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If 6^k is a factor of (40!), what is the greatest possible value of k? [#permalink]
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10 Aug 2011, 08:00
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If 6^k is a factor of (40!), what is the greatest possible value of k? Solution: 6 is not a prime number but we make a 6 by multiplying 2 with 3. To get the number of 6s in 40!, we just need to find the number of 3s because the number of 3s will be fewer than the number of 2s. If you are a little confused, don’t worry. Look at the solution given below. Let’s find the number of 2s in 40! Step 1: 40/2 = 20 Step 2: 20/2 = 10 Step 3: 10/2 = 5 Step 4: 5/2 = 2 Step 5: 2/2 = 1 Step 6: 20 + 10 + 5 + 2 + 1 = 38 Total number of 2s is 38. Let’s find the number of 3s now. Step 1: 40/3 = 13 Step 2: 13/3 = 4 Step 3: 4/3 =1 Step 4: 13 + 4 + 1 = 18 Total number of 3s is 18. with kind regards why do i have to make the process of the 2s if only the highest prime factor(here 3) is important. why not finding only the 3s and of course they will be less than the 2s.



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Re: If 6^k is a factor of (40!), what is the greatest possible value of k? [#permalink]
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10 Aug 2011, 09:39
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dimri10 wrote: Question: If 6^k is a factor of (40!), what is the greatest possible value of k? Solution: 6 is not a prime number but we make a 6 by multiplying 2 with 3. To get the number of 6s in 40!, we just need to find the number of 3s because the number of 3s will be fewer than the number of 2s. If you are a little confused, don’t worry. Look at the solution given below. Let’s find the number of 2s in 40! Step 1: 40/2 = 20 Step 2: 20/2 = 10 Step 3: 10/2 = 5 Step 4: 5/2 = 2 Step 5: 2/2 = 1 Step 6: 20 + 10 + 5 + 2 + 1 = 38 Total number of 2s is 38. Let’s find the number of 3s now. Step 1: 40/3 = 13 Step 2: 13/3 = 4 Step 3: 4/3 =1 Step 4: 13 + 4 + 1 = 18 Total number of 3s is 18.
with kind regards why do i have to make the process of the 2s if only the highest prime factor(here 3) is important.
why not finding only the 3s and of course they will be less than the 2s. Guess, you have a point here. I'd solve it just the same way you did. Just find it for the largest prime factor. There may be some catch though.
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Re: If 6^k is a factor of (40!), what is the greatest possible value of k? [#permalink]
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10 Aug 2011, 11:38
that's the approach i generally follow..take the highest of the prime factors and get the answer...i don't think there is a catch



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Re: If 6^k is a factor of (40!), what is the greatest possible value of k? [#permalink]
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22 Oct 2014, 05:07
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fluke wrote: dimri10 wrote: Question: If 6^k is a factor of (40!), what is the greatest possible value of k? Solution: 6 is not a prime number but we make a 6 by multiplying 2 with 3. To get the number of 6s in 40!, we just need to find the number of 3s because the number of 3s will be fewer than the number of 2s. If you are a little confused, don’t worry. Look at the solution given below. Let’s find the number of 2s in 40! Step 1: 40/2 = 20 Step 2: 20/2 = 10 Step 3: 10/2 = 5 Step 4: 5/2 = 2 Step 5: 2/2 = 1 Step 6: 20 + 10 + 5 + 2 + 1 = 38 Total number of 2s is 38. Let’s find the number of 3s now. Step 1: 40/3 = 13 Step 2: 13/3 = 4 Step 3: 4/3 =1 Step 4: 13 + 4 + 1 = 18 Total number of 3s is 18.
with kind regards why do i have to make the process of the 2s if only the highest prime factor(here 3) is important.
why not finding only the 3s and of course they will be less than the 2s. Guess, you have a point here. I'd solve it just the same way you did. Just find it for the largest prime factor. There may be some catch though. Finally, please tell what is answer ? Thanks



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Re: If 6^k is a factor of (40!), what is the greatest possible value of k? [#permalink]
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22 Oct 2014, 06:09
anupamadw wrote: fluke wrote: dimri10 wrote: Question: If 6^k is a factor of (40!), what is the greatest possible value of k? Solution: 6 is not a prime number but we make a 6 by multiplying 2 with 3. To get the number of 6s in 40!, we just need to find the number of 3s because the number of 3s will be fewer than the number of 2s. If you are a little confused, don’t worry. Look at the solution given below. Let’s find the number of 2s in 40! Step 1: 40/2 = 20 Step 2: 20/2 = 10 Step 3: 10/2 = 5 Step 4: 5/2 = 2 Step 5: 2/2 = 1 Step 6: 20 + 10 + 5 + 2 + 1 = 38 Total number of 2s is 38. Let’s find the number of 3s now. Step 1: 40/3 = 13 Step 2: 13/3 = 4 Step 3: 4/3 =1 Step 4: 13 + 4 + 1 = 18 Total number of 3s is 18.
with kind regards why do i have to make the process of the 2s if only the highest prime factor(here 3) is important.
why not finding only the 3s and of course they will be less than the 2s. Guess, you have a point here. I'd solve it just the same way you did. Just find it for the largest prime factor. There may be some catch though. Finally, please tell what is answer ? Thanks The answer is 18. For similiar questions check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory. Hope this helps.
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Re: If 6^k is a factor of (40!), what is the greatest possible value of k? [#permalink]
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30 Oct 2014, 16:25
Why are we not taking any other number than 40,10,20,5,2 for (2) and 40,13,4 for (3). If we have to get 2's and 3's. Why cant we take 39/3, 3/3,18/3 or 25/3, 19/3 can somebody help to make me understand it .



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Re: If 6^k is a factor of (40!), what is the greatest possible value of k? [#permalink]
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30 Oct 2014, 18:59
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In 40!, its obvious that number of 3's would be less than the number of 2's.......... We require to find value of k in 6^k = 2^k * 3^k Counting 3's from 40 to 1 39 ......... 36 .......... 33 .......... 30 ........ 27 ........... 24 ......... 21 ......... 18 ....... 15 .......... 12 .......... 9 ........6 .... 3 1 + 2 + 1 + 1 + 3 + 1 + 1 + 2 + 1 + 1 + 2 + 1 = 8 + 5+5 = 18
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Re: If 6^k is a factor of (40!), what is the greatest possible value of k? [#permalink]
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14 Aug 2015, 13:36
Please tag divisibility/factors/multiples dimri10 wrote: If 6^k is a factor of (40!), what is the greatest possible value of k? Solution: 6 is not a prime number but we make a 6 by multiplying 2 with 3. To get the number of 6s in 40!, we just need to find the number of 3s because the number of 3s will be fewer than the number of 2s. If you are a little confused, don’t worry. Look at the solution given below. Let’s find the number of 2s in 40! Step 1: 40/2 = 20 Step 2: 20/2 = 10 Step 3: 10/2 = 5 Step 4: 5/2 = 2 Step 5: 2/2 = 1 Step 6: 20 + 10 + 5 + 2 + 1 = 38 Total number of 2s is 38. Let’s find the number of 3s now. Step 1: 40/3 = 13 Step 2: 13/3 = 4 Step 3: 4/3 =1 Step 4: 13 + 4 + 1 = 18 Total number of 3s is 18. with kind regards why do i have to make the process of the 2s if only the highest prime factor(here 3) is important. why not finding only the 3s and of course they will be less than the 2s.
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Re: If 6^k is a factor of (40!), what is the greatest possible value of k? [#permalink]
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30 Aug 2016, 04:44
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dimri10 wrote: If 6^k is a factor of (40!), what is the greatest possible value of k? Solution: 6 is not a prime number but we make a 6 by multiplying 2 with 3. To get the number of 6s in 40!, we just need to find the number of 3s because the number of 3s will be fewer than the number of 2s. If you are a little confused, don’t worry. Look at the solution given below. Let’s find the number of 2s in 40! Step 1: 40/2 = 20 Step 2: 20/2 = 10 Step 3: 10/2 = 5 Step 4: 5/2 = 2 Step 5: 2/2 = 1 Step 6: 20 + 10 + 5 + 2 + 1 = 38 Total number of 2s is 38. Let’s find the number of 3s now. Step 1: 40/3 = 13 Step 2: 13/3 = 4 Step 3: 4/3 =1 Step 4: 13 + 4 + 1 = 18 Total number of 3s is 18. with kind regards why do i have to make the process of the 2s if only the highest prime factor(here 3) is important. why not finding only the 3s and of course they will be less than the 2s. Please find the detailed and quickest solution as attached
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Re: If 6^k is a factor of (40!), what is the greatest possible value of k? [#permalink]
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24 Jan 2017, 19:12
1) We need to find the number of prime factors 2*3 in the 40! 2) In a consecutive sequence there will always be more 2's than 3's, so the task is to find the number of 3's 3) 40/3=13; 40/9=4; 40/27=1; 13+4+1=18
The correct answer is 18



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Re: If 6^k is a factor of (40!), what is the greatest possible value of k? [#permalink]
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25 Feb 2017, 10:27
I am getting answer 21 by thinking in form as  40! means 39*38*37*....*2*1. So I find numbers in this range which are divisible by 6 which are 6,12,18,24,30,36. Now counting number of time 6 appears in this number we get 1+2+3+4+5+6 = 21. Why is this wrong. What am I missing any ideas ?
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Re: If 6^k is a factor of (40!), what is the greatest possible value of k? [#permalink]
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25 Feb 2017, 10:40
fluke wrote: dimri10 wrote: Question: If 6^k is a factor of (40!), what is the greatest possible value of k? Solution: 6 is not a prime number but we make a 6 by multiplying 2 with 3. To get the number of 6s in 40!, we just need to find the number of 3s because the number of 3s will be fewer than the number of 2s. If you are a little confused, don’t worry. Look at the solution given below. Let’s find the number of 2s in 40! Step 1: 40/2 = 20 Step 2: 20/2 = 10 Step 3: 10/2 = 5 Step 4: 5/2 = 2 Step 5: 2/2 = 1 Step 6: 20 + 10 + 5 + 2 + 1 = 38 Total number of 2s is 38. Let’s find the number of 3s now. Step 1: 40/3 = 13 Step 2: 13/3 = 4 Step 3: 4/3 =1 Step 4: 13 + 4 + 1 = 18 Total number of 3s is 18.
with kind regards why do i have to make the process of the 2s if only the highest prime factor(here 3) is important.
why not finding only the 3s and of course they will be less than the 2s. Guess, you have a point here. I'd solve it just the same way you did. Just find it for the largest prime factor. There may be some catch though. Can anyone explain the solution in more detail. I still dont get the reasons behind doing 40/3 = 13. Then for the next division 13/3 = 4 and then 4/3 =1 and finally doing sum of 13+4+1 = 18. I am surely missing a concept. Help appreciated. Thanks.
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If 6^k is a factor of (40!), what is the greatest possible value of k? [#permalink]
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22 Aug 2017, 22:09
6 is not a prime number but we make a 6 by multiplying 2 with 3 which are prime individually. To get the number of 6s in 40!, we just need to find the number of 3s because the number of 3s will be fewer than the number of 2s. The power of 3 in 40! is 18.
So option with 18.




If 6^k is a factor of (40!), what is the greatest possible value of k?
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