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.
What is the least positive integer n such that n! is divisible by 1000?
3 answers
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
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 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 $\boxed{15}$.