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Product of 1 to 30

Expert replies
by kumar720 » Sat Sep 27, 2008 8:38 pm
If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3k is a factor of p?
A. 10
B. 12
C. 14
D. 16
E. 18

OA[spoiler]:C[/spoiler]

Help me solving the problem
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Source: — Problem Solving |

Re: Product of 1 to 30

by sudhir3127 » Sat Sep 27, 2008 9:47 pm
kumar720 wrote:If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3k is a factor of p?
A. 10
B. 12
C. 14
D. 16
E. 18

OA[spoiler]:C[/spoiler]

Help me solving the problem
i go with C as well...

30/3 + 30/9 + 30/27

10 + 3 + 1 = 14

hence C .

hope that helps...
Join the discussion

by manulath » Sat Sep 27, 2008 10:12 pm
Did the original question asked k when 3^k?

or was it 3*k?
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by kumar720 » Sat Sep 27, 2008 11:38 pm
Hi Manulath, it was 3^k, typo..
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Re: Product of 1 to 30

by 4meonly » Sun Sep 28, 2008 2:24 am
sudhir3127 wrote:
kumar720 wrote:If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3k is a factor of p?
A. 10
B. 12
C. 14
D. 16
E. 18

OA[spoiler]:C[/spoiler]

Help me solving the problem
i go with C as well...

30/3 + 30/9 + 30/27

10 + 3 + 1 = 14

hence C .

hope that helps...

Can you please specify your logic?
You mean that 30! has 10 multiplies of 3, 2 multiplies of 9 and 1 of 27?
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Re: Product of 1 to 30

by Morgoth » Sun Sep 28, 2008 6:18 am
4meonly wrote: Can you please specify your logic?
You mean that 30! has 10 multiplies of 3, 2 multiplies of 9 and 1 of 27?
That is exactly what Sudhir did.
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by VP_Jim » Sun Sep 28, 2008 9:22 am
Essentially, what you want to do here is determine how many times 3 is a factor of 30!. The most foolproof way of doing this just to write it out:

30: 3x10
27: 3x3x3
24: 3x8
21: 3x7
18: 3x3x2
15: 3x5
12: 3x4
9: 3x3
6: 3x2
3: 3x1

Count up your threes, and you'll see 14 of them, so the answer is C.

Hope this helps!
Jim S. | GMAT Instructor | Veritas Prep
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by sumithshah » Tue Sep 30, 2008 4:57 am
Can someone explain what and how Sudir did what he did
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by nitin86 » Tue Sep 30, 2008 9:21 am
sumithshah wrote:Can someone explain what and how Sudir did what he did
That is a method to calculate what highest power of a number will divide a factorial.

For eg, what highest power of 2 will divide 30!

So, what we do is take all powers of 2, which are small than the factorial number ( in this case 30 )

So, we have powers of 2 as (2 , 4 , 8, 16 ) Note - its till 16 because 32 is greater than 30

Now, we divide 30 by each power and take the sum of quotient of the division

so 30/ 2 = 15
30/4 = 7
30/8 = 3
30/16 = 1

where (15, 7, 3, 1) are quotient of the division.

Now, take the sum of these, which is 26. Hence, highest power of 2 that divides 30! is 24

Hope it was helpful
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by sumithshah » Tue Sep 30, 2008 10:17 am
u bet it was! someone should make a document with all these shourtcuts. Until now I was writing down all events from 1-30 and all that jazz!
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by singalong » Sat Nov 19, 2011 1:40 am
nitin86 wrote:
sumithshah wrote:Can someone explain what and how Sudir did what he did
That is a method to calculate what highest power of a number will divide a factorial.

For eg, what highest power of 2 will divide 30!

So, what we do is take all powers of 2, which are small than the factorial number ( in this case 30 )

So, we have powers of 2 as (2 , 4 , 8, 16 ) Note - its till 16 because 32 is greater than 30

Now, we divide 30 by each power and take the sum of quotient of the division

so 30/ 2 = 15
30/4 = 7
30/8 = 3
30/16 = 1

where (15, 7, 3, 1) are quotient of the division.

Now, take the sum of these, which is 26. Hence, highest power of 2 that divides 30! is 24

Hope it was helpful
If I apply this method to the above question for the powers of 3 I get (10, 3, 9). Sum of these is 22. How would I arrive at 14?
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by GMATGuruNY » Sat Nov 19, 2011 4:29 am
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by Abhishek009 » Sat Nov 19, 2011 10:17 am
kumar720 wrote:If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3k is a factor of p?
A. 10
B. 12
C. 14
D. 16
E. 18

OA[spoiler]:C[/spoiler]

Help me solving the problem
p is the product of the integers from 1 to 30 means P is the factorial of 30....

p = 30!


Greatest integer k for which 3k is a factor of p means - the number must contain 3 and a product of k.

So 30! / 3 = 10

10/3 = 3

3/3 = 1

Total 10 + 3 + 1 = 14...
Abhishek
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by Poisson » Sun Jun 26, 2016 2:56 pm
nitin86 wrote:
sumithshah wrote:Can someone explain what and how Sudir did what he did
That is a method to calculate what highest power of a number will divide a factorial.

For eg, what highest power of 2 will divide 30!

So, what we do is take all powers of 2, which are small than the factorial number ( in this case 30 )

So, we have powers of 2 as (2 , 4 , 8, 16 ) Note - its till 16 because 32 is greater than 30

Now, we divide 30 by each power and take the sum of quotient of the division

so 30/ 2 = 15
30/4 = 7
30/8 = 3
30/16 = 1

where (15, 7, 3, 1) are quotient of the division.

Now, take the sum of these, which is 26. Hence, highest power of 2 that divides 30! is 24

Hope it was helpful
Why would the highest power of 2 be 24 and not 26?

Thanks for your help
Join the discussion

by Poisson » Fri Jul 01, 2016 6:28 am
Poisson wrote:
nitin86 wrote:
sumithshah wrote:Can someone explain what and how Sudir did what he did
That is a method to calculate what highest power of a number will divide a factorial.

For eg, what highest power of 2 will divide 30!

So, what we do is take all powers of 2, which are small than the factorial number ( in this case 30 )

So, we have powers of 2 as (2 , 4 , 8, 16 ) Note - its till 16 because 32 is greater than 30

Now, we divide 30 by each power and take the sum of quotient of the division

so 30/ 2 = 15
30/4 = 7
30/8 = 3
30/16 = 1

where (15, 7, 3, 1) are quotient of the division.

Now, take the sum of these, which is 26. Hence, highest power of 2 that divides 30! is 24

Hope it was helpful
Why would the highest power of 2 be 24 and not 26?

Thanks for your help
Bumping for clarification. Thanks again
Join the discussion