The number 21! = 51,090,942,171,709,440,000 has over 60,000 positive

21!= 2^18 * 3^9*5^4 * 7^3 * 11* 13*17*19
Total number of divisors= 19*10*5*4*2*2*2*2
Odd divisors= 10*5*4*2*2*2*2
Probability of choosing odd divisors = (10*5*4*2*2*2*2)/(19*10*5*4*2*2*2*2) = 1/19

We can first work out the power on the '2' in the prime factorization of 21!. First, looking only at the even numbers in the product 21*20*19*18*..., we have

(20)(18)(16)(14)(12)(10)(8)(6)(4)(2)

and extracting only the 2's from each of the above (ignoring any odd prime divisors) the above gives us

(2^2)(2^1)(2^4)(2^1)(2^2)(2^1)(2^3)(2^1)(2^2)(2^1) = 2^18

So 21! = (2^18)(a lot of odd primes)

Notice then that for any one odd divisor d we have of 21!, we can make 18 even divisors, namely 2d, (2^2)d, (2^3)d, ... (2^17)d and (2^18)d. So the ratio of odd divisors to even divisors of 21! is 1 to 18, and picking one divisor at random, the probability it is odd is 1/19.

We can let 21! = 2^a x 3^b x 5^c x 7^d x 11^e x 13^f x 17^g x 19^h.

Therefore, the number of factors 21! has is (a + 1)(b + 1)(c + 1)(d + 1)(e + 1)(f + 1)(g + 1)(h + 1) and the number of odd factors 21! has is (b + 1)(c + 1)(d + 1)(e + 1)(f + 1)(g + 1)(h + 1). So the probability a random chosen factor is odd is:

[(b + 1)(c + 1)(d + 1)(e + 1)(f + 1)(g + 1)(h + 1)]/[(a + 1)(b + 1)(c + 1)(d + 1)(e + 1)(f + 1)(g + 1)(h + 1)] = 1/(a + 1)

In other words, if we can find the number of twos 21! has, then we can determine the answer to the question.

21/2 = 10 (ignore remainder)

10/2 = 5

5/2 = 2 (ignore remainder)

2/2 = 1

So 21! has 10 + 5 + 2 + 1 = 18 twos. In other words, a = 18 and thus the probability is 1/(18 + 1) = 1/19.

Alternate Solution:

Let’s determine the greatest power of 2 that divides 21!:

21/2 = 10 (ignore remainder)

10/2 = 5

5/2 = 2 (ignore remainder)

2/2 = 1

So, the greatest power of 2 that divides 21! is 10 + 5 + 2 + 1 = 18.

Let x be an odd factor of 21!. Then, 2x, (2^2)x, (2^3)x, ... , (2^18)x are all factors of 21!. In other words, for every odd factor of 21!, there are 18 other even factors. Then, the number of even factors of 21! is 18 times the number of odd factors of 21!. Thus, the probability that a randomly chosen factor is odd is 1/(18 + 1) = 1/19.

Answer: B
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Hello ScottTargetTestPrep!!!

Why do we have to sum 1 to 18? (red part)

Kind regards!

It was explained earlier in the solution, in the part "Therefore, the number of factors 21! has is (a + 1)(b + 1)(c + 1)(d + 1)(e + 1)(f + 1)(g + 1)(h + 1) and the number of odd factors 21! has is (b + 1)(c + 1)(d + 1)(e + 1)(f + 1)(g + 1)(h + 1)."

It depends on the following fact: If the prime factorization of a number is (p^n)*(q^m)*(s^k)*..., then the number of factors of that number is (n + 1)(m + 1)(k + 1)... . Notice that any factor of 21! containing the prime 2 is even and any factor not containing 2 is odd. The total number of factors is found using the formula I explained earlier. The number of odd factors is found using the same formula, but leaving out the exponent a (since a is the exponent of 2). Since, when we take the ratio of the number of odd factors to the number of every factor of 21!, every number besides a + 1 cancel out, we are left with a + 1 in the denominator.

In short, that plus one is coming from the formula for the number of factors of an integer given its prime factor decomposition.
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Hi,
I don't understand why you add 1 to the exponents for the total divisors and for the odd numbers?

Dear Sir,

Could you please elaborate on this method of finding twos?

21/2 = 10 (ignore remainder)

10/2 = 5

5/2 = 2 (ignore remainder)

2/2 = 1

So 21! has 10 + 5 + 2 + 1 = 18 twos.

Thanks in advance.

Maybe I can help with your query.
We add 1 to each of the exponents, because we consider that every prime factor can assume any value from 0 to the exponent value. For example, we have 2^18, then we can say that for each factor of 21!, 2's exponent can attain a value from 0 to 18 , thus it can attain 19 values. Similarly if we take example of 3, we can say that 3's exponent can take any value from 0 to 9, thus the number of values it can attain will be 10.

Hope it helps!!

This method is just an elaborated form of trailing zeroes. Like if we want to know how many zeroes will be there in 20! We will calculate total no. Of pair of 2*5 in 10!
So 20! Have 10/2 = 5 ----------(1)
5/2 = 2 --------(2)
2/2 = 1 --------(3)
Adding the right hand digits = 5+2+1 = 8 2's
And 10! Will have
10/5 = 2 5's
[As 2/5 {can't be divided further}]
So no. Of trailing zeroes = 2 (because we only have 2 pair of 2 & 5)

Similarly we can use this method to define prime factors of the number too. (2&5 are also co primes)

21! Have 18 (2's), 9 (3's) , 4 (5's) , 3 (7's), 1 (11's) ,1 (13's), 1 (17's), 1 (19's)
Where 2,3,5,7,11,13,17,& 19 are co prime.

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Explanation:

The first quotient 21/2 = 10 gives us the number of factors that are divisible by 2 (i.e. the factors 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20; there are 10 in total). Notice that some of these factors have only one 2 (such as 2, 6, 10 etc.) while some of them have more than one 2 (such as 4, 8, 12 etc.).

To determine the correct number of factors of 2 in 21!, we should next determine the factors that have at least two 2s. These are all factors which are a multiple of 4, and the number of such factors can be found by dividing the quotient from the previous step by 2; i.e. 10/2 = 5 factors are divisible by 4. Notice that these factors are 4, 8, 12, 16 and 20. As a matter of fact, we could also find the same number by dividing 21 by 4 (and ignoring the remainder).

Next, we find the number of factors that have at least three 2s. These are the factors that are a multiple of 8, and the number of such factors can be found by dividing the quotient from the previous step by 2; i.e. 5/2 = 2 (ignore the remainder). The two factors with at least three 2s are 8 and 16. As in the previous case, we could have also found the same number by dividing 21 by 8 (and ignoring the remainder).

Finally, we find the number of factors which have four 2s. These are the factors that are a multiple of 16, and there is only one such factor, namely 16. We could have arrived at this number either by dividing the quotient from the previous case by 2 (i.e. 2/2 = 1) or dividing 21 by 16 (and ignoring the remainder).

We notice that there are no factors which contain five 2s, since the smallest such factor is 2^5 = 32.

Now, to determine the number of 2s in 21!, we simply add the above numbers: there are 10 + 5 + 2 + 1 = 18 twos in 21!. The reason this sum gives us the total number of factors of 2 in 21! is as follows: the first 10 factors have at least one 2 and 5 factors have at least two 2s; therefore, 10 - 5 = 5 factors have exactly one 2. Since 5 factors have at least two 2s and 2 factors have at least three 2s, 5 - 2 = 3 factors have exactly two 2s. Similarly, 2 - 1 = 1 factor has exactly three 2s, and 1 factor has exactly four 2s. Thus, the total number of factors of 2 in 21! is:

5 + 2 * 3 + 3 * 1 + 4 * 1 = 18

Hope this helps!
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Probability that a Factor of 21! that is chose at Random is an ODD Factor =

Number of ODD Factors
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Total No. of Factors Possible


Prime Factorization of 21! = 2'18 * 3'9 * 5'4 * 7'3 * 11'1 * 13'1 * 17'1 * 19'2


DEN = Total No. of Factors Possible = 19 * 10 * 5 * 4 * 2 * 2 * 2 * 2


to Find All the Different ODD Factors of 21!, we need to find all the Different Combinations of Prime Bases that do NOT include a Prime Base of 2

For the Prime Base 2: Only Option we can use is: 2'0 = 1 Option

For the Prime Base 3: 10 Options
Prime Base 5: 5 Options
Prime Base 7: 4 Options
Prime Base 11: 2 Options
Prime Base 13: 2 Options
Prime Base 17: 2 Options
Prime Base 19: 2 Options

No. of ODD Factors of 21! = 1 * 10 * 5 * 4 * 2 * 2 * 2 * 2
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Total No. of Factors of 21! = 19 * 10 * 5 * 4 * 2 * 2 * 2 * 2

After Canceling the Factors in the NUM and DEN ---- Probability of Selecting an ODD Factor = 1/19

-B-

Don't we also need to consider 5^4. Though it will be cancelled and the end result will be same.
But it will be good to have 5^4 just like other factors

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