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

Question

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

young_gun · Answer

d, 18...i think we had this problem just recently.

KillerSquirrel · Answer

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 the answer is (D)

jimjohn · Answer

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.

KillerSquirrel · Answer

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

StartupAddict · Answer

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

kevincan · Answer

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

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

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