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# Math: Number Theory

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31 Jan 2010, 10:00
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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31 Jan 2010, 10:19
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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31 Jan 2010, 11:06
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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31 Jan 2010, 11:34
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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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01 Feb 2010, 16:46
Hey Bunuel, great post so far, just wondering when the Percent notes will go up in this section.

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01 Feb 2010, 20:19
Got it, thank you both - walker and Bunuel.

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21 Feb 2010, 06:09
This is just what i've been looking for !

Thanks

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05 Mar 2010, 01:22
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The topic is done. At last!

I'll break it into several smaller ones in a day or two.

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05 Mar 2010, 14:06
Bunuel wrote:
The topic is done. At last!

I'll break it into several smaller ones in a day or two.

What Topic are we talking abt??
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05 Mar 2010, 14:41
I guess Bunuel meant Number Theory
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08 Mar 2010, 21:28
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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17 Mar 2010, 14:43
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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17 Mar 2010, 15:07
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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17 Mar 2010, 15:29
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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01 Apr 2010, 10:03
this is a big help. thanks

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05 Apr 2010, 21:45
This is amazing...thanks for all the great work guys!!
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06 Apr 2010, 10:12
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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07 Apr 2010, 02:12
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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?

$$50!=900^xa=(2^2*3^2*5^2)^x*a$$, 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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30 Apr 2010, 14:19
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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:
$$\sqrt{2}\approx{1.41}$$
$$\sqrt{3}\approx{1.73}$$
$$\sqrt{5}\approx{2.24}$$
$$\sqrt{6}\approx{2.45}$$
$$\sqrt{7}\approx{2.65}$$
$$\sqrt{8}\approx{2.83}$$
$$\sqrt{10}\approx{3.16}$$

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30 Apr 2010, 14:30
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AloneAndInsufficient wrote:
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:
$$\sqrt{2}\approx{1.41}$$
$$\sqrt{3}\approx{1.73}$$
$$\sqrt{5}\approx{2.24}$$
$$\sqrt{6}\approx{2.45}$$
$$\sqrt{7}\approx{2.65}$$
$$\sqrt{8}\approx{2.83}$$
$$\sqrt{10}\approx{3.16}$$

Thanks. Edited. +1 for spotting this.
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Re: Math: Number Theory   [#permalink] 30 Apr 2010, 14:30

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