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What is the largest power of 55 that can divide 274!

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Director
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What is the largest power of 55 that can divide 274! [#permalink] New post 15 Sep 2006, 03:28
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What is the largest power of 55 that can divide 274! ?

Last edited by Futuristic on 15 Sep 2006, 06:02, edited 1 time in total.
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 [#permalink] New post 15 Sep 2006, 05:11
The largst power of 55 that devides 274! is that of 11 ...am i right ?

still i cant approach the problem but am i going the right direction or not?
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 [#permalink] New post 15 Sep 2006, 05:30
hmmm.. I seem to be missing something.

Given

1x2x3.... 274 = 55^k x m

where m is some integer.

As 55 = 11 x 5

The maximum value of k can be the maximum power of 11 (as 11 > 5).

Therefore # of multiples of 11 from 1 to 274
= (264-11)/11 +1 = 24

11 x 22 x 33 x 44... x 264
= 11^24 x 2 x 3 x 4 x .. 24
= 11^24 x 11^2 x K (as there are two more multiples of 11 from 2..24)

=11^26.

This should be 26.. but all options are three digits.

Not sure...

Help futuristic!!! :(
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 [#permalink] New post 15 Sep 2006, 06:05
Sorry guys, I sincerely apologize....got 2 questions mixed up in my sleep. Should not post at 4am in the morning....haas your answer is correct:)
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 [#permalink] New post 15 Sep 2006, 06:12
Thanks for clarifying Futuristic..all of us have done it at one time or another.

Get some sleep though :sleep :sleep :sleep .. :P

Futuristic wrote:
Sorry guys, I sincerely apologize....got
2 questions mixed up in my sleep. Should not post at 4am in the morning....haas your answer is correct:)
:wall
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 [#permalink] New post 15 Sep 2006, 20:27
haas_mba07 wrote:
hmmm.. I seem to be missing something.

Given

1x2x3.... 274 = 55^k x m

where m is some integer.

As 55 = 11 x 5

The maximum value of k can be the maximum power of 11 (as 11 > 5).

Therefore # of multiples of 11 from 1 to 274
= (264-11)/11 +1 = 24

11 x 22 x 33 x 44... x 264
= 11^24 x 2 x 3 x 4 x .. 24
= 11^24 x 11^2 x K (as there are two more multiples of 11 from 2..24)

=11^26.

This should be 26.. but all options are three digits.

Not sure...

Help futuristic!!! :(


As unrelenting as you can be...Just awesome
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 [#permalink] New post 16 Sep 2006, 00:45
Folks, I'll tell u how to calculate highest power of a number in any factorial.

Taking the above example 55 = 11x5
Since the highes power of 11 in any factorial will be less than the highest power of 5 it is just enough to calculate highest power of 11 in 274!

I think someone has already suggested a method to calculate highest power.

Let's look at one more way (The logic is same as mentioned earlier)

274/11 = 24 (don't worry about the remainder)
24/11 = 2
Add up now 24+2 = 26 will be the highest power of 11 in 274!

Let's look at one more example.
Highest power of 5 in 350!
350/5 = 70
70/5 = 14
14/5 = 2
So 70+14+2 = 86 is the highes power of 5 in 350!.
I guess the explanation is lucid
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Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)


Last edited by cicerone on 25 Sep 2008, 00:09, edited 1 time in total.
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 [#permalink] New post 16 Sep 2006, 06:32
Thanks Cicerone... thats a great and quick way to solve a problem like this... this should help shorten the time to calculate.

Can you post a link relevant to this calculation?

Thanks/

cicerone wrote:
Folks, I'll tell u how to calculate highest power of a number in any factorial.

Taking the above example 55 = 11x5
Since the highes power of 11 in any factorial will be less than the highest power of 5 it is just enough to calculate highest power of 11 in 274!

I think someone has already suggested a method to calculate highest power.

Let's look at one more way (The logic is same as mentioned earlier)

274/11 = 24 (don't worry about the remainder)
24/11 = 2
Add up now 24+2 = 26 will be the highest power of 11 in 274!

Let's look at one more example.
Highest power of 5 in 350!
350/5 = 70
70/5 = 14
14/5 = 2
So 70+14+2 = 86 is the highes power of 5 in 350!.
I guess the explanation is lucid
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 [#permalink] New post 16 Sep 2006, 11:50
The method mentioned here is what I learnt as well...check out 4gmat.com number properties book.
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Re: PS: Largest power [#permalink] New post 16 Sep 2006, 18:15
Futuristic wrote:
What is the largest power of 55 that can divide 274! ?


55= 11*5

#multiples of 11 in 274! = 11,22,33 etc... 264 -->24 in total
#multiples of 5 in 274! appearing once = 5,10,15,....270 --> 54 in total

We dont have to worry about 5 appearing twice since largest power of 55 will have to be the largest # of 11's =24

Hence 24 ?

Heman

Last edited by heman on 16 Sep 2006, 18:37, edited 1 time in total.
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Re: PS: Largest power [#permalink] New post 16 Sep 2006, 18:20
There are two more multiples of 11 between the multiples of 11 you outlined below...

heman wrote:
Futuristic wrote:
What is the largest power of 55 that can divide 274! ?


55= 11*5

#multiples of 11 in 274! = 11,22,33 etc... 264 -->24 in total
#multiples of 5 in 276! appearing once = 5,10,15,....275 --> 55 in total

We dont have to worry about 5 appearing twice since largest power of 55 will have to be the largest # of 11's =24

Hence 24 ?

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 [#permalink] New post 16 Sep 2006, 18:32
trivikram wrote:
haas_mba07 wrote:
hmmm.. I seem to be missing something.

Given

1x2x3.... 274 = 55^k x m

where m is some integer.

As 55 = 11 x 5

The maximum value of k can be the maximum power of 11 (as 11 > 5).

Therefore # of multiples of 11 from 1 to 274
= (264-11)/11 +1 = 24

11 x 22 x 33 x 44... x 264
= 11^24 x 2 x 3 x 4 x .. 24
= 11^24 x 11^2 x K (as there are two more multiples of 11 from 2..24)

=11^26.

This should be 26.. but all options are three digits.

Not sure...

Help futuristic!!! :(


As unrelenting as you can be...Just awesome


Can you explain this the two more multiples of 11 between 2and 24 are 11& 22 But arent they already accounted for in 11^24 ?

Heman
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 [#permalink] New post 16 Sep 2006, 18:48
No they are not...

For the multiples of 11 between 1 and 274,

we have 11, 22... 99, 121, 132, 143, 154, 165, 176, 187, 198, 201, 209, 220, 231, 242, 253, 264

From this we take 11 common to get 11^24

i.e. 11^24 x (1,2,3,4,5,6,7,8,9,11,12,13....,24)

In this we have 11 and 22

i.e. 11x24 x 11^2 ( 1,2..5,6...9,10,12,....20,21,23,24)

which gives 11^26.



heman wrote:
trivikram wrote:
haas_mba07 wrote:
hmmm.. I seem to be missing something.

Given

1x2x3.... 274 = 55^k x m

where m is some integer.

As 55 = 11 x 5

The maximum value of k can be the maximum power of 11 (as 11 > 5).

Therefore # of multiples of 11 from 1 to 274
= (264-11)/11 +1 = 24

11 x 22 x 33 x 44... x 264
= 11^24 x 2 x 3 x 4 x .. 24
= 11^24 x 11^2 x K (as there are two more multiples of 11 from 2..24)

=11^26.

This should be 26.. but all options are three digits.

Not sure...

Help futuristic!!! :(


As unrelenting as you can be...Just awesome


Can you explain this the two more multiples of 11 between 2and 24 are 11& 22 But arent they already accounted for in 11^24 ?

Heman
  [#permalink] 16 Sep 2006, 18:48
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