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If k is an integer and 33!/22! is divisible by 6^k
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22 Dec 2012, 07:52
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If k is an integer and 33!/22! is divisible by 6^k, what is the maximum possible value of k? (A) 3 (B) 4 (C) 5 (D) 6 (E) 7
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Re: If k is an integer and 33!/22! is divisible by 6^k
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22 Dec 2012, 10:08
+1 D \(\frac{33!}{22!} = 33*32*...24*23\) Therefore we have to identify how many 6s has the product of \(33*32*...24*23\) The number of 6s will indicate us the highest value of k. Let´s see how many 6s we can find in the elements of the product \(33*32*...24*23\): For example, the number 30 = 6*5..........one 6 Also 24 = 6*4......... another 6 But remember that 6 = 3*2 So, we can create 6s with the 2s and 3s of that product. Be careful of not duplicating the 6 you have identified. For example: 33 has a 3, and 32 has a 2, there is another 6. In other words, you have to identify how many 6 you can build. I found six 6s. The correct answer. I think I deserve kudos




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Re: If k is an integer and 33!/22! is divisible by 6^k
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22 Dec 2012, 10:22
basically count the no of 3s you can collect from 2333 !! there are plenty of 2s so it doest atter much!! but MISSED OUT THE FACT THAT 27 has 3  3s!



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Re: If k is an integer and 33!/22! is divisible by 6^k
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22 Dec 2012, 10:25
daviesj wrote: If k is an integer and 33!/22! is divisible by 6^k, what is the maximum possible value of k?
(A) 3 (B) 4 (C) 5 (D) 6 (E) 7 \(33!/22!=23*24*25*26*27*28*29*30*31*32*33\) \(24=2*2*2*3\) \(25=5*5\) \(26=2*13\) \(27=3*3*3\) \(28=2*2*7\) \(30=2*3*5\) \(32=2*2*2*2*2\) \(33=3*11\) I am not considering the numbers 23, 29 and 31 because they do not contain any factor other than the number itself and 1. Now in this question, our motive is to count the number of pairs of \(2\) and \(3\). The # of 3s is less than # of 2s. So our critical number is 3. The maximum number of k will be the number of pairs of \(2*3\) which is equal to number of 3s. Answer is 6. +1D
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Re: If k is an integer and 33!/22! is divisible by 6^k
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22 Dec 2012, 14:12



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Re: If k is an integer and 33!/22! is divisible by 6^k
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22 Dec 2012, 20:12
Hi carcass, this question is from manhattan gmat advanced quant book...how can it be out of scope? Posted from my mobile device
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Re: If k is an integer and 33!/22! is divisible by 6^k
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22 Dec 2012, 21:17



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Re: If k is an integer and 33!/22! is divisible by 6^k
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22 Dec 2012, 21:58
6^k =2*3 ^k
now lets take 33! number of power of 2 is 33/2 + 33/4 + 33/8 + 33/16 + 33/32 = 16+8+4+2+1=31 number of power of 3 is 33/3 + 33/9 + 33/27 = 11+3+1=15
Similarly for 22! number of power of 2 is 22/2 + 22/4 +22/8 + 22/16 = 11+5+2+1=19 number of power of 3 is 22/3 + 22/9 = 7+2=9 33!/22! contains 2^12 * 3^6 since we need maximum power of 6 we are limited by power of 3 above which is 6.... so k=6 let me know if you want me to be more clear.....



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Re: If k is an integer and 33!/22! is divisible by 6^k
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22 Dec 2012, 22:11



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Re: If k is an integer and 33!/22! is divisible by 6^k
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22 Dec 2012, 22:15
Hi Amateur, That one is good method to find power of a certain number. You just earned one kudo for that. But is it trustable method?
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Re: If k is an integer and 33!/22! is divisible by 6^k
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22 Dec 2012, 22:25
daviesj wrote: Hi Amateur, That one is good method to find power of a certain number. You just earned one kudo for that. But is it trustable method? Thank You.... Yes it is..... you can refer to factors part in the following link.... mathnumbertheory88376.html



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Re: If k is an integer and 33!/22! is divisible by 6^k
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22 Dec 2012, 22:36
Yeh, got it..thanks again! *here's the method: Finding the number of powers of a prime number p, in the n!. The formula is: \(\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3} ... till p^x\leq{n}\) What is the power of 2 in 25!? \(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\) Finding the power of nonprime in n!: How many powers of 900 are in 50! Make the prime factorization of the number: \(900=2^2*3^2*5^2\), then find the powers of these prime numbers in the n!. Find the power of 2: \(\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47\) = \(2^{47}\) Find the power of 3: \(\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22\) =\(3^{22}\) Find the power of 5: \(\frac{50}{5}+\frac{50}{25}=10+2=12\) =\(5^{12}\) We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!.
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Re: If k is an integer and 33!/22! is divisible by 6^k
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23 Dec 2012, 06:22
daviesj wrote: Hi carcass, this question is from manhattan gmat advanced quant book...how can it be out of scope? Posted from my mobile device wait a moment. I said that in my opinion after months of study (forum, blog, books, millions of sources and so on) those question are a bit out of scope. this does not mean that MGMAT is one of the best prep company out there. Often the question are more difficult to teach and learn the logic, the strategy, the techniques to tackle the most difficult questions during the exam and this is awesome, thanks prep company The same thing said stacey koprince (thanks Stacy for your articles) I do not want to be misunderstood . In concrete, they are made more difficult than normal
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Re: If k is an integer and 33!/22! is divisible by 6^k
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23 Dec 2012, 06:24



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Re: If k is an integer and 33!/22! is divisible by 6^k
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23 Dec 2012, 07:01
Hii Carcass. Can you please explain how could have we implemented the technique in the original question i.e. 33!/22!?
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Re: If k is an integer and 33!/22! is divisible by 6^k
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23 Dec 2012, 07:20



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Re: If k is an integer and 33!/22! is divisible by 6^k
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23 Dec 2012, 07:55
Marcab wrote: Hii Carcass. Can you please explain how could have we implemented the technique in the original question i.e. 33!/22!? I consider it to be a tough one...... involving meticulous approach when solving.... what if the question is 40!/14!? It would take more than 2min for the approach.... So I think my approach is a better way of solving in the time frame.....



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Re: If k is an integer and 33!/22! is divisible by 6^k
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23 Dec 2012, 08:27
Amateur wrote: Marcab wrote: Hii Carcass. Can you please explain how could have we implemented the technique in the original question i.e. 33!/22!? I consider it to be a tough one...... involving meticulous approach when solving.... what if the question is 40!/14!? It would take more than 2min for the approach.... So I think my approach is a better way of solving in the time frame..... Amateur the key is to learn as much techniques as possible. One is better in a certain situation, and in another one do not. Gmat for this reason in quite challenging because do not have a preconceived strategy.
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Re: If k is an integer and 33!/22! is divisible by 6^k
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23 Dec 2012, 10:23
carcass wrote: Amateur wrote: Marcab wrote: Hii Carcass. Can you please explain how could have we implemented the technique in the original question i.e. 33!/22!? I consider it to be a tough one...... involving meticulous approach when solving.... what if the question is 40!/14!? It would take more than 2min for the approach.... So I think my approach is a better way of solving in the time frame..... Amateur the key is to learn as much techniques as possible. One is better in a certain situation, and in another one do not. Gmat for this reason in quite challenging because do not have a preconceived strategy. I agree, I never faulted your approach... it is perfectly fine but in this situation I am just proposing a one which takes less time.... but it all depends on ones comfort.....



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Re: If k is an integer and 33!/22! is divisible by 6^k
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23 Dec 2012, 10:25
I guess, we have deviated a bit. Can anyone please let me know how to solve 33!/22! with alternative method.
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Re: If k is an integer and 33!/22! is divisible by 6^k &nbs
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