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Futuristic
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yezz
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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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haas_mba07
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Thanks for clarifying Futuristic..all of us have done it at one time or another.

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

Futuristic
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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haas_mba07
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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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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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
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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Futuristic
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The method mentioned here is what I learnt as well...check out 4gmat.com number properties book.
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Futuristic
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
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There are two more multiples of 11 between the multiples of 11 you outlined below...

heman
Futuristic
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 ?

Heman
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trivikram
haas_mba07
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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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
trivikram
haas_mba07
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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