Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 19 Dec 2013, 00:43

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Math: Number Theory

Author Message
TAGS:
Math Expert
Joined: 02 Sep 2009
Posts: 15204
Followers: 2558

Kudos [?]: 15808 [0], given: 1572

Re: Math: Number Theory [#permalink]  30 Apr 2010, 13:30
Expert's post
AloneAndInsufficient wrote:
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}

Thanks. Edited. +1 for spotting this.
_________________
 Kaplan Promo Code Knewton GMAT Discount Codes Manhattan GMAT Discount Codes
Intern
Joined: 26 Jan 2010
Posts: 25
Followers: 1

Kudos [?]: 6 [0], given: 0

Re: Math: Number Theory [#permalink]  04 May 2010, 04:21
Thank you
Manager
Joined: 20 Apr 2010
Posts: 154
Location: I N D I A
Followers: 3

Kudos [?]: 14 [2] , given: 16

Re: Math: Number Theory [#permalink]  10 May 2010, 01:41
2
KUDOS
Example: A company received $2 million in royalties on the first$10 million in sales and then $8 million in royalties on the next$100 million in sales. By what percent did the ratio of royalties to sales decrease from the first $10 million in sales to the next$100 million in sales?

Solution: Percent decrease can be calculated by the formula above: Percent=\frac{Change}{Original}*100=\frac{\frac{2}{10}-\frac{10}{100}}{\frac{2}{10}}*100=50%, so the royalties decreased by 50%.

I could not get this , i think there is some error... Plzz explain..

as the same Q in Percent Part of Math book is giving an answer of 60 %..

Last edited by sag on 10 Jun 2010, 20:15, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 15204
Followers: 2558

Kudos [?]: 15808 [0], given: 1572

Re: Math: Number Theory [#permalink]  10 May 2010, 15:51
Expert's post
sag wrote:
Example: A company received $2 million in royalties on the first$10 million in sales and then $8 million in royalties on the next$100 million in sales. By what percent did the ratio of royalties to sales decrease from the first $10 million in sales to the next$100 million in sales?

Solution: Percent decrease can be calculated by the formula above: Percent=\frac{Change}{Original}*100=\frac{\frac{2}{10}-\frac{10}{100}}{\frac{2}{10}}*100=50%, so the royalties decreased by 50%.

I could not get this , i think there is some error... Plzz explain..

as the same Q in Percent Part of Math book is giving an answer of 60 %..

There was a typo. I edited it in Percent section and forgot to edit it here. Now it's OK. Thanks. +1 for spotting this.
_________________
Current Student
Joined: 09 Mar 2010
Posts: 58
Schools: HBS - Class of 2012, INSEAD,
Followers: 5

Kudos [?]: 5 [0], given: 4

Re: Math: Number Theory [#permalink]  12 May 2010, 12:53
sag wrote:
Example: A company received $2 million in royalties on the first$10 million in sales and then $8 million in royalties on the next$100 million in sales. By what percent did the ratio of royalties to sales decrease from the first $10 million in sales to the next$100 million in sales?

Solution: Percent decrease can be calculated by the formula above: Percent=\frac{Change}{Original}*100=\frac{\frac{2}{10}-\frac{10}{100}}{\frac{2}{10}}*100=50%, so the royalties decreased by 50%.

I could not get this , i think there is some error... Plzz explain..

as the same Q in Percent Part of Math book is giving an answer of 60 %..

2 million royalties on 10 million in sales is equivalent to 20 million royalties on 100 million sales (multiply both number by 10). Going down from 20 million royalties to 8 million royalties is a decrease of 60%.
Intern
Joined: 25 Apr 2010
Posts: 13
Location: Mumbai
Followers: 1

Kudos [?]: 0 [0], given: 3

Re: Math: Number Theory [#permalink]  15 May 2010, 23:24
Thanks Bunuel for all the efforts put in creating this. Really appreciate.
Intern
Joined: 14 May 2010
Posts: 5
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Math: Number Theory [#permalink]  10 Jun 2010, 12:14
If a is a factor of bc, and gcd(a,b)=1, then a is a factor of c.

Can anyone please explain this rule??? I'm not sure what it means by gcd(a,b)=1.

Thanks a bunch and great summary !!!!!
Math Expert
Joined: 02 Sep 2009
Posts: 15204
Followers: 2558

Kudos [?]: 15808 [0], given: 1572

Re: Math: Number Theory [#permalink]  10 Jun 2010, 14:00
Expert's post
bely202 wrote:
If a is a factor of bc, and gcd(a,b)=1, then a is a factor of c.

Can anyone please explain this rule??? I'm not sure what it means by gcd(a,b)=1.

Thanks a bunch and great summary !!!!!

gcd(a,b)=1 means that greatest common divisor of a and b is 1, or in other words they are co-prime, the don't share any common factor but 1. So if we are told that a is a factor of bc and a and b don't share any common factors, then it must be true that a is a factor of only c.

So if a=3, b=5 (a and b don't share any common factors but 1, gcd(a,b)=1), c=6 bc=30 --> a=3 is a factor of c=6.
_________________
Intern
Joined: 14 May 2010
Posts: 5
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Math: Number Theory [#permalink]  10 Jun 2010, 14:08
Thanks a lot of the detailed explanation !!!!
Manager
Joined: 21 Feb 2010
Posts: 214
Followers: 1

Kudos [?]: 12 [0], given: 1

Re: Math: Number Theory [#permalink]  12 Jun 2010, 18:11
thanks for sharing!!
Intern
Joined: 23 Jun 2010
Posts: 9
Followers: 0

Kudos [?]: 3 [0], given: 2

Re: Math: Number Theory [#permalink]  23 Jun 2010, 10:21
My only problem is that I was not able to spot such a good site earlier.
Great effort by Bunuel. Thanks a lot.
Intern
Joined: 28 May 2010
Posts: 37
Followers: 0

Kudos [?]: 3 [0], given: 15

Re: Math: Number Theory [#permalink]  21 Jul 2010, 17:38
Thank you so much for compiling this! I have a few questions on the reasoning behind the rules... Sorry for the very long post.

Why is:

Perfect Square
• The sum of distinct factors of a perfect square is ALWAYS ODD.
• A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors.
• Perfect square always has even number of powers of prime factors.

Consecutive Integers
• The product of n consecutive integers is always divisible by n!.
Given n=4 consecutive integers: \{3,4,5,6\}. The product of 3*4*5*6 is 360, which is divisible by 4!=24.

Evenly Spaced
• If the first term is a_1 and the common difference of successive members is d, then the n_{th} term of the sequence is given by:
a_ n=a_1+d(n-1)

Terminating Decimal
Why must the denominator be 2^n5^m?

Exponents
Why are:
a^n-b^n is ALWAYS divisible by a-b.
a^n + b^n is divisible by a+b if n is odd, and not divisible by a+b if n is even.
Intern
Joined: 28 May 2010
Posts: 37
Followers: 0

Kudos [?]: 3 [0], given: 15

Re: Math: Number Theory [#permalink]  21 Jul 2010, 19:33
I'm so sorry to make you explain them, but I have a really hard time memorizing rules that I don't understand... thanks again!
Manager
Joined: 27 Jun 2010
Posts: 55
Location: India
WE 1: 1 Year Research & Development
WE 2: 7.5 Years IT Consultant
Followers: 0

Kudos [?]: 4 [0], given: 21

Re: Math: Number Theory [#permalink]  23 Jul 2010, 05:01
Thanks for this post.
Manager
Joined: 24 Jan 2010
Posts: 167
Location: India
Schools: ISB
Followers: 2

Kudos [?]: 11 [0], given: 14

Re: Math: Number Theory [#permalink]  25 Jul 2010, 20:43
Hi Bunuel,

Thanks for this wonderful post on number properties.

I was just wondering if we can add topics related question numbers from official book as we have done for Triangles.

_________________

_________________
If you like my post, consider giving me a kudos. THANKS!

Intern
Joined: 05 Nov 2009
Posts: 32
Followers: 0

Kudos [?]: 4 [2] , given: 3

Re: Math: Number Theory [#permalink]  11 Aug 2010, 16:12
2
KUDOS
Bunuel wrote:
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?

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.

Does this relationship breakdown at some point? I thought this was great and was just experimenting and looked at 21! (calculated in excel) and it ends with 5 zeros. Using the methodology you described above it should have 4 zeros. Am I missing something or did I make a mistake somewhere?
Math Expert
Joined: 02 Sep 2009
Posts: 15204
Followers: 2558

Kudos [?]: 15808 [2] , given: 1572

Re: Math: Number Theory [#permalink]  11 Aug 2010, 17:07
2
KUDOS
Expert's post
utfan2424 wrote:
Does this relationship breakdown at some point? I thought this was great and was just experimenting and looked at 21! (calculated in excel) and it ends with 5 zeros. Using the methodology you described above it should have 4 zeros. Am I missing something or did I make a mistake somewhere?

You made everything right 21! ends with 21/5=4 zeros. It's excel: it makes rounding with such a huge numbers thus giving incorrect result.
_________________
Intern
Joined: 03 Mar 2010
Posts: 20
Followers: 0

Kudos [?]: 0 [0], given: 2

Re: Math: Number Theory [#permalink]  21 Aug 2010, 22:03
Great Collection.. kudos..
Manager
Joined: 06 Apr 2010
Posts: 58
Followers: 0

Kudos [?]: 21 [0], given: 13

Re: Math: Number Theory [#permalink]  07 Sep 2010, 20:20
_________________

If you liked my post, please consider thanking me with Kudos! I really appreciate it!

Senior Manager
Joined: 20 Jul 2010
Posts: 274
Followers: 2

Kudos [?]: 31 [0], given: 9

Re: Math: Number Theory [#permalink]  09 Sep 2010, 11:16
great compilation.......
_________________

If you like my post, consider giving me some KUDOS !!!!! Like you I need them

Re: Math: Number Theory   [#permalink] 09 Sep 2010, 11:16
Similar topics Replies Last post
Similar
Topics:
number theory 2 01 Apr 2004, 10:08
Number theory 6 06 Jul 2004, 15:44
1 NUMBER THEORY 2 02 Feb 2009, 10:38
8 Math: Number Theory (broken into smaller topics) 7 10 Mar 2010, 05:20
27 Math: Number Theory - Percents 42 22 Mar 2010, 14:24
Display posts from previous: Sort by