12 Days of Christmas GMAT Competition - Day 9: If one of the positive

Question

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If one of the positive factors of 10! is picked at random, what is the probability that it will be odd?

A. 1/8
B. 1/9
C. 1/10
D. 1/11
E. 1/12

Schinchan_bh · Accepted Answer

Let's first factorize 10!=(2^8)*(3^4)*(5^2)*(7)
Number of factors=9*5*3*2=270
Number of odd factors=5*3*2=30
Probability=30/270=1/9 (B) imo

DongTan · Answer

To find the probability that a randomly chosen factor of 10! (10 factorial) is odd, we need to determine the number of odd factors and divide it by the total number of factors.

First, let's find the prime factorization of 10!:
10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 2^8 × 3^4 × 5^2 × 7

For a factor to be odd, it must not contain any 2 as a prime factor. This means we can only consider the powers of 3, 5, and 7. We have the following possibilities for each prime factor:

3: 3^0, 3^1, 3^2, 3^3, 3^4 (5 possibilities)
5: 5^0, 5^1, 5^2 (3 possibilities)
7: 7^0, 7^1 (2 possibilities)

To find the number of odd factors, we can choose any combination of the powers of 3, 5, and 7. Thus, the total number of odd factors is:
5 (possibilities for 3) × 3 (possibilities for 5) × 2 (possibilities for 7) = 30

The total number of factors of 10! is found by adding 1 to the exponents of the prime factorization and multiplying them together:
(8 + 1) × (4 + 1) × (2 + 1) × (1 + 1) = 9 × 5 × 3 × 2 = 270

Therefore, the probability of randomly choosing an odd factor is:
Number of odd factors / Total number of factors = 30 / 270 = 1 / 9

Hence, the correct answer is (B) 1/9.

Bunuel · Answer

GMAT Club Official Explanation:

If one of the positive factors of 10! is picked at random, what is the probability that it will be odd?

A. 1/8
B. 1/9
C. 1/10
D. 1/11
E. 1/12

To determine the probability, first calculate the highest power of 2 in 10! by summing the  quotients  of 10 divided by powers of 2. This is computed as 10/2 + 10/4 + 10/8 = 5 + 2 + 1 = 8. Therefore, 10! includes a factor of 2^8.

Now, consider that EACH odd factor of 10! can be combined with powers of 2, ranging from 2^0, which is 1, to 2^8. This results in 9 possible combinations for each odd factor:

(odd factor)*2^0, (odd factor)*2^1, (odd factor)*2^2, ..., (odd factor)*2^8.

Among these combinations, only the first one, (odd factor)*1, is odd. Thus, the probability of selecting an odd factor from the factors of 10! is 1 out of these 9 options, or 1/9.

Answer: B.

mendax · Answer

Total numbers of instances of 2 in 10!= 10/2+10/2^2+10/2^3= 5+2+1=8
Total instances of 3 in 10!= 10/3+10/3^2=3+1=4
Instances of 5= 10/5=2
Instances of 7= 10/7=1
Total number of factors= (8+1)*(4+1)*(2+1)*(1+1)=9*5*3*2=9*30
Total number f odd factors(remove instances of 2)= 5*3*2
Answer= 30/9*30= 1/9

Correct answer is B

Kinshook · Answer

Asked:  If one of the positive factors of 10! is picked at random, what is the probability that it will be odd?

10! = 2^8*3^4*5^2*7 
Total factors of 10! = (1+8)(1+4)(1+2)(1+1) = 9*5*3*2 = 270
Total odd factors of 10! = (1+4)(1+2)(1+1) = 5*3*2 = 30

If one of the positive factors of 10! is picked at random, what is the probability that it will be odd = 30/270 = 1/9

IMO B

Evronredor · Answer

Hello Sir,

I am a bit confused and I wonder if you could help.

The question is asking us the probability of picking a positive factor of 10! and not just the prime factors. 1 is also a factor, hence why didn't we consider it?

Thanks,

Bunuel · Answer

Check detailed explanation here: https://gmatclub.com/forum/12-days-of-c ... l#p3324111

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12 Days of Christmas GMAT Competition - Day 9: If one of the positive

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