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Math: Number Theory

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Re: Math: Number Theory [#permalink]  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]  31 Jan 2010, 10:34
Expert's post
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]  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]  01 Feb 2010, 19:19
Got it, thank you both - walker and Bunuel.
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Re: Math: Number Theory [#permalink]  21 Feb 2010, 05:09
This is just what i've been looking for !

Thanks
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Re: Math: Number Theory [#permalink]  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]  05 Mar 2010, 13:41
Expert's post
I guess Bunuel meant Number Theory
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Re: Math: Number Theory [#permalink]  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]  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]  17 Mar 2010, 14:07
Expert's post
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]  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]  01 Apr 2010, 09:03
this is a big help. thanks
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Re: Math: Number Theory [#permalink]  05 Apr 2010, 20:45
This is amazing...thanks for all the great work guys!!
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Re: Math: Number Theory [#permalink]  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?

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Re: Math: Number Theory [#permalink]  04 May 2010, 04:21
Thank you
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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.
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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%.
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Re: Math: Number Theory [#permalink]  15 May 2010, 23:24
Thanks Bunuel for all the efforts put in creating this. Really appreciate.
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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 !!!!!
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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 $$c=6$$.
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Re: Math: Number Theory   [#permalink] 10 Jun 2010, 14:00

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