Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2601:30 AM EDT
-02:30 AM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more.
Apr 2512:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
18 Nov 2018, 14:41
Kudos
Bookmarks
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
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
04 Oct 2022, 13:09
Kudos
Bookmarks
Automated notice from GMAT Club BumpBot:
A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.
This post was generated automatically.
A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.
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!
