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Math: Number Theory

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Re: Math: Number Theory [#permalink] New post 27 Feb 2011, 10:55
If a is a factor of bc, and gcd(a,b)=1, then a is a factor of c.

any particular example.
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Re: Math: Number Theory [#permalink] New post 03 Mar 2011, 04:14
Nice work mates....very informative source to kickkkk start my Prep..... :-D
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Re: Math: Number Theory [#permalink] New post 06 Mar 2011, 04:50
Can someone please explain the following:

1. "Special Cases" section in "Evenly Spaced Integers"

2. Last digit of a power

Thanks.
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Re: Math: Number Theory [#permalink] New post 27 May 2011, 07:21
thank you for the great post. I currently use the GMAT Toolkit app, which I highly recommend, when can I expect this update? In addition, when will the Manhattan GMAT books be updated to the app?

Thanks,
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Re: Math: Number Theory [#permalink] New post 27 May 2011, 08:24
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Hi,

Thanks for your words! 1.6.0 update is available for download. Just get it, go to Store and you can buy any of 10 famous Manhattan GMAT books.

Let me know if you have any questions.

OrenY wrote:
thank you for the great post. I currently use the GMAT Toolkit app, which I highly recommend, when can I expect this update? In addition, when will the Manhattan GMAT books be updated to the app?

Thanks,

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Re: Math: Number Theory [#permalink] New post 17 Jul 2011, 09:18
is this always true?
The product of n consecutive integers is always divisible by n!.
Given consecutive integers: . The product of 3*4*5*6 is 360, which is divisible by 4!=24
.

for example, n=10 and the first number starts at 100000, then this rule doesn't hold.
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Re: Math: Number Theory [#permalink] New post 05 Sep 2011, 00:15
Hi GMAT CLUB,

Thank you for this super book.

I am completely following it for my exam in November.

I am going through the chapter on Number Theory, under heading "Finding the Number of Factors of an Integer".

Please can someone explain how you arrived at the following rule. Is there any proof for it.



For an integer n=a^p*b^q*c^r, where a, b, and c are prime factors of n and p, q, and r are their powers.

The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1).

NOTE: this will include 1 and n itself.

Thank You for your help.
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Re: Math: Number Theory [#permalink] New post 10 Sep 2011, 22:04
This one piece is awesome of all on the math book!
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Re: Math: Number Theory [#permalink] New post 10 Sep 2011, 22:08
gujralam wrote:
Hi GMAT CLUB,

Thank you for this super book.

I am completely following it for my exam in November.

I am going through the chapter on Number Theory, under heading "Finding the Number of Factors of an Integer".

Please can someone explain how you arrived at the following rule. Is there any proof for it.



For an integer n=a^p*b^q*c^r, where a, b, and c are prime factors of n and p, q, and r are their powers.

The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1).

NOTE: this will include 1 and n itself.

Thank You for your help.


Please search for "unique-prime-factorization theorem" on web, you should be able to get what you are looking for!
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Re: Math: Number Theory [#permalink] New post 21 Sep 2011, 01:37
Thanks a lot Bunuel.. truly awesome...
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Re: Math: Number Theory [#permalink] New post 24 Nov 2011, 17:23
This is THE BEST thing anyone has ever posted. THANK YOU SO MUCH.





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Re: Math: Number Theory [#permalink] New post 17 Dec 2011, 07:52
Hi, thanks for the great summary. BTW, do you have a list of questions (just question number) in OG12 + Quant Review 2nd edition to practice, just like the Triangles and Circle section?

Thanks again!
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Re: Math: Number Theory [#permalink] New post 06 Feb 2012, 14:55
Bunnel you simply ROCK!!!! :)
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Re: Math: Number Theory [#permalink] New post 05 Mar 2012, 00:29
Breathtaking post! (Literally!)
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Re: Math: Number Theory [#permalink] New post 06 Mar 2012, 12:00
Bunuel wrote:
NUMBER THEORY

EXPONENTS

Exponents and divisibility:
a^n-b^n is ALWAYS divisible by a-b.
a^n-b^n is divisible by a+b if n is even.
a^n + b^n is divisible by a+b if n is odd, and not divisible by a+b if n is even.



Hello, Bunuel. Great post!

Do you have an example problem in which this applies. I plugged in numbers to understand the concept I was just curious about the application and seeing this in action. Thanks.
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Re: Math: Number Theory [#permalink] New post 06 Mar 2012, 12:03
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destroyerofgmat wrote:
Bunuel wrote:
NUMBER THEORY

EXPONENTS

Exponents and divisibility:
a^n-b^n is ALWAYS divisible by a-b.
a^n-b^n is divisible by a+b if n is even.
a^n + b^n is divisible by a+b if n is odd, and not divisible by a+b if n is even.



Hello, Bunuel. Great post!

Do you have an example problem in which this applies. I plugged in numbers to understand the concept I was just curious about the application and seeing this in action. Thanks.


Check this: if-n-is-an-integer-1-is-3-n-2-n-divisible-by-84992.html
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Re: Math: Number Theory [#permalink] New post 06 Mar 2012, 13:32
Bunuel wrote:
destroyerofgmat wrote:
Bunuel wrote:
NUMBER THEORY

EXPONENTS

Exponents and divisibility:
a^n-b^n is ALWAYS divisible by a-b.
a^n-b^n is divisible by a+b if n is even.
a^n + b^n is divisible by a+b if n is odd, and not divisible by a+b if n is even.



Hello, Bunuel. Great post!

Do you have an example problem in which this applies. I plugged in numbers to understand the concept I was just curious about the application and seeing this in action. Thanks.


Check this:


Awesome! Thanks. That's definitely above my level but good practice no doubt.
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Re: Math: Number Theory [#permalink] New post 17 May 2012, 01:03
Thanks a Tonn for the detailed post and help !!!!!
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Re: Math: Number Theory [#permalink] New post 01 Jun 2012, 12:27
Thanks a lot for the detailed post ! This is sure to help with my recurrent mistakes on number properties :)
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Re: Math: Number Theory [#permalink] New post 04 Sep 2012, 18:34
Bunuel wrote:
NUMBER THEORY

Trailing zeros:
Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}, where k must be chosen such that 5^k<n.

It's easier if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!?
\frac{32}{5}+\frac{32}{5^2}=6+1=7 (denominator must be less than 32, 5^2=25 is less)

Hence, there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.




I noticed in case the number (n) is multiple of 5^k and we have to find number of trailing zero zeroes, then it will be 5^k<=n rather 5^k<n

no of trailing zeros in 25! =6

\frac{25}{5}+\frac{25}{5^2}= 5+1;
Please correct me, clarify if i'm wrong. Thanks :)
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Re: Math: Number Theory   [#permalink] 04 Sep 2012, 18:34
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