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What is the smallest value of n for which the product of

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What is the smallest value of n for which the product of [#permalink]

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New post 19 Aug 2008, 06:15
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What is the smallest value of n for which the product of 1*2*3*4*....n is divisible by \(2^{27}\)?

This is my own question, not a question of GMAT authors. Not sure how likely this type of question is to be on the real thing, but the principles behind the answer are rather interesting. Let me know if you want me to clarify something about the question.
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Re: Interesting divisibility problem. [#permalink]

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New post 19 Aug 2008, 06:34
jallenmorris wrote:
What is the smallest value of n for which the product of 1*2*3*4*....n is divisible by \(2^{27}\)?

This is my own question, not a question of GMAT authors. Not sure how likely this type of question is to be on the real thing, but the principles behind the answer are rather interesting. Let me know if you want me to clarify something about the question.



Is that 2^27 or 2^2 * 7??
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New post 19 Aug 2008, 06:35
Not sure if I understood the question correctly.

If you refer to 2^(27), then I will get 2s from even numbers of 1*2*3*4*....till I reach 27 or more 2s. The first time I get 27 (or more) 2s is when n is 32.

(2^1)*(2^2)*(2^1)*(2^3)*(2^1)*(2^2)*(2^1)*(2^4)*(2^1)*(2^2)*(2^1)*(2^3)*(2^1)*(2^2)


However, if you meant (2^2)*7, then obviously, n will be 7.

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New post 19 Aug 2008, 06:36
\(2^{27}\) or 2^(27) - sorry about that.
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Re: Interesting divisibility problem. [#permalink]

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New post 19 Aug 2008, 06:38
jallenmorris wrote:
What is the smallest value of n for which the product of 1*2*3*4*....n is divisible by \(2^27\)?

This is my own question, not a question of GMAT authors. Not sure how likely this type of question is to be on the real thing, but the principles behind the answer are rather interesting. Let me know if you want me to clarify something about the question.


count 2s and stop when total > 27
number - 2s - total
2 - 1 - 1
4 - 2 - 3
6 - 1 - 4
8 - 3 - 7
10 - 1 - 8
12 - 2 - 10
14 - 1 - 11
16 - 4 - 15
18 - 1 - 16
20 - 2 - 18
22 - 1 - 19
24 - 3 - 22
26 - 1 - 23
28 - 2 - 24
30 - 1 - 25
32 - 5 - 30

n = 32

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Re: Interesting divisibility problem. [#permalink]

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New post 19 Aug 2008, 06:41
jallenmorris wrote:
What is the smallest value of n for which the product of 1*2*3*4*....n is divisible by \(2^{27}\)?

This is my own question, not a question of GMAT authors. Not sure how likely this type of question is to be on the real thing, but the principles behind the answer are rather interesting. Let me know if you want me to clarify something about the question.



I assume it is 2^27

1*2*3*4*....n . = 2^1* 2^2 * 2^3 ... 1*3*5*9..
= 2^ (1+2+3....x)...

(1+2+3....x)... we have to find x such that sum is the smallest possible value greater than > 27

x(x+1)/2 >27 when x=7 x(x+1)/2 = 28.

so x has to be 7
max term is 2^7= 128

n=128

oopss!! I didnt consider.. 6 * 12*..10*.14*...
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Re: Interesting divisibility problem. [#permalink]

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New post 19 Aug 2008, 06:49
I've created a spreadsheet to try this out thinking that it would be simple to count the number of 2's and that is the answer. Whatever n! has the right number of 2s, then that's your answer, but Excel is telling me otherwise and I'm not sure why.

Excel is telling me there are 27-2's in 23!
---------25852016738885000000000.0 <- This is 23! I then divide by 2. the 1---is the count of 2's
1--------12926008369442500000000.0
2--------6463004184721240000000.0
3--------3231502092360620000000.0
4--------1615751046180310000000.0
5--------807875523090156000000.0
6--------403937761545078000000.0
7--------201968880772539000000.0
8--------100984440386269000000.0
9--------50492220193134700000.0
10-------25246110096567400000.0
11-------12623055048283700000.0
12-------6311527524141840000.0
13-------3155763762070920000.0
14-------1577881881035460000.0
15-------788940940517730000.0
16-------394470470258865000.0
17-------197235235129433000.0
18-------98617617564716300.0
19-------49308808782358100.0
20-------24654404391179100.0
21-------12327202195589500.0
22-------6163601097794770.0
23-------3081800548897380.0
24-------1540900274448690.0
25-------770450137224346.0
26-------385225068612173.0
27-------192612534306086.0
28-------96306267153043.2 <- Decimal. No longer another 2?
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Re: Interesting divisibility problem. [#permalink]

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New post 19 Aug 2008, 07:03
It should be 32.

32! has 32/2 numbers divisible by 2, 32/4 numbers divisible by 4, ... 32/16 divisible by 16. So, 32/2=16 means we have 16 even numbers; 32/4=8 means in addition to those 16 even numbers we have 8 more twos and so on. You can add those for a total of 30 2s. Or am I doing something wrong?
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Re: Interesting divisibility problem. [#permalink]

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New post 20 Aug 2008, 21:02
durgesh79 wrote:
jallenmorris wrote:
What is the smallest value of n for which the product of 1*2*3*4*....n is divisible by \(2^27\)?

This is my own question, not a question of GMAT authors. Not sure how likely this type of question is to be on the real thing, but the principles behind the answer are rather interesting. Let me know if you want me to clarify something about the question.


count 2s and stop when total > 27
number - 2s - total
2 - 1 - 1
4 - 2 - 3
6 - 1 - 4
8 - 3 - 7
10 - 1 - 8
12 - 2 - 10
14 - 1 - 11
16 - 4 - 15
18 - 1 - 16
20 - 2 - 18
22 - 1 - 19
24 - 3 - 22
26 - 1 - 23
28 - 2 - 24.......25
30 - 1 - 25...26
32 - 5 - 30....31

n = 32



its trail & error

lets try at 30!:
30/2 = 15
30/4=7
30/8=3
30/16=1
30/32=0
total number of 2's in 30! = 26. we need 1 more 2. so definitely it is 32!.
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Re: Interesting divisibility problem. [#permalink]

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New post 21 Aug 2008, 05:10
I know you're all saying it should be 32! or something way higher than I am, but:

What is the value of \(2^{27}\)? 134217728

What is the value of 32!? 263130836933694000000000000000000000

Yes, 32! is evenly divisible by \(2^{27}\), but it's not the smallest number that is divisible by \(2^{27}\).

23! = 25,852,016,738,885,000,000,000

1) 25852016738885000000000<--Start and divide by 2 = the next line
2)12926008369442500000000 <--again divide by 2
3)6463004184721240000000 <--again divide by 2 and down the line
4)3231502092360620000000
5)1615751046180310000000
6)807875523090156000000
7)403937761545078000000
8)201968880772539000000
9)100984440386269000000
10) 50492220193134700000
11)25246110096567400000
12)12623055048283700000
13)6311527524141840000
14)3155763762070920000
15)1577881881035460000
16)788940940517730000
17)394470470258865000
18)197235235129433000
19)98617617564716300
20)49308808782358100
21)24654404391179100
22)12327202195589500
23)6163601097794770
24)3081800548897380
25)1540900274448690
26)770450137224346
27)385225068612173

This is from Excel and I'm kind of thinking it's messed up because when I divide 385225068612173 by 2, it gives me 192612534306086
Which of course isn't correct if the units digit is 3!
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Re: Interesting divisibility problem. [#permalink]

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New post 21 Aug 2008, 05:19
I also tried with Excel and arrived at the same result as you... Could it be that Excel is actually not giving an integer value and hides some digit located at the 10,000,000 mark after the decimal point?
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New post 21 Aug 2008, 05:42
It's possible, but if we apply the rules we know about multiplication of odd/even numbers, the process we're using should be exactly right.

My only other question with this is once we've pulled out all the 2's in say 23!, we should be left with only odd numbers.

An odd*odd = an odd product.

23*22*21*20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2
Prime factorization of 23!
23*(2*11)*(3*7)*(2*2*5)*19*(2*3*3)*17*(2*2*2*2)*(3*5)*(2*7)*13*(2*2*3)*11*(2*5)*(3*3)*(2*2*2)*7*(2*3)*5*(2*2)*3*2

There are 19-2's in 23!

So, the largest power of 2 divisible into 23! would be 2^19, right?

The odd primes remaining are 23*11*3*7*5*19*3*3*17*3*5*7*13*3*11*5*3*3*7*3*5*3

Someone correct me if I'm wrong:

The only way would could get any additional 2's out of the product of the odd primes is if we multiplied some of them and then multiplied that product by 2 (in order to get an even number because odd*odd = odd) and then that even number would have to be divisible by at least 4 or else we would not have "discovered" any additional 2's.

This just isn't going to happen. Sometime has to be wrong with how excel processes the huge numbers. The method of prime factorization and counting 2's has to be the best method.
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Re: Interesting divisibility problem. [#permalink]

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New post 21 Aug 2008, 08:50
Nerdboy wrote:
I also tried with Excel and arrived at the same result as you... Could it be that Excel is actually not giving an integer value and hides some digit located at the 10,000,000 mark after the decimal point?


exactly. excel doesnot give the exact result for values beyond certain digit.
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Re: Interesting divisibility problem.   [#permalink] 21 Aug 2008, 08:50
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