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# Math: Number Theory

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Intern
Joined: 12 Sep 2017
Posts: 5

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22 Sep 2017, 04:11
Thank you Bunuel and everyone for their contribution here
Intern
Joined: 01 Apr 2017
Posts: 6
GMAT 1: 720 Q48 V40

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14 Oct 2017, 03:51
Nothing is better than this
Intern
Status: hope for the best, prepare for the worst
Joined: 13 Dec 2015
Posts: 6
Location: Pakistan
GPA: 3.19

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26 Dec 2017, 14:39
How many powers of 900 are in 50!? the statement written in following question wa
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!
what does it mean?
Math Expert
Joined: 02 Sep 2009
Posts: 43896

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26 Dec 2017, 20:11
abbas57 wrote:
How many powers of 900 are in 50!? the statement written in following question wa
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!
what does it mean?

I tried to elaborate this in the posts below:
https://gmatclub.com/forum/math-number- ... ml#p678298
https://gmatclub.com/forum/math-number- ... ml#p710819
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Intern
Joined: 14 Jul 2015
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28 Dec 2017, 12:13
Finding the power of non-prime in n!:

How many powers of 900 are in 50!

Make the prime factorization of the number: 900=22∗32∗52900=22∗32∗52, then find the powers of these prime numbers in the n!.

Find the power of 2:
502+504+508+5016+5032502+504+508+5016+5032=25+12+6+3+1=47=25+12+6+3+1=47

= 247247

Find the power of 3:
503+509+5027=16+5+1=22503+509+5027=16+5+1=22

=322322

Find the power of 5:
505+5025=10+2=12505+5025=10+2=12

=512512

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!.

Hi Bunuel,
I read the explanation you gave to defoue but couldn't apply it to the above example of 900. Could you please elaborate?

Thanks.
Intern
Joined: 14 Jul 2015
Posts: 4

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28 Dec 2017, 12:18
I just read the remaining explanations...got it! Sorry
Intern
Joined: 05 Jan 2018
Posts: 1

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05 Jan 2018, 05:18
A die is rolled 14 times.
a) What is the probability that we obtain exactly one 6?
b) What is the probability that we obtain 4 times 4, 2 times 6 and 8 times 3?
Intern
Joined: 18 May 2016
Posts: 12

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22 Feb 2018, 20:16
Bunuel

Excellent Post!

Edit suggestions:
1. If p is a prime number and p is a factor of ab, then p is a factor of a or p is a factor of b.

If p = a = b = 2, the above point fails
as p will be a factor of ab and a and b

2. Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.

I think you should change the underlined "number" to digits for better understanding
and then provide an example of the working
eg. 203/7
Last digit -> 3
Double it -> 3*2 = 6
Rest of the digits = 20 -> 20-6 = 14
Answer (14) -> Divisible by 7 -> Yes

3. Verifying the primality (checking whether the number is a prime) of a given number n can be done by trial division ...

Wouldn't it be easier & faster to just use another method that you've mentioned earlier in the post?
all prime numbers above 3 are of the form 6n−1 or 6n+1
-> Divide a number by 6 and +1/-1 to check if it is a prime

Thanks
P
Math Expert
Joined: 02 Sep 2009
Posts: 43896

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22 Feb 2018, 21:30
1
KUDOS
Expert's post
ppnimkar wrote:
Bunuel

Excellent Post!

Edit suggestions:
1. If p is a prime number and p is a factor of ab, then p is a factor of a or p is a factor of b.

If p = a = b = 2, the above point fails
as p will be a factor of ab and a and b

2. Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.

I think you should change the underlined "number" to digits for better understanding
and then provide an example of the working
eg. 203/7
Last digit -> 3
Double it -> 3*2 = 6
Rest of the digits = 20 -> 20-6 = 14
Answer (14) -> Divisible by 7 -> Yes

3. Verifying the primality (checking whether the number is a prime) of a given number n can be done by trial division ...

Wouldn't it be easier & faster to just use another method that you've mentioned earlier in the post?
all prime numbers above 3 are of the form 6n−1 or 6n+1
-> Divide a number by 6 and +1/-1 to check if it is a prime

Thanks
P

1. x or y means x or y or both. So, everything is correct there.

2. I don't think it would be better.

3. Any prime number p, which is greater than 3, could be expressed as $$p=6n+1$$ or $$p=6n+5$$ or $$p=6n-1$$, where n is an integer greater than 1.

Any prime number p, which is greater than 3, when divided by 6 can only give the remainder of 1 or 5 (remainder cannot be 2 or 4 as in this case p would be even and the remainder cannot be 3 as in this case p would be divisible by 3).

So, any prime number p, which is greater than 3, could be expressed as $$p=6n+1$$ or $$p=6n+5$$ or $$p=6n-1$$, where n is an integer greater than 1.

But:
Not all number which yield a remainder of 1 or 5 upon division by 6 are primes, so vise-versa of the above property is not true. For example 25 yields the remainder of 1 upon division be 6 and it's not a prime number.

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Re: Math: Number Theory   [#permalink] 22 Feb 2018, 21:30

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