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• 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!

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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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

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.

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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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|

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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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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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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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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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: \(\sqrt{2}\approx{1.41}\) \(\sqrt{3}\approx{1.73}\) \(\sqrt{5}\approx{2.24}\) \(\sqrt{6}\approx{2.45}\) \(\sqrt{7}\approx{2.65}\) \(\sqrt{8}\approx{2.83}\) \(\sqrt{10}\approx{3.16}\)

gmatclubot

Re: Math: Number Theory
[#permalink]
30 Apr 2010, 14:19

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