VeritasPrepKarishma wrote:
whiplash2411 wrote:
Would that also be 18? Since 12 is 6*2 and 6 would be the limiting factor? I feel like I might be missing something
Also kudos! Your explanation was fantastic.
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Yes, greatest power of 12 in 40! will also be 18 because
12= 3*2^2
Total number of 3s = 13 + 4 + 1 = 18 (as shown above)
Total number of 2s in 40! is 20+10 + 5 + 2 + 1 = 38 (as shown above)
So you can make 19 4s. The limiting factor is still 3.
So rxs0005, bhushan288 and whiplash2411, you all had correct answers.
The interesting thing is the maximum power of 12 in 30!
30/2 = 15
15/2 = 7
7/2 = 3
3/2 = 1
Total 2s = 15 + 7 + 3 + 1 = 26 So you can make 13 4s
30/3 = 10
10/3 = 3
3/3 = 1
Total 3s = 10 + 3 + 1 = 14!
The maximum power of 12 is 13, not 14.
Here, the limiting factor is the number of 4s (i.e. half of the number of 2s). Of course the number of 2s is more but that number gets divided by 2 to make 4s. It becomes the limiting factor.
In such cases, you will need to check for both 2 and 3.
Applauds to both karishma and Bunuel for such wonderful way to attack these problems
karishma's way seemed easier at first , until I encountered highest power of 12 in 30!
somehow I understood : that I found the highest power of 2's then halved it to get 13
no. of 3's is 14 , so as explained the answer is 13 and not 14 because in this case the highest prime is not the limiting factor
but rather 2^2 is the limiting factor. Till here it was clear.
but I got stuck when I came to bunuels example of highest power of 900 in 50!
Karishma how to do it by your method? ( want to grasp both methods and then decide , which i'd like to use )
Highest no of 2's as shown by bunuel is 47 , now I cannot half it to find the highest power of 4 , that will give me a non integer.
Also highest power of greatest prime ( 5) is 12 and answer to this question is 6
so my concern , what is the limiting factor here , or how to approach this problem ,lets say Karishma's way .