From the set of positive factors of 840 one factor is chosen at random

Question

From the set of positive factors of 840 one factor is chosen at random. What is the probability that it is divisible by 15?

A. 1/2
B. 1/4
C. 7/32
D. 1/16
E. 1/32

udaypratapsingh99 · Accepted Answer

840 in prime factors form:   (2^3) * 3 * 5 * 7

Therefore, No. of factors =  4*2*2*2 =  32

Also, for factors to be divisible by 15, it must be divisible by 3 & 5. Keeping 3&5 fixed, factors divisible by 15 = 4*2 =  8 .

Therefore required probability =   (8/32) = (1/4)  
  Correct Answer is B

DinoPen · Answer

There are 32 factors from 840: (2^3)(3)(5)(7)

There is only one factor divisible by 15. (15)

Therefore probability is 1/15.

IMHO E

Kinshook · Answer

Asked: From the set of positive factors of 840 one factor is chosen at random. What is the probability that it is divisible by 15?

840=2^3×3×5×7
Total number of factors of 840 = 4×2×2×2= 32
Factors divisible by 15 = 4×2=8
Probability = 8/32 =1/4

IMO B

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Taikiqi · Answer

Thanks for the explaination. Can you please help to elaborate on the factors to be divisble by 15? I don’t know how to get the 8 factors.

Thank you.

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Omerniazi · Answer

UdayPratapSingh99  can you please tell me how did you calculate factors divisible by 15?

udaypratapsingh99 · Answer

Hi  Taikiqi   Omerniazi , for factors to be divisible by 15, it must be divisible by 3 & 5

Therefore, out of  (2^3), 3, 5, 7 , I fixed 3&5 as these will be there as a foctor for any no. divisible by by 15, and then I took all the combination of others, ie all such factors are  ( 2^0 + 
2^1 + 2^2 + 2^3)( 7^0 + 7^1)( 5^1) ( 3^1)

And therefore, their count is 4*2*1*1

Go through this calculation and think what's going on and see that I have not taken  5^0 & 3^0  , Why?
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unraveled · Answer

Taikiqi
You have to break 840 in its prime factors form i.e. \(a^p*b^q*c^r...\) form where a,b and c are primes(2,3,5,7,11,13...) and p,q and r are powers(times any prime is used in that number - here 840).
So, \(840 = 2^3*3^1*5^1*7^1\) [(3+1)*(1+1)*(1+1)*(1+1) = 32 factors in total]
We have 2, 3, 5 and 7 only as primes using which got 840.

Now, if we need to find the factors which are divisible by 15 we need to do this:
\(15*b^q*c^r... = 3*5*b^q*c^r...\)
Eventually, we need to find \(b^q*c^r...\) i.e. what are number of factors that when multiplied with 15 result in 840 or its factors.

Hence, after leaving 3 and 5, we have only 2 and 7, which are used in their various forms(like \(2^0, 2^1, 2^2...\)). Each of these forms multiply with 15 to get us a factor of 840 or 840.
15 * \((2^0+2^1+2^2+2^3)*(3^0+3^1)\)
15 * (2 in 4 ways)*(3 in 2 ways)

\(\implies\) 4*2 = 8 is the number of factors that are divisible by 15.

Following links may help you overall understand the concepts more.
https://gmatclub.com/forum/how-many-fac ... l#p2431743
https://gmatclub.com/forum/if-a-and-b-a ... l#p2355197

\(P_{15} = \frac{8}{32} = \frac{1}{4}\)

Answer B.

WIZARD1234 · Answer

why not 7^0 and 7^1 in 2 ways ??

Bunuel · Answer

Yes, 7^0 and 7^1 should be counted in 2 ways.

The quoted solution has a typo here:

15 * (2^0 + 2^1 + 2^2 + 2^3) * ( 3^0 + 3^1 )

It should be:

15 * (2^0 + 2^1 + 2^2 + 2^3) * (7^0 + 7^1)

Since 15 already supplies the required 3 and 5, the remaining choices come from powers of 2 and 7. So there are 4 * 2 = 8 favorable factors out of 32 total factors, giving 8/32 = 1/4.

Answer: B.

Adit_ · Answer

840 = 2^3 * 3 * 5 * 7
Keep 3*5 intact and now we have possible powers of:
0/1 -> for 7.
0/1/2/3 -> for 2.
Thus 2*4 possibilities = 8.
Total factors = 4*2*2*2 = 32.

Probability = 8/32 =  1/4 .

Answer:  Option B

From the set of positive factors of 840 one factor is chosen at random

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