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18 Nov 2018, 14:41
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This is my very first post, I looked around with the search function and I found many results that didn't clarify very well the concept to me.
Finding the number of powers of non-prime number \(q\) in \(n!\)
So, this is the process:
1. Prime factorization of the non-prime number.
2. Use this formula to find the number of powers for each prime factor in n!
\(\frac{n}{{p^1}}+\frac{{n}}{{p^2}}+\frac{{n}}{{p^3}}+....+\frac{{n}}{{p^x}}\) such that \(p^x <n\), where \(p\) is the prime factor.
After this point I saw many diverging suggestions, "grab the exponent of the highest prime factor, that's your answer" was the one that prompted me to write this post.
It's not as mechanical as it seems for every case.
Your "goal" is to find an item in the form of \((p_1*p_2*p_3)^x\) that multiplies an integer \(m\) and gives back \(n!\), where \(p_1*p_2*p_3 = q\), your original non-prime number.
The above generalisation fails in many cases.
Example:
How many powers / What's the highest power of 900 in 50! ? If in the multiple choices there were both 12 and 6 with the information I found in many topics, many would have answered 12.
1. Prime factorization of \(900 = 2^2*3^2*5^2\)
2.
How many powers of 2 in 50! :
\(\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32} = 25 + 12 + 6 + 3 +1 = 47\)
How many powers of 3 in 50! :
\(\frac{50}{3}+\frac{50}{9}+\frac{50}{27} = 16 + 5 + 1 = 22\)
How many powers of 5 in 50! :
\(\frac{50}{5}+\frac{50}{25} = 10 + 2 = 12\)
Now we have:
\(50! = (2^{47} * 3^{22} * 5^{12}) * m\)
Do I pick 12 as answer? Absolutely no. Remember that the goal is top find \(x\) in here \(50! = 900^x * m\)
So we have to tweak a little our \((2^{47} * 3^{22} * 5^{12})\) so that it becomes \(900^x\) which is \((2^2*3^2*5^2)^x\)
If all exponents of the prime factors are equal, pick the exponent of the highest prime number in the prime factorization, you want to keep that.
So now we have \((2^{47} * 3^{22} * 5^{2})^x\), what's the x we're looking for? \(2 * x = 12\), \(x=6\).
And 6 is the final answer for the example, however you want to arrive at \((2^{2} * 3^{2} * 5^{2})^6 * something = 50!\) if the question is about that "something".
So we have to toy with exponents: in order to get \(2^2\) inside we have: \((2^2)^6 = 2^{12}\), how many more to reach \(2^{47}\)? \(2^{35}\)
In order to get \(3^2\) inside: \((3^2)^6 = 3^{12}\), how many more to reach \(3^{22}\)? \(3^{10}\)
So that \(something\) will be made of \(2^{35} * 3^{10} * m\) and our final result will be:
\((2^{2} * 3^{2} * 5^{2})^6 * 2^{35} * 3^{10} * m = 50!\)
Hopefully I didn't make any silly mistakes and the post is correct + helpful to understand better the particular cases.
Finding the number of powers of non-prime number \(q\) in \(n!\)
So, this is the process:
1. Prime factorization of the non-prime number.
2. Use this formula to find the number of powers for each prime factor in n!
\(\frac{n}{{p^1}}+\frac{{n}}{{p^2}}+\frac{{n}}{{p^3}}+....+\frac{{n}}{{p^x}}\) such that \(p^x <n\), where \(p\) is the prime factor.
After this point I saw many diverging suggestions, "grab the exponent of the highest prime factor, that's your answer" was the one that prompted me to write this post.
It's not as mechanical as it seems for every case.
Your "goal" is to find an item in the form of \((p_1*p_2*p_3)^x\) that multiplies an integer \(m\) and gives back \(n!\), where \(p_1*p_2*p_3 = q\), your original non-prime number.
The above generalisation fails in many cases.
Example:
How many powers / What's the highest power of 900 in 50! ? If in the multiple choices there were both 12 and 6 with the information I found in many topics, many would have answered 12.
1. Prime factorization of \(900 = 2^2*3^2*5^2\)
2.
How many powers of 2 in 50! :
\(\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32} = 25 + 12 + 6 + 3 +1 = 47\)
How many powers of 3 in 50! :
\(\frac{50}{3}+\frac{50}{9}+\frac{50}{27} = 16 + 5 + 1 = 22\)
How many powers of 5 in 50! :
\(\frac{50}{5}+\frac{50}{25} = 10 + 2 = 12\)
Now we have:
\(50! = (2^{47} * 3^{22} * 5^{12}) * m\)
Do I pick 12 as answer? Absolutely no. Remember that the goal is top find \(x\) in here \(50! = 900^x * m\)
So we have to tweak a little our \((2^{47} * 3^{22} * 5^{12})\) so that it becomes \(900^x\) which is \((2^2*3^2*5^2)^x\)
If all exponents of the prime factors are equal, pick the exponent of the highest prime number in the prime factorization, you want to keep that.
So now we have \((2^{47} * 3^{22} * 5^{2})^x\), what's the x we're looking for? \(2 * x = 12\), \(x=6\).
And 6 is the final answer for the example, however you want to arrive at \((2^{2} * 3^{2} * 5^{2})^6 * something = 50!\) if the question is about that "something".
So we have to toy with exponents: in order to get \(2^2\) inside we have: \((2^2)^6 = 2^{12}\), how many more to reach \(2^{47}\)? \(2^{35}\)
In order to get \(3^2\) inside: \((3^2)^6 = 3^{12}\), how many more to reach \(3^{22}\)? \(3^{10}\)
So that \(something\) will be made of \(2^{35} * 3^{10} * m\) and our final result will be:
\((2^{2} * 3^{2} * 5^{2})^6 * 2^{35} * 3^{10} * m = 50!\)
Hopefully I didn't make any silly mistakes and the post is correct + helpful to understand better the particular cases.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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04 Oct 2022, 13:09
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This post was generated automatically.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
