2007 October 30th, 03:31 AM
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#2 (permalink) |
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TestMagic Guru-in-Training
 
Join Date: Jan 2005
Location: Toronto
Posts: 897
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There are seven odd prime numbers between 1 and 20 inclusive.
3, 5, 7, 11, 13, 17, 19 To find the highest power of each: [20/3] + [20/9] = 8 [20/5] = 4 [20/7] = 2 [20/11] = 1 (13, 17, and 19 also have a highest power of 1) 9*5*3*(2^4) = 2160 Answer is C. Last edited by lsr : 2007 October 30th at 04:46 AM. |
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2007 October 30th, 05:02 AM
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#5 (permalink) |
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Within my grasp!
Join Date: Mar 2006
Location: Atlanta
Posts: 284
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IMO it should be 9c1+9c2+9c3+9c4+.....+9c9 (although no choice looks close, so this in all probabilities is incorrect) because there are 9 odd numbers between 1 and 20 and any combination of these numbers when multiplied will yield an odd divisor. The question does not say prime divisors. I could not understand lsr's solution, but it sure looks very short. i would be interested in knowing its details.
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2007 October 30th, 05:03 AM
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#6 (permalink) |
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Eager!
Join Date: Jul 2007
Posts: 95
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how abt this strategy for getting to the answer choice (but not know what the actual answer is)....
primes 3,5,7,11,13,17,19 answer must be less than 7! any answer close to that is the right choice!! Hence (c) do note that if the answer choices were not that widely ranged, you must follow a different strategy.... Let me know if you like this strategy? |
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2007 October 30th, 05:21 AM
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#7 (permalink) |
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Eager!
Join Date: Jul 2007
Posts: 95
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pandeyrav, you cannot do what you suggest for odd numbers but for prime numbers only...
remember, 3,5 are both prime divisors which via combination yield 15 another odd (though not prime) divisor... hence, by your solution, the answer would be [7C1+....+7C7 + 1]/2 (dont forget 1 as it is also an odd divisor) but this results in 128/2 odd divisors... hence the answer (E) now, I think what we need is the OA? |
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2007 October 30th, 10:28 AM
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#8 (permalink) |
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Within my grasp!
Join Date: Sep 2007
Posts: 175
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Guys i can Explain the Logic. We have to find out Number of Odd factors of 20!. So Let us start by Expressing 20! in terms of Prime factors. Since
N = 20 X 19 X18...................1 = 2^m + 3^n + 5^p + 7^q. Our First goal is to find the Highest Power of 2, 3, 5, 7, 11, 17, 19 which divide 20!. Now Here is the Method Which ISR has shown, but in a complex form. Highest Power of 2 Which will divide 20! is = 20/2 + 20/4 + 20/8 + 20/16 = 10 + 5 + 2 + 1 = 18. Remember only Quotients need to be considered By similar method the highest Power of 3, 5, 7,11,13,17,19 is 8 ,4,2,1,1,1,1 respectively. There fore now we can express 20! as 20! = 2^18 + 3^8 + 5^4 + 7^2 + 11^1 + 13^1 + 17^1 +19^1. Since We have to find the odd divisors we will neglect powers of 2. Total no of odd Divisors is (8+1) x ( 4+1) x ( 2+1) X (1+1) x(1+1) x(1+1)x(1+1) = 9 x 5x3x16 = 2160 answer I guess it is clear now. |
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2007 October 30th, 10:06 PM
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#10 (permalink) |
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Within my grasp!
Join Date: Oct 2007
Posts: 165
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If you are asking about why the + 1 when multiplying:
Because the 3 ^ 0 should also be counted in determining the number of factors. Similarly, the powers for other primes have a +1. If the question is about why they are multiplied: To determine the total no. of distinct factors (ways they can be combined) that can be formed from the given set of prime factors (like a combination problem). |
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