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# Math

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Math [#permalink]  18 Apr 2010, 12:40
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If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer?

[Reveal] Spoiler:
7

How to solve it? Pls. with explaination, thanks!
Current Student
Status: Nothing comes easy: neither do I want.
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Kudos [?]: 1216 [1] , given: 235

Re: Math [#permalink]  18 Apr 2010, 12:52
1
KUDOS
$$\frac{N}{18^k} = \frac{N}{2^k * 3^{2k}} = 2^{26} * \frac{3^{15} * M}{2^k * 3^{2k}}$$

take the powers of 2 and 3 from N! and leaving M as the other value.

for this to be an integer 15 < 2k
max value of k is 7
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Last edited by gurpreetsingh on 18 Apr 2010, 12:55, edited 1 time in total.
Senior Manager
Joined: 24 Jul 2009
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Re: Math [#permalink]  18 Apr 2010, 12:55
2
KUDOS
sudai wrote:
If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer?

[Reveal] Spoiler:
7

How to solve it? Pls. with explaination, thanks!

In these type of question. Proceed in the following way..
Now n= 30!
and 18 = $$3^2 * 2$$

Find the maximum power in the numerator(30!) of all the prime no. available in the denominator.

The maximum power of 2 can be determined in the following way..

30!/2 =15
30!/4 = 7
30!/8 = 3
30!/16 = 1
30!/32 = 0

so the maximum power of 2 available is 15+7+3+1= 26

The maximum power of 3 can be determined in the following way..

30!/3 = 10
30!/9 = 3
30!/27 = 1
30!/81 = 0

so the maximum power of 3 available is 10+3+1= 14

Now we have lot of 2's and we have few 3's.. and if we need to make maximum 18's, we need at least two 3's and and one 2.

so even if we have 14 threes, we can make only seven 18's.

So the maximum power (k) is 7...!!
Manager
Joined: 13 Dec 2009
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Re: Math [#permalink]  18 Apr 2010, 18:24
sudai wrote:
If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer?

[Reveal] Spoiler:
7

How to solve it? Pls. with explaination, thanks!

basically we need to find number of 18's available in product of 1 to 31 i.e. in 31!
or we need to find set of ( 1 two and 2 three ) in 31!
2's will be sufficiently available, so just need to count 3's
form 1,31 3's are available in 3,6,9,12,15,18,21,24,27,30
total 3's are 14 =>number of sets =14/2 (as each set require 2 3's) =7

hence there are 7 sets of (1 two and 2 three ) available in 31!
or in other words there are 7 18's available in 31! => k=7
Manager
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Re: Math [#permalink]  18 Apr 2010, 18:33
nverma wrote:
sudai wrote:
If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer?

[Reveal] Spoiler:
7

How to solve it? Pls. with explaination, thanks!

In these type of question. Proceed in the following way..
Now n= 30!
and 18 = $$3^2 * 2$$

Find the maximum power in the numerator(30!) of all the prime no. available in the denominator.

The maximum power of 2 can be determined in the following way..

30!/2 =15
30!/4 = 7
30!/8 = 3
30!/16 = 1
30!/32 = 0

so the maximum power of 2 available is 15+7+3+1= 26

The maximum power of 3 can be determined in the following way..

30!/3 = 10
30!/9 = 3
30!/27 = 1
30!/81 = 0

so the maximum power of 3 available is 10+3+1= 14

Now we have lot of 2's and we have few 3's.. and if we need to make maximum 18's, we need at least two 3's and and one 2.

so even if we have 14 threes, we can make only seven 18's.

So the maximum power (k) is 7...!!

vow i like the way you have done to find 2's and 3's

The maximum power of 2 can be determined in the following way..

30!/2 =15
30!/4 = 7
30!/8 = 3
30!/16 = 1
30!/32 = 0

so the maximum power of 2 available is 15+7+3+1= 26

The maximum power of 3 can be determined in the following way..

30!/3 = 10
30!/9 = 3
30!/27 = 1
30!/81 = 0

so the maximum power of 3 available is 10+3+1= 14

+1 from my side

just a quick question can we get number on any number in any factorials or it is only for 2 and 3.
thanks
Re: Math   [#permalink] 18 Apr 2010, 18:33
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# Math

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