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power of 8

by sanju09 » Mon Oct 11, 2010 10:29 pm
What power of 8 exactly divides 25!?
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9


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by Testluv » Mon Oct 11, 2010 11:41 pm
For 25! to be divisble by 8^n, 25! needs to contain all of the prime factors in 8^n. 8's prime factorization is 2*2*2 or 2^3. (Or 8^n = 2^(3n)).

So, how many 2s does 25! have? 22.

Thus, 25! is divisble by 2^1, 2^2,....2^22. In other words, 25! can be divided by powers of 8 that have 22 or fewer 2s in them.

Looking at the choices:

8^5 is 2^15, which will divide 25!

8^6 is 2^18, which will divide 25!

8^7 is 2^21, which will divide 25!

8^8 is 2^24, which won't divide 25!

Thus, answer choices A, B, and C are all correct.
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by uwhusky » Mon Oct 11, 2010 11:46 pm
Testluv wrote:So, how many 2s does 25! have? 22.
Is there a short cut to figure this fact out?
Yep.
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by Geva@EconomistGMAT » Mon Oct 11, 2010 11:54 pm
No power of 8 exactly divides 25! as far as I can tell - there's a superfluous 2. See below.

25! =25*24*23*22...1
This includes 8, 16 and 24 but that's not the only powers of 8 in there. For example, 4 and 2 multiplied make another power, as do 6 and 12 (6*12 is 72, which another multiple of 8). Instead of hunting for combinations, work systematically.

1) break down 8 to its prime factors: to make an 8, we need 2^3. This means we need to focus on the 2s in 25! - all the odd integers are of no concern.

2) List the even integers in 25!: 2*4*6*8*10*12*14*16*18*20*22*24
In order to know how many powers of 8 divide this, we need to find how many powers of 2 are in these integers? count the powers of 2, taking care around those even integers that have more than 1 power such as 4:

2 - one power
4 = 2^2 - 2 powers
6 = 3*2 - 1 power
8 = 2^3 - 3 powers
10 = 5*2 - 1 power
12 = 3*4 = 3*2^2 - 2 powers
14 = 7*2 - 1 power
16 = 2^4 - 4 powers
18 = 9*2 - 1 power
20 = 5*4 = 5*2^2 - 2 powers
22 = 11*2 = 1 power
24 = 3*8 = 3*2^3 = 3 powers

total of 22 powers of 2. Since 8=2^3, we need 3 powers of 2 to make a single power of 8, so we can 'make' 7 powers of 8 from these 22 powers of 2, and we're left with 1 power of 2 uncounted for. Answer is C 7
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by Testluv » Tue Oct 12, 2010 12:01 am
Of course since 25! contains 22 2s, no power of 8 will "exactly" divide it. That's why I ignored "exactly" in the question. If we were to interpret "exactly" as meaning that every 2 in 8^n participates in the division, then we can just as easily argue that the answer is 5 or 6 with more than 1 power of 2 unaccounted for.

In my opinion, this is not a good question.
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by sanju09 » Tue Oct 12, 2010 3:15 am
Testluv wrote:Of course since 25! contains 22 2s, no power of 8 will "exactly" divide it. That's why I ignored "exactly" in the question. If we were to interpret "exactly" as meaning that every 2 in 8^n participates in the division, then we can just as easily argue that the answer is 5 or 6 with more than 1 power of 2 unaccounted for.

In my opinion, this is not a good question.
You are correct at it, Testluv. The correct wordings of the question should have been, "What greatest power of 8 exactly divides 25!?". And Geva Stern, you're stunning!
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by Testluv » Tue Oct 12, 2010 8:58 am
Thanks Sanju, and I'll partially agree with you. IMO, the best wording for the question would have been:

"Which of the following is the greatest power of 8 that divides 25!?"

or

"Which of the following is the greatest power of 8 by which 25! is divisible?"

That is, I would suggest that the word "exactly" is unnecessary. For example, if we say "x divides n", it is the same as saying "n is divisible by x", and it is understood that x successfully divides n into smaller integer groups of integers.
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by uwhusky » Tue Oct 12, 2010 9:45 am
So I guess there isn't shortcut to figuring out factors of 25! other than factoring all the even numbers in 25!.
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by Testluv » Tue Oct 12, 2010 9:56 am
uwhusky wrote:So I guess there isn't shortcut to figuring out factors of 25! other than factoring all the even numbers in 25!.
Oh, sorry. No, I did it the same way Geva did. As Geva said, just write down the even numbers in the appropriate range, then start tallying up how many 2s you have, being careful at numbers like 4 and 8 and 12, etc.
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by selango » Tue Oct 12, 2010 9:57 am
uwhusky wrote:
Testluv wrote:So, how many 2s does 25! have? 22.
Is there a short cut to figure this fact out?
If u need to find out how many a's occurs in n!, divide n by increasing power of 'a' until the highest power of 'a' which is less than n.

For ex: to find number of 2's in 25!

25/2+25/2^2+25/2^3+25/2^4 [Note that take only quotient]

12+6+3+1=22

Another example:To find number of 5's in 50!

50/5+50/5^2=10+2=12

Hope this clarify!!!
Last edited by selango on Tue Oct 12, 2010 10:30 am, edited 1 time in total.
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by uwhusky » Tue Oct 12, 2010 9:59 am
Wow, very interesting approach! Thanks!
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by sanju09 » Tue Oct 12, 2010 8:17 pm
selango wrote:
uwhusky wrote:
Testluv wrote:So, how many 2s does 25! have? 22.
Is there a short cut to figure this fact out?
If u need to find out how many a's occurs in n!, divide n by increasing power of 'a' until the highest power of 'a' which is less than n.

For ex: to find number of 2's in 25!

25/2+25/2^2+25/2^3+25/2^4 [Note that take only quotient]

12+6+3+1=22

Another example:To find number of 5's in 50!

50/5+50/5^2=10+2=12

Hope this clarify!!!
That's very useful, hope this would save lot of time in figuring out the many powers of 'a' embedded in n!, where a ≤ n. Great info Salengo, and thanks for sharing it with us.
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