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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.
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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. "

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?

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)

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!!

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!!

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.

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.

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

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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