Math: Number Theory : GMAT Quantitative Section - Page 8
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# Math: Number Theory

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01 May 2014, 06:50
Thanks for the post!! So helpful!
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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 integer prime 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.
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24 Jul 2014, 15:03
new to here. precious compilation. Great job
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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. "
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14 Aug 2014, 19:25
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 $$a^m^n=a^{(m*n)}$$?

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15 Aug 2014, 00:44
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 $$a^m^n=a^{(m*n)}$$?

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).
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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 $$a^m^n=a^{(m*n)}$$?

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?
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17 Sep 2014, 21:09
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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)

10000x -100x = 2512.(12) - 25.(12) = 2487
9900x = 2487
x = 2487/9900

That is the logic. Hope that helps.
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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?
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20 Sep 2014, 12:31
converge wrote:
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?

Because 900 = 2^2*3^2*5^2 = 2*2*3*3*5*5.
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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?
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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?
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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.
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04 Nov 2014, 04:11
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.

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14 Nov 2014, 15:57
this is great
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06 Dec 2014, 08:23
Bunuel, Can't we also add a small point to this theory, viz. that NO Perfect Square end with the unit digits as 2,3,7,8.
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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
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15 Jan 2015, 00:08
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

For similiar questions check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.

Hope this helps.
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19 Jan 2015, 21:10
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Last edited by 10046844 on 25 Oct 2016, 10:55, edited 1 time in total.
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11 Feb 2015, 08:08
Bunuel

Finding the Number of Factors of an Integer

First make prime factorization of an integer n=a^p*b^q*c^r, where a, b, and c are prime factors of n and p, q, and r are their powers.

The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: 450=2^1*3^2*5^2

Total number of factors of 450 including 1 and 450 itself is (1+1)*(2+1)*(2+1)=2*3*3=18 factors.

On GMAT do we count -ive factors as well , in which case the total factors will be 18*2 ?

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Re: Math: Number Theory   [#permalink] 11 Feb 2015, 08:08

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