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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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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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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Kudos [?]: 615 [0], given: 32 SVP Joined: 07 Nov 2007 Posts: 1792 Kudos [?]: 1062 [0], given: 5 Location: New York Re: Interesting divisibility problem. [#permalink] ### Show Tags 19 Aug 2008, 06:34 jallenmorris wrote: What is the smallest value of n for which the product of 1*2*3*4*....n is divisible by $$2^{27}$$? This is my own question, not a question of GMAT authors. Not sure how likely this type of question is to be on the real thing, but the principles behind the answer are rather interesting. Let me know if you want me to clarify something about the question. Is that 2^27 or 2^2 * 7?? _________________ Your attitude determines your altitude Smiling wins more friends than frowning Kudos [?]: 1062 [0], given: 5 SVP Joined: 17 Jun 2008 Posts: 1534 Kudos [?]: 279 [0], given: 0 Re: Interesting divisibility problem. [#permalink] ### Show Tags 19 Aug 2008, 06:35 Not sure if I understood the question correctly. If you refer to 2^(27), then I will get 2s from even numbers of 1*2*3*4*....till I reach 27 or more 2s. The first time I get 27 (or more) 2s is when n is 32. (2^1)*(2^2)*(2^1)*(2^3)*(2^1)*(2^2)*(2^1)*(2^4)*(2^1)*(2^2)*(2^1)*(2^3)*(2^1)*(2^2) However, if you meant (2^2)*7, then obviously, n will be 7. Kudos [?]: 279 [0], given: 0 SVP Joined: 30 Apr 2008 Posts: 1867 Kudos [?]: 615 [0], given: 32 Location: Oklahoma City Schools: Hard Knocks Re: Interesting divisibility problem. [#permalink] ### Show Tags 19 Aug 2008, 06:36 $$2^{27}$$ or 2^(27) - sorry about that. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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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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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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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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**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Kudos [?]: 615 [0], given: 32 Senior Manager Joined: 16 Jul 2008 Posts: 289 Kudos [?]: 16 [0], given: 4 Schools: INSEAD Dec'10 Re: Interesting divisibility problem. [#permalink] ### Show Tags 19 Aug 2008, 07:03 It should be 32. 32! has 32/2 numbers divisible by 2, 32/4 numbers divisible by 4, ... 32/16 divisible by 16. So, 32/2=16 means we have 16 even numbers; 32/4=8 means in addition to those 16 even numbers we have 8 more twos and so on. You can add those for a total of 30 2s. Or am I doing something wrong? _________________ http://applicant.wordpress.com/ Kudos [?]: 16 [0], given: 4 SVP Joined: 29 Aug 2007 Posts: 2472 Kudos [?]: 843 [0], given: 19 Re: Interesting divisibility problem. [#permalink] ### Show Tags 20 Aug 2008, 21:02 durgesh79 wrote: jallenmorris wrote: What is the smallest value of n for which the product of 1*2*3*4*....n is divisible by $$2^27$$? This is my own question, not a question of GMAT authors. Not sure how likely this type of question is to be on the real thing, but the principles behind the answer are rather interesting. Let me know if you want me to clarify something about the question. count 2s and stop when total > 27 number - 2s - total 2 - 1 - 1 4 - 2 - 3 6 - 1 - 4 8 - 3 - 7 10 - 1 - 8 12 - 2 - 10 14 - 1 - 11 16 - 4 - 15 18 - 1 - 16 20 - 2 - 18 22 - 1 - 19 24 - 3 - 22 26 - 1 - 23 28 - 2 - 24.......25 30 - 1 - 25...26 32 - 5 - 30....31 n = 32 its trail & error lets try at 30!: 30/2 = 15 30/4=7 30/8=3 30/16=1 30/32=0 total number of 2's in 30! = 26. we need 1 more 2. so definitely it is 32!. _________________ Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html GT Kudos [?]: 843 [0], given: 19 SVP Joined: 30 Apr 2008 Posts: 1867 Kudos [?]: 615 [0], given: 32 Location: Oklahoma City Schools: Hard Knocks Re: Interesting divisibility problem. [#permalink] ### Show Tags 21 Aug 2008, 05:10 I know you're all saying it should be 32! or something way higher than I am, but: What is the value of $$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}$$. 23! = 25,852,016,738,885,000,000,000 1) 25852016738885000000000<--Start and divide by 2 = the next line 2)12926008369442500000000 <--again divide by 2 3)6463004184721240000000 <--again divide by 2 and down the line 4)3231502092360620000000 5)1615751046180310000000 6)807875523090156000000 7)403937761545078000000 8)201968880772539000000 9)100984440386269000000 10) 50492220193134700000 11)25246110096567400000 12)12623055048283700000 13)6311527524141840000 14)3155763762070920000 15)1577881881035460000 16)788940940517730000 17)394470470258865000 18)197235235129433000 19)98617617564716300 20)49308808782358100 21)24654404391179100 22)12327202195589500 23)6163601097794770 24)3081800548897380 25)1540900274448690 26)770450137224346 27)385225068612173 This is from Excel and I'm kind of thinking it's messed up because when I divide 385225068612173 by 2, it gives me 192612534306086 Which of course isn't correct if the units digit is 3! _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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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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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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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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