For any integer P greater than 1, P! denotes the product of all intege

Question

For any integer P greater than 1, P! denotes the product of all integers from 1 to P, inclusive. A number is chosen at random from the list of factors of 8!. What is the probability that the chosen number is a multiple of 28?

(A) 0
(B) 1/16
(C) 3/8
(D) 7/16
(E) 7/8

(A) 0
(B) 1/16
(C) 3/8
(D) 7/16
(E) 7/8

nick1816 · Accepted Answer

8! =  2^7*3^2*5*7

Any factor of 8! can be written as  2^a*3^b*5^c*7^d , where a, b, c and d are integers

a can take 8 values (0 to7)
b can take 3 values (0 to 2)
c can take 2 values (0 to 1)
d can take 2 values (0 to 1)

Number of factors = 8*3*2*2

If the factor of 8! is multiple of 28, in that case-

a can take 6 values (2 to 7)
b can take 3 values (0 to 2)
c can take 2 values (0 to 1)
d can take 1 value (only 1)

Number of multiples of 28 = 6*3*2*1

Probability =  \frac{6*3*2*1}{8*3*2*2} = \frac{3}{8}

vikasg27 · Answer

Fators of 8! - 2^7 * 3^2 * 5^1 * 7^1 --- 96 factors

Factors of 28 - 2^2 * 7^1

Remaining factors are 2^5 * 3^2 * 5^1 ---- 36 (which can be the multiple of 28)

soln - 36/96 -- 6/16 --- ( 3/8 )

Fdambro294 · Answer

Step 1: Found the Total No. of Factors in 8!

the Prime Factorization of 8! = 2^7 * 3^2 * 5^1 * 7^1

Total No. of +Positive Factors = 8 * 3 * 2 * 2 = 96 Total + Pos. Factors

Step 2: Probability = No. of Fav. Outcomes in which the Factor includes 2^2 * 7^1 / 96 Total Factors

In order for 1 of the Factors to be a Multiple of 28, it must include: 2^2 and 7^1

this leaves the Prime Factors of: 2^5 * 3^2 * 5^1

How many different Combinations can we have for each Prime Factor Above given that the 0th Power can be included for each (if they are all to the 0th Power, then the Factor just = 2^2 * 7^1 = 28). All these different Combinations can be combined with 2^2 and 7^1 to create Multiples of 28

How many Options for the Prime Factor 2?

6 options

How many Options for the Prime Factor 3?

3 options

How many Options for the Prime Factor 5?

2 options

6 * 3 * 2 = 36 Total Multiples

Answer: Probability = 36 / 96 = 3/8

Answer C

EDIT: just saw the post above mine. Same exact concept. Sorry lol

chetan2u · Answer

We have to find the factors of 8! and  \frac{8!}{28} , and to find the factors we have to get the numbers in their prime factorization.

8! would consist of all the prime factors till 8, so would contain 2, 3, 5, 7 and their powers.
 8!=2^a*3^b*5^c*7^d 
Factors of 2 in 8! = \frac{8}{2} ]+ \frac{8}{2^2} ]+ \frac{8}{8} ], where   gives the integer value less than or equal to a.
Factors of 2 = 4+2+1=7
Factors of 3 in 8! = \frac{8}{3} ]+ \frac{8}{3^2} ]=2+0=2
Factors of 5 in 8! = \frac{8}{5} ]+ \frac{8}{5^2} ]=1+0=1
Factors of 7 in 8! = \frac{8}{7} ]+ \frac{8}{7^2} ]=1+0=1
So  8!=2^7*3^2*5*7 , where number of factors = (7+1)*(2+1)*(1+1)*(1+1)=8*3*2*2

Now  \frac{8!}{28}  would be  2^5*3^2*5^1 , where number of factors = (5+1)*(2+1)*(1+1)=6*3*2

Probability =  \frac{6*3*2}{8*3*2*2}=\frac{3}{8}

C

IanStewart · Answer

8! = (2^7)(3^2)(5)(7)

To find a random factor of 8!, we can generate a number that looks like this: 2^a * 3^b * 5^c * 7^d, where a can take any value from 0 to 7 inclusive, b any value from 0 to 2 inclusive, and c and d can be 0 or 1. If we want our random factor to be a multiple of 2^2*7, we need d to equal 1, which will happen 1/2 the time. We also need a to be at least 2, which will happen 6/8 of the time (six of the eight possible values of a are 2 or greater). We don't care about b or c. So the answer is (1/2)(6/8) = 3/8.

edited - i originally had a typo at the very end

DelademAnku · Answer

Why do you use the remaining factors instead of its factors itself????
Thank you in advance

Gemmie · Answer

8! = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8
    = (2 * 4 * 8 * 2) * (3 * 3) *  5 * 7 
    = (2 * 2^2 * 2^3 * 2) * 3^2 * 5 * 7 
    = 2^7 * 3^2 * 5 * 7 
    = (2^2 * 7) * 2^5 * 3^2 * 5

Total number of factors: 8 * 3 * 2 * 2 
Number of actors being a multiple of 28: 6 * 3 * 2

Probability =  \frac{6}{8 * 2} = \frac{3}{8}

Kinshook · Answer

@ProfChaosGiven: For any integer P greater than 1, P! denotes the product of all integers from 1 to P, inclusive. A number is chosen at random from the list of factors of 8!.

Asked: What is the probability that the chosen number is a multiple of 28?
 28 = 2ˆ2*7

8! = 2ˆ(4+2+1) * 3ˆ(2)*5*7 = 2ˆ7*3ˆ2*5*7

Factors of 8! = 8*3*2*2 
Factors of 8! such that the chosen number is a multiple of 28 = 6*3*2*1

The probability that the chosen number is a multiple of 28 = 6*3*2*1/8*3*2*2 = 3/8

​​​​​​​IMO C

KritiJain0805 · Answer

why did u add +1, like after 2^, you added 7+1= 8

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For any integer P greater than 1, P! denotes the product of all intege

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