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Math: Number Theory

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Re: Math: Number Theory [#permalink] New post 26 Jan 2010, 21:57
Thanks for sharing.
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Re: Math: Number Theory [#permalink] New post 31 Jan 2010, 09:00
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!
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Re: Math: Number Theory [#permalink] New post 31 Jan 2010, 09:19
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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
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Re: Math: Number Theory [#permalink] New post 31 Jan 2010, 10:06
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?
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Re: Math: Number Theory [#permalink] New post 31 Jan 2010, 10:34
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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.
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Re: Math: Number Theory [#permalink] New post 01 Feb 2010, 15:46
Hey Bunuel, great post so far, just wondering when the Percent notes will go up in this section.
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Re: Math: Number Theory [#permalink] New post 01 Feb 2010, 19:19
Got it, thank you both - walker and Bunuel.
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Re: Math: Number Theory [#permalink] New post 21 Feb 2010, 05:09
This is just what i've been looking for !

Thanks
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Re: Math: Number Theory [#permalink] New post 05 Mar 2010, 00:22
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Re: Math: Number Theory [#permalink] New post 05 Mar 2010, 13:06
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??
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Re: Math: Number Theory [#permalink] New post 05 Mar 2010, 13:41
Expert's post
I guess Bunuel meant Number Theory
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Re: Math: Number Theory [#permalink] New post 08 Mar 2010, 20:28
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.
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Re: Math: Number Theory [#permalink] New post 17 Mar 2010, 13:43
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?
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Re: Math: Number Theory [#permalink] New post 17 Mar 2010, 14:07
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gurpreetsingh wrote:
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?


I don't think that these theorems are needed for GMAT.
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Re: Math: Number Theory [#permalink] New post 17 Mar 2010, 14:29
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?
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Re: Math: Number Theory [#permalink] New post 01 Apr 2010, 09:03
this is a big help. thanks
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Re: Math: Number Theory [#permalink] New post 05 Apr 2010, 20:45
This is amazing...thanks for all the great work guys!!
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Re: Math: Number Theory [#permalink] New post 06 Apr 2010, 09:12
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!
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Re: Math: Number Theory [#permalink] New post 07 Apr 2010, 01:12
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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.
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Re: Math: Number Theory [#permalink] New post 30 Apr 2010, 13:19
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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, 13:19
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