Power of prime in prime factorization of a factorial.

Question

Please advise on how to arrive at solution for determining the power of 17 in the prime factorization of 2890!

Also, is there a short-cut?

So far I know: Prime factorization of 2890 = 2 x 5 x 17^2

Thank you!

Answer 1

Hint: use your factorization to answer the following questions.

How many multiples of $17$ are there in the numbers up to $2890$ (which are multiplied to get $2890!$) ?

How many multiples of $17^2$ are there? (Each of these has an additional factor of $17$.)

Are there any multiples of $17^n$ for $n\ge3$?

Answer 2

What I would do is think about all of the numbers between 1 and 2890 that are divisible by 17. Hence, those numbers would have 17 as a prime factor of themselves. Next, I would think about the numbers that could be represented as 17^2 x 2 x 5 or smaller while still having 17^2 as a prime factor. There are only 10 total numbers like that. Now you take the total number of numbers divisible by 17, subtract 10 from that,and add that number to 2 times the number of numbers divisible by 17^2 and that should be the power of 17 in the prime factorization for 2890!. I am not sure if this is correct and this is the first answer I have every answered but I hope it helps!

Answer 3

Between $1$ and $2890$, there are $170$ multiples of $17$, each of which contributes at least $1$ each to the exponent. Therefore, in $2890! = 17^x m^y$ (where $m$ is coprime to $17$, and we're not concerned with the value of $y$), it must be the case that $x \geq 170$.

Notice that in your factorization of $2890$ you have $17^2$, which is $289$. And since $17^3 > 2890$, each multiple of $17$ can't contribute more than $2$ to the exponent. Now, how many multiples of $289$ are there between $1$ and $2890$? Each of those contributes $2$ to the exponent, but we have already counted $1$ each in $x \geq 170$. So now we only need to add $1$ for each multiple of $289$.

If you have any doubt, go to Wolfram Alpha.