.
To determine the least positive integer n such that n! (n factorial) is divisible by 1000, we need to find the prime factorization of 1000.
The prime factorization of 1000 can be expressed as 2^3 * 5^3. In other words, 1000 = 2 * 2 * 2 * 5 * 5 * 5.
Since 1000 = 2^3 * 5^3, in order for n! to be divisible by 1000, we need n! to have at least three factors of 2 and three factors of 5.
Now, let's consider the factors of 2 and 5 in the factorial of any positive integer n. Since there are significantly more factors of 2 than 5 in any factorial, it is sufficient to consider the factors of 5.
To determine the least positive integer n such that n! has at least three factors of 5, we can start by checking small values of n.
For n=1: 1! = 1 (No factors of 5)
For n=2: 2! = 2 (No factors of 5)
For n=3: 3! = 6 (No factors of 5)
For n=4: 4! = 24 (No factors of 5)
For n=5: 5! = 120 (One factor of 5)
For n=6: 6! = 720 (One factor of 5)
At n=6, the factorial reaches 720 which has one factor of 5.
Now, to find the smallest value of n such that n! is divisible by 1000, we need the number of factors of 5 to be at least three. So, we need to check further.
For n=7: 7! = 5040 (Two factors of 5)
For n=8: 8! = 40320 (Two factors of 5)
For n=9: 9! = 362880 (Two factors of 5)
For n=10: 10! = 3628800 (Two factors of 5)
At n=10, the factorial reaches 3,628,800 which still has only two factors of 5. Hence, n=10 is not the answer.
To find the least positive integer n such that n! is divisible by 1000, we continue checking larger values of n.
For n=11: 11! = 39,916,800 (Two factors of 5)
For n=12: 12! = 479,001,600 (Two factors of 5)
For n=13: 13! = 6,227,020,800 (Three factors of 5)
At n=13, the factorial reaches 6,227,020,800 which has three factors of 5.
Therefore, the least positive integer n such that n! is divisible by 1000 is n=13.
We can consider the prime factorization of 1000: $$1000=10^3=(2^3)(5^3).$$ To get a $2^3$, we need a 2 from one of the integers in the factorial, because there are not three factors of 2 within 1 through $n$ for any positive integer $n$.
To get three 5s, we need 5 from the integer in the factorial. We should not consider any other prime factors of the integers from 1 through $n$ because 1000 will always have more factors of 2 and 5 than any other prime number.
Thus $n!$ is divisible by 1000 when $\boxed{n=5}$ because we can use 5 as the integer.