|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 22 Jan 2010
Posts: 125
Schools: UCLA, INSEAD
Followers: 2
Kudos [?]:
4
[0], given: 15
|
Re: Math: Number Theory [#permalink]
 26 Jan 2010, 22:57
Thanks for sharing.
|
|
|
|
|
|
|
Current Student
Joined: 12 Nov 2008
Posts: 368
Schools: Ross (R2), Cornell (R3) , UNC (R3) , INSEAD (R1 Jan)
WE 1: Advisory (2 yrs)
WE 2: FP & Analysis (2 yrs at matriculation)
Followers: 20
Kudos [?]:
79
[0], given: 45
|
Re: Math: Number Theory [#permalink]
31 Jan 2010, 10:00
Bunuel wrote: • All prime numbers except 2 and 5 end in 1, 3, 7 or 9, since numbers ending in 0, 2, 4, 6 or 8 are multiples of 2 and numbers ending in 0 or 5 are multiples of 5. Similarly, all prime numbers above 3 are of the form 6n-1 or 6n+1, because all other numbers are divisible by 2 or 3.
Awesome post, thank you so much! +1 What is the quickest way to figure out whether a number is prime? I usually check if it's odd or even, then sum its digits to figure out if it's divisible by 3, then look if it ends in 5 and if all else fails divide it by 7. Is this the recommended approach? What might be a bit confusing is that while all prime numbers are of the form 6n-1 or 6n+1, not all numbers of that form are in fact prime. I think this is crucial. For instance, the number 49 is 6n+1, but is not prime. Any insight on a quicker check (if one exists) would be much appreciated and thank you again for your efforts. They make a real difference!
|
|
|
|
|
|
CEO
Joined: 17 Nov 2007
Posts: 3594
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 231
Kudos [?]:
1300
[0], given: 346
|
Re: Math: Number Theory [#permalink]
31 Jan 2010, 10:19
ariel wrote: What is the quickest way to figure out whether a number is prime? Unfortunately, there is no such quick way to say that this number is prime. You can remember all numbers till 50 and then use rule: Rule: To check whether a number is prime or not, we try to divide it by 2, 3, 5 and so on. You can stop at \sqrt{number} - it is enough. Why? Because if there is prime divisor greater than \sqrt{number}, there must be another prime divisor lesser than \sqrt{number}. Example, n = 21 -- > \sqrt{21}~ 4-5 So, we need to check out only 2,3 because for 7, for instance, we have already checked out 3. n = 101 --> 2,3,5 is out (the last digit is not even or 5 and sum of digits is not divisible by 3). we need to check out only 7
_________________
iOS/Android: GMAT ToolKit - The bestselling GMAT prep app | GMAT Club (free) | PrepGame | GRE ToolKit | LSAT ToolKit
PROMO: Are you an exiting GMAT ToolKit (iOS) user? Get GMAT ToolKit 2 (iOS) for free* (read more)
Math: GMAT Math Book ||| General: GMATTimer ||| Chicago Booth: Slide Presentation
The People Who Are Crazy Enough to Think They Can Change the World, Are the Ones Who Do.
|
|
|
|
|
|
Current Student
Joined: 12 Nov 2008
Posts: 368
Schools: Ross (R2), Cornell (R3) , UNC (R3) , INSEAD (R1 Jan)
WE 1: Advisory (2 yrs)
WE 2: FP & Analysis (2 yrs at matriculation)
Followers: 20
Kudos [?]:
79
[0], given: 45
|
Re: Math: Number Theory [#permalink]
31 Jan 2010, 11:06
Appreciate the very prompt response, walker. To your point re divisibility by 7: I'm having a hard time proving this algebraically, is it a fair statement to say that the only non-prime numbers of the form 6n-1 and 6n+1 are the ones that are divisible by 7?
If so, a quick way to check whether a big number is prime would be to: 1) check whether it's of the form 6n-1 or 6n+1 2) check whether it's divisible by 7
Is this correct?
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11566
Followers: 1796
Kudos [?]:
9577
[0], given: 826
|
Re: Math: Number Theory [#permalink]
31 Jan 2010, 11:34
ariel wrote: Appreciate the very prompt response, walker. To your point re divisibility by 7: I'm having a hard time proving this algebraically, is it a fair statement to say that the only non-prime numbers of the form 6n-1 and 6n+1 are the ones that are divisible by 7?
If so, a quick way to check whether a big number is prime would be to: 1) check whether it's of the form 6n-1 or 6n+1 2) check whether it's divisible by 7
Is this correct? Not so. Divisibility by 7 does not check whether the number is prime or not. Actually this issue is covered in the post. First you should know that all prime numbers except 2 and 5 end in 1, 3, 7 or 9. So if it ends in some other digit it's not prime. Next, if the above didn't help (meaning that number ends in 1, 3, 7 or 9) there is a way to check whether the number is prime or not. Walker gave an example how to do this, but here it is again: Verifying the primality of a given number n can be done by trial division, that is to say dividing n by all integer numbers smaller than \sqrt{n}, thereby checking whether n is a multiple of m<\sqrt{n}. Examples: Verifying the primality of 161: \sqrt{161} is little less than 13. We should check 161 on divisibility by numbers from 2 to 13. From integers from 2 to 13, 161 is divisible by 7, hence 161 is not prime. Verifying the primality of 149: \sqrt{149} is little more than 12. We should check 149 on divisibility by numbers from 2 to 12, inclusive. 149 is not divisible by any of the integers from 2 to 12, hence 149 is prime. Verifying the primality of 73: \sqrt{73} is little less than 9. We should check 73 on divisibility by numbers from 2 to 9. 73 is not divisible by any of the integers from 2 to 9, hence 149 is prime. Hope it helps.
_________________
PLEASE READ AND FOLLOW: 11 Rules for Posting!!!
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
What are GMAT Club Tests?
25 extra-hard Quant Tests
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Intern
Joined: 06 Oct 2009
Posts: 36
Schools: Ryerson University
Followers: 1
Kudos [?]:
7
[0], given: 7
|
Re: Math: Number Theory [#permalink]
01 Feb 2010, 16:46
Hey Bunuel, great post so far, just wondering when the Percent notes will go up in this section.
|
|
|
|
|
|
Current Student
Joined: 12 Nov 2008
Posts: 368
Schools: Ross (R2), Cornell (R3) , UNC (R3) , INSEAD (R1 Jan)
WE 1: Advisory (2 yrs)
WE 2: FP & Analysis (2 yrs at matriculation)
Followers: 20
Kudos [?]:
79
[0], given: 45
|
Re: Math: Number Theory [#permalink]
01 Feb 2010, 20:19
Got it, thank you both - walker and Bunuel.
|
|
|
|
|
|
Manager
Joined: 17 Aug 2009
Posts: 222
Followers: 3
Kudos [?]:
14
[0], given: 18
|
Re: Math: Number Theory [#permalink]
21 Feb 2010, 06:09
This is just what i've been looking for !
Thanks
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11566
Followers: 1796
Kudos [?]:
9577
[1] , given: 826
|
Re: Math: Number Theory [#permalink]
05 Mar 2010, 01:22
1
This post received
KUDOS
|
|
|
|
|
|
Senior Manager
Joined: 22 Dec 2009
Posts: 368
Followers: 9
Kudos [?]:
136
[0], given: 47
|
Re: Math: Number Theory [#permalink]
05 Mar 2010, 14:06
Bunuel wrote: The topic is done. At last!
I'll break it into several smaller ones in a day or two.
Any comments, advises and/or corrections are highly appreciated. What Topic are we talking abt??
_________________
Cheers!
JT...........
If u like my post..... payback in Kudos!! 
|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|
~~Better Burn Out... Than Fade Away~~
|
|
|
|
|
|
CEO
Joined: 17 Nov 2007
Posts: 3594
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 231
Kudos [?]:
1300
[0], given: 346
|
Re: Math: Number Theory [#permalink]
05 Mar 2010, 14:41
I guess Bunuel meant Number Theory
_________________
iOS/Android: GMAT ToolKit - The bestselling GMAT prep app | GMAT Club (free) | PrepGame | GRE ToolKit | LSAT ToolKit
PROMO: Are you an exiting GMAT ToolKit (iOS) user? Get GMAT ToolKit 2 (iOS) for free* (read more)
Math: GMAT Math Book ||| General: GMATTimer ||| Chicago Booth: Slide Presentation
The People Who Are Crazy Enough to Think They Can Change the World, Are the Ones Who Do.
|
|
|
|
|
|
Intern
Joined: 06 Mar 2010
Posts: 4
Followers: 0
Kudos [?]:
0
[0], given: 1
|
Re: Math: Number Theory [#permalink]
08 Mar 2010, 21:28
Maybe a suggestion for the reciprocal section, this is a question I got tricked on in an early GMAT quant review-
In which of the following pairs are the two numbers reciprocals of each other?
i. 3 and 1/3
ii. 1/17 and -1/17
iii. sqrt3 and sqrt3/3
a) i only
b) ii only
c) i and ii
d) i and iii
e) ii and iii
OA is D.
|
|
|
|
|
|
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2758
Location: Malaysia
Concentration: Marketing, Entrepreneurship
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 123
Kudos [?]:
634
[0], given: 221
|
Re: Math: Number Theory [#permalink]
17 Mar 2010, 14:43
Hi Bunnel, I m confused about the extent of level for number properties.. do we have to remmeber eculer's, fermat's,wilson's theorem on prime number. Actually I found their application to be quite useful but m not sure whther there are other ways to solve the questions as well. eg difficult remainder questions and questions on HCF like if HCF of 2 numbers is 13 and their sum is 2080, how many such pairs are possible? do we see such questions on gmat?
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight
Money Saved is the Money Earned 
Jo Bole So Nihaal , Sat Shri Akaal
 Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
Find out what's new at GMAT Club - latest features and updates
Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11566
Followers: 1796
Kudos [?]:
9577
[0], given: 826
|
Re: Math: Number Theory [#permalink]
17 Mar 2010, 15:07
|
|
|
|
|
|
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2758
Location: Malaysia
Concentration: Marketing, Entrepreneurship
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 123
Kudos [?]:
634
[0], given: 221
|
Re: Math: Number Theory [#permalink]
17 Mar 2010, 15:29
So is there any way we can solve the above HCF question? Also does the number theory stated here is sufficient to cover the concepts asked?
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
Find out what's new at GMAT Club - latest features and updates
Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html
|
|
|
|
|
|
Intern
Joined: 29 Mar 2010
Posts: 19
Followers: 0
Kudos [?]:
2
[0], given: 0
|
Re: Math: Number Theory [#permalink]
01 Apr 2010, 10:03
this is a big help. thanks
|
|
|
|
|
|
Manager
Joined: 24 Mar 2010
Posts: 102
Followers: 1
Kudos [?]:
29
[0], given: 11
|
Re: Math: Number Theory [#permalink]
05 Apr 2010, 21:45
This is amazing...thanks for all the great work guys!!
_________________
Please do consider giving kudos if you like my posts
|
|
|
|
|
|
Intern
Joined: 21 Feb 2010
Posts: 33
Location: Ukraine
Followers: 1
Kudos [?]:
1
[0], given: 9
|
Re: Math: Number Theory [#permalink]
06 Apr 2010, 10:12
Thanks! It was very very helpful! Kudos!
But I have a question:
How many powers of 900 are in 50!
Make the prime factorization of the number: 900=2^2*3^2*5^2, then find the powers of these prime numbers in the n!.
Find the power of 2:
\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47
= 2^{47}
Find the power of 3:
\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22
=3^{22}
Find the power of 5:
\frac{50}{5}+\frac{50}{25}=10+2=12
=5^{12}
We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!.
Why do we take just 5 from {2,3,5} and why do we need divide 12 by 2 to get the result?
Thanks in advance!
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11566
Followers: 1796
Kudos [?]:
9577
[5] , given: 826
|
Re: Math: Number Theory [#permalink]
07 Apr 2010, 02:12
5
This post received
KUDOS
fruit wrote: Thanks! It was very very helpful! Kudos!
But I have a question:
How many powers of 900 are in 50!
Make the prime factorization of the number: 900=2^2*3^2*5^2, then find the powers of these prime numbers in the n!.
Find the power of 2:
\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47
= 2^{47}
Find the power of 3:
\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22
=3^{22}
Find the power of 5:
\frac{50}{5}+\frac{50}{25}=10+2=12
=5^{12}
We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!.
Why do we take just 5 from {2,3,5} and why do we need divide 12 by 2 to get the result?
Thanks in advance! 50!=900^xa=(2^2*3^2*5^2)^x*a, where x is the highest possible value of 900 and a is the product of other multiples of 50!. 50!=2^{47}*3^{22}*5^{12}*b=(2^2*3^2*5^2)^6*(2^{35}*3^{10})*b=900^{6}*(2^{35}*3^{10})*b, where b is the product of other multiples of 50!. So x=6. Below is another example: Suppose we have the number 18! and we are asked to to determine the power of 12 in this number. Which means to determine the highest value of x in 18!=12^x*a, where a is the product of other multiples of 18!. 12=2^2*3, so we should calculate how many 2-s and 3-s are in 18!. Calculating 2-s: \frac{18}{2}+\frac{18}{2^2}+\frac{18}{2^3}+\frac{18}{2^4}=9+4+2+1=16. So the power of 2 (the highest power) in prime factorization of 18! is 16. Calculating 3-s: \frac{18}{3}+\frac{18}{3^2}=6+2=8. So the power of 3 (the highest power) in prime factorization of 18! is 8. Now as 12=2^2*3 we need twice as many 2-s as 3-s. 18!=2^{16}*3^8*a=(2^2)^8*3^8*a=(2^2*3)^8*a=12^8*a. So 18!=12^8*a --> x=8.
_________________
PLEASE READ AND FOLLOW: 11 Rules for Posting!!!
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
What are GMAT Club Tests?
25 extra-hard Quant Tests
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Intern
Joined: 30 Apr 2010
Posts: 28
Followers: 2
Kudos [?]:
12
[2] , given: 1
|
Re: Math: Number Theory [#permalink]
30 Apr 2010, 14:19
2
This post received
KUDOS
Bunuel wrote: NUMBER THEORY
• For GMAT it's good to memorize following values:
\sqrt{2}\approx{1.41}
\sqrt{3}\approx{1.73}
\sqrt{5}\approx{2.24}
\sqrt{7}\approx{2.45}
\sqrt{8}\approx{2.65}
\sqrt{10}\approx{2.83}
Anyone else notice that these are wrong? They should be: • For GMAT it's good to memorize following values: \sqrt{2}\approx{1.41}\sqrt{3}\approx{1.73}\sqrt{5}\approx{2.24}\sqrt{6}\approx{2.45}\sqrt{7}\approx{2.65}\sqrt{8}\approx{2.83}\sqrt{10}\approx{3.16}
|
|
|
|
|
|
|
Re: Math: Number Theory
[#permalink]
30 Apr 2010, 14:19
|
|
|
|
|
|
|
|
|
|
|
close
