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If n is the product of integers from 1 to 20 inclusive, what

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If n is the product of integers from 1 to 20 inclusive, what [#permalink]

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26 Oct 2007, 11:43
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If n is the product of integers from 1 to 20 inclusive, what is the greatest k for which 2^k is a factor of n?

A. 10
B. 12
C. 15
D. 18
E. 20
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26 Oct 2007, 11:48
d, 18...i think we had this problem just recently.
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26 Oct 2007, 11:48
20! = 2*3*(2^2)*5*(3*2)*7*(2^3)*(3^2)*(2*5)*11*(2*3*2)*13*(7*2)*(5*3)*(2^4)*17*(3^2*2)*19*(2*5*2)

count the twos ---> 2^18

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26 Oct 2007, 12:00
ok so i understand how you broke it up into primes multiplied by each other. but can someone plz explain how does counting up the 2's give you the right answer then. thanks.
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26 Oct 2007, 12:05
jimjohn wrote:
ok so i understand how you broke it up into primes multiplied by each other. but can someone plz explain how does counting up the 2's give you the right answer then. thanks.

assume the same problem for 4!

4! = 2*3*(2*2)

(2*3*(2*2))/(2*2*2) ---> 3 ---> integer

the most 2 you can use is 2^3

same way in 20!/2^18 ---> integer

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27 Oct 2007, 12:02
18

what the question is asking is how many times is 2 multiplied in the product? so we want to write out the product (20 numbers), then count the # of 2's

quick way to do it (30 seconds):

write out the numbers:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

then just scan and do the factor sin your head and keep track fo the # of 2's

so

20 = 4x5 = 2^2 x 5 = 2 2's
18 = 2x9 = one 2
16 = 2^4 = 4 2's
12 = 3x4 = 3*2^2 = 2 2's
10 = 2x5 = one 2
8 = 2^3 = three 2's
6 = 2x3 = 1 2
4 = 2^2 = 2 2's
2 = one 2

1 + 2 + 1 + 3 + 1 + 2 + 4 + 1 + 2 = 18

2^18 is the biggest factor in regards to 2^k
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27 Oct 2007, 16:04
Also, there are 10 multiples of 2 among 1-20, 5 multiples of 4, 2 multiples of 8 and 1 multiple of 16

10+5+2+1=18

Thus 20! =2^18 *k, where k is an odd number
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