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Re: Math: Number Theory [#permalink]
21 Jun 2014, 03:08

Hi Bunuel, You can change this section to.

Verifying the primality (checking whether the number is a prime) of a given number n can be done by trial division, that is to say dividing n by all integerprime numbers smaller than \sqrt{n}, thereby checking whether n is a multiple of m<\sqrt{n}. Example: Verifying the primality of 161: \sqrt{161} is little less than 13, from integers from 2 to 13, 161 is divisible by 7, hence 161 is not prime. _________________

Re: Math: Number Theory [#permalink]
14 Aug 2014, 16:17

Hi -

Can somebody explain where the '00' part of the 9900 in the denominator is coming from? I understand where the '99' is coming from. I copied this from the number theory section.

Thanks

" Example #2: Convert 0.2512(12) to a fraction. 1. The number consisting with non-repeating digits and repeating digits is 2512; 2. Subtract 25 (non-repeating number) from above: 2512-25=2487; 3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25): 2487/9900=829/3300. "

Re: Math: Number Theory [#permalink]
15 Aug 2014, 03:06

Bunuel wrote:

prudhviram wrote:

Bunuel,

I am confused regarding the following exponent rule in the Math Book: \(a^m^n=a^{(m^n)}\) and not \((a^m)^n\)

Isn't \(a^m^n=a^{(m*n)}\)?

Thanks in advance!

No.

\((a^m)^n=a^{mn}\)

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\) (if exponentiation is indicated by stacked symbols, the rule is to work from the top down).

I guess when the exponentiation is stacked, its not much of a confusion.

I was tripped by the following problem: If n = 10^10 and (n^n) = (10^d), what is the value of d? (I know there is a different thread for this problem, but I wanted to refer it here for the sake of clarity)

Here the exponentiation 10^10^10^10 is not explicitly stacked. How do I approach such problems?

Re: Math: Number Theory [#permalink]
17 Sep 2014, 21:09

1

This post received KUDOS

lbnyc13 wrote:

Hi -

Can somebody explain where the '00' part of the 9900 in the denominator is coming from? I understand where the '99' is coming from. I copied this from the number theory section.

Thanks

" Example #2: Convert 0.2512(12) to a fraction. 1. The number consisting with non-repeating digits and repeating digits is 2512; 2. Subtract 25 (non-repeating number) from above: 2512-25=2487; 3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25): 2487/9900=829/3300. "

let x =0.2512(12) 100x = 25.(12) 10000x = 2512.(12)

Re: Math: Number Theory [#permalink]
20 Sep 2014, 07:44

Guys can anyone explain this line in factorial factorization?

"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 the primes have to be represented twice? _________________

GMATPrep 1: 410 Q26 V20 (20 and 19 incorrect) GMATPrep 2: 620 Q44 V32 (12 and 14 incorrect) GMATPrep 3: 690 Q48 V37 (12 and 12 incorrect)

Most important lessons learned so far: 1) Your first prep is NOT indicative of your true level. 2) First 10 questions ARE important (notice that between preps 2 and 3 I had only 2 less incorrect questions on Quant, but a 70 point total score difference)

Re: Math: Number Theory [#permalink]
03 Nov 2014, 11:02

As a part of an HR initiative, each day the MD of Telair Ltd. decides to have breakfast with 4 other employees, such that each day there is a different set of 4 employees. If he has 15 employees in all, how many ways MD can have breakfast with the employees? please help me with this!!

Re: Math: Number Theory [#permalink]
03 Nov 2014, 11:03

As a part of an HR initiative, each day the MD of Telair Ltd. decides to have breakfast with 4 other employees, such that each day there is a different set of 4 employees. If he has 15 employees in all, how many ways MD can have breakfast with the employees? please help me with this!!

Re: Math: Number Theory [#permalink]
03 Nov 2014, 11:04

In how many ways can a committee of 10 be selected with at least 4 women and atleast 4 men from 7 women and 8 men if Ms X refuses to work with Mr. Y. How to approach such questions ?? please help.

Re: Math: Number Theory [#permalink]
04 Nov 2014, 04:11

Expert's post

kittyy wrote:

In how many ways can a committee of 10 be selected with at least 4 women and atleast 4 men from 7 women and 8 men if Ms X refuses to work with Mr. Y. How to approach such questions ?? please help.

Re: Math: Number Theory [#permalink]
14 Jan 2015, 17:38

Bunuel wrote:

defoue wrote:

Hi buddies, going through this awesome contribution, I got some issues to understand the following : would you please help to understand by giving some extra examples?

1. Finding the power of non-prime in n!:

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!.

2. I did not get the tips 3 or 4 regarding the pefect square whare it says a perfect square has an odd nbr of odd powers and an even nbr of even power what aboute 36 which is 2^{2} x 3^{2} I am sure I am missing something here..

Thx guys

1. It's highly unlikely that this concept will be tested in GMAT. But still:

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 \(18!=12^8*a\) --> \(x=8\).

I am still struggling with this somewhat. Could you by chance off another practice problem with explanation? Or explain the problem from the guide? I don't understand how we determine that 5 can only provide us with 6 pairs when we have 5^12

Re: Math: Number Theory [#permalink]
15 Jan 2015, 00:08

Expert's post

mmallison wrote:

Bunuel wrote:

defoue wrote:

Hi buddies, going through this awesome contribution, I got some issues to understand the following : would you please help to understand by giving some extra examples?

1. Finding the power of non-prime in n!:

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!.

2. I did not get the tips 3 or 4 regarding the pefect square whare it says a perfect square has an odd nbr of odd powers and an even nbr of even power what aboute 36 which is 2^{2} x 3^{2} I am sure I am missing something here..

Thx guys

1. It's highly unlikely that this concept will be tested in GMAT. But still:

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 \(18!=12^8*a\) --> \(x=8\).

I am still struggling with this somewhat. Could you by chance off another practice problem with explanation? Or explain the problem from the guide? I don't understand how we determine that 5 can only provide us with 6 pairs when we have 5^12

Re: Math: Number Theory [#permalink]
19 Jan 2015, 21:10

Are there practice problems related to number there in the OG for 12th edition or any edition after that? I see the practice problems after the percents, but those all relate to percent questions.

Anything questions in the OGs that anyone knows of related to number theory and these rules would be great practice. Thanks!

gmatclubot

Re: Math: Number Theory
[#permalink]
19 Jan 2015, 21:10

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