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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

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.

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?? _________________

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|

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? _________________

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. _________________

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? _________________

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. _________________

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%.

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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