GMAT Club World Cup 2022 (DAY 7): If p is a prime number and p! has mo

Prime number
2,3,5,7,11,17


P=7
7! = 7*6*5*4*3*2*1
odd factors 7*3*2*5*2^2*3*2*1
2*2*2*2*2 ; 2^5 ; 32

for P =11 11! has odd factors
11*10*9*8*7*6*5*4*3*2*1
11*2*5*3^2*2^3*7*3*2*5*2^2*2*1
odd factors
11*5*3^2*7*3*5*3*1
factors will be
2*2*3*2*2*2*2*2
2^7*3 384 factors
Next Prime is 13 ; include its 2 more factors + 2 factors of ( 3 with 12)
13*2^2*3 * 11!
2*2 *384 1536

Only at P = 11 we get p! has more than 100 but less than 1,000 positive odd factors
OPTION A is correct

Number of odd factors is the product of (power+1) of each of the prime odd factors.

\(5! = 5*4*3*2*1 = 2^3*3*5\)
There are two prime odd factors, 3 and 5. 3 is to the first power and 5 is to the first power, so we have (1+1)*(1+1) = 2*2 = 4

\(7! = 7*6*above = 2^4*3^2*5*7\)
3*2*2 = 12

Lets start omitting the 2's since they don't matter.

\(11! = 11*10*9*8*above = 3^4*5^2*7*11\)
5*3*2*2 = 60

\(13! = 13*12*above = 3^5*5^2*7*11*13\)
6*3*2*2*2 = 144

\(17! = 17*16*15*14*above = 3^6*5^3*7^2*11*13*17\)
7*4*3*2*2*2 = 672

If we add 19^1, we will need to multiply 672 by at least 2, thereby exceeding 1000.

13! and 17! are our only options.

Answer choice B.

My solution is probably incorrect and *very* long. Anyone has a shorter way to do it?

Asked: If p is a prime number and p! has more than 100 but less than 1,000 positive odd factors, how many different values can p take ?

Prime factorization of 11! = 2^8*3^4*5^2*7*11; Number of positive odd factors of 11! = (4+1)(2+1)(1+1)(1+1)= 5*3*2*2 = 60
Prime factorization of 13! = 2^10*3^5*5^2*7*11*13; Number of positive odd factors of 13! = (5+1)(2+1)(1+1)(1+1)(1+1)= 6*3*2*2*2 = 144
Prime factorization of 17! = 2^15*3^6*5^3*7*11*13*17; Number of positive odd factors of 17! = (6+1)(3+1)(1+1)(1+1)(1+1)(1+1)= 7*4*2*2*2*2 = 448
Prime factorization of 19! = 2^16*3^8*5^3*7*11*13*17*19; Number of positive odd factors of 19! = (8+1)(3+1)(1+1)(1+1)(1+1)(1+1)(1+1)= 9*4*2*2*2*2*2 = 1152

p can take values = 13 & 17

IMO B

Correct answer : Choice B

Let N as a number.

N in prime factorization = a^p × b^q × c^r
No. of factors of N = (p+1)(q+1)(r+1)
Now suppose that b and c are odd prime numbers in prime factorization of N
Now to find even no of factors you have to find odd no of factors first.
Odd no of factors = (q+1)(r+1)

So taking 23 as a prime number to check :
odd factors of 23! = 23 * 22 * 21 * 20 * 19 * 18 ...................
= 23 to the power 1 * 11 to the power 1 * 2 power 1 * 7 to the power 1 * 3 to the power 1 .......
= on doing this , we get that the factors are greater than 1000. So this does not work.

going to the next prime number below 23 , which is 19:
Applying the same process above to 19!, = on doing this , we get that the factors are greater than 1000. So this does not work

Applying the same process above to 17!, we get the count of odd factors which is above 100 and below 1000, so that works

Applying the same process above to 13!, we get the count of odd factors which is above 100 and below 1000, so that works

Applying the same process above to 11!, we get the count of odd factors which is below 100, so that does not work.

hence we have two numbers that work, 13 and 17.
Correct answer : Choice B

only 2 prime no 13 and 17

As we need to find odd factors, we cannot include any power of 2 in the calculation.

We know that p is prime, so the total factors of p! will consists of the powers of all prime number greater than 2 and less than or equal to p.

Lets assume p = 13

Power of 3 in 13 ! = 5
Power of 5 in 13 ! = 2
Power of 7 in 13 ! = 1
Power of 11 in 13 ! = 1
Power of 13 in 13 ! = 1

Total number of odd factors = 6 * 3 * 2 * 2 * 2 = 144. Hence 13 ! is vaild

Lets assume p = 17

Power of 3 in 17 ! = 6
Power of 5 in 17 ! = 3
Power of 7 in 17 ! = 2
Power of 11 in 17 ! = 1
Power of 13 in 17 ! = 1
Power of 17 in 17 ! = 1

Total number of odd factors = 7 * 4 * 3 * 2 * 2 * 2 lies in the range. Hence valid.

From 19! onwards the range exceeds. Hence only valid values are 13 & 17

IMO B

If p is a prime number and p! has more than 100 but less than 1,000 positive odd factors, how many different values can p take ?

A. 1
B. 2
C. 3
D. 5
E. 7

two digit p is sufficient for 100 positive odd factors, 3 digit p can take more that 100 odd factors
answer C

Please refer to the attachment for the solution.

If p is a prime number and p! has more than 100 but less than 1,000 positive odd factors, how many different values can p take ?

Since p is prime number and question is asking about odd factors only.

Number of odd factors in 11! = highest power of 3,5,7,11 = 3^4x5^2X7^1x11^1 = 5*3*2*2=60
Number of odd factors in 19! = Highest power of 3,5,7,11,13,17,19=3^8x5^3X7^2x11X13X17= 9*4*3*16 =1728


So P can take 13,15,17 only.
Answer 3

If p is a prime number and p! has more than 100 but less than 1,000 positive odd factors, how many different values can p take ?

A. 1
B. 2
C. 3
D. 5
E. 7

Suppose if we consider p= 7 Then the odd factors of 7! are 1,3,5,7
and along with that product of any 2 or more odd factors from the set {3,5,7} are also odd factors of 7! .We are excluding 1 from the set as 1 multiplied with any number of odd factors is going to be same.
Thus total number of odd factors would be =Number of subsets of {3,5,7} - 1(excluding the empty set) +1 (1 as the first odd factor)
So for p ,Number of odd numbers between 1 and p= (p+1)/2
excluding 1 from here =(p+1)/2-1=(p-1)/2
Number of subsets formed by (p-1)/2 elements =2^(p-1)/2 -1
Thus as per the statement
100<2^(p-1)/2<1000
As 2^7 =128 ,2^9=512
2^10 =1024
(p-1)/2 will hold true for {7,8,9}
and p will be {15,17,19}.As 15 is not a prime number thus possible values of p are {17,19}
Thus p can take 2 different values.
Answer B

IMO option C is the answer.

The list of prime numbers are 2,3,5,7,11,13,17.......

Positive odd factor of 2! is just 1 . The total number of positive odd factor 2! = \(2^0\) =1

Positive odd factors of 3! are 1 and 3, The total number of positive odd factor of 3! is = \(2^1\) =2

Positive odd factors of 5! are 1,3, 5 and 15. The total number of positive odd factor of 5! is = \(2^2\) =4

Positive odd factors of 7! are 1,3,5,7, 15,21, 35 and 105. The total number of positive odd factor of 7! is = \(2^3\) =8

Clearly a pattern has been established.

The total number of positive odd factor of 11! is = \(2^4\) =16
The total number of positive odd factor of 13! is = \(2^5\) =32
The total number of positive odd factor of 17! is = \(2^6\) =64
The total number of positive odd factor of 19! is = \(2^7\) =128
The total number of positive odd factor of 23! is = \(2^8\) =256
The total number of positive odd factor of 29! is = \(2^9\) =512
The total number of positive odd factor of 31! is = 2^10 =1024

From the conditions in the question, p can be 19, 23 or 29. Therefore IMO option C = 3 is the answer.

Imo B

P is a prime number.
P can take 2,3,5,7,11,13
Now we have to find the odd factors of p! such that it lies between 100 to 1000

Lets start from some higher number such as 11!, 13!, and 17!
Odd factors of 11! means powers of all the prime numbers except 2.
Power of 3 in 11! = 3+1 = 4
Power of 5 in 11! = 2
Power of 7 in 11! = 1
Power of 11 in 11! = 1
Odd factors = (4+1) * (2+1) * (1 +1) * (1+1) = 60, which is less than 100.

Odd factors of 13! means powers of all the prime numbers except 2.
Power of 3 in 13! = 4+1 = 5
Power of 5 in 13! = 2
Power of 7 in 13! = 1
Power of 11 in 13! = 1
Power of 13 in 13! = 1
Odd factors = (5+1) * (2+1) * (1 +1) * (1+1) * (1+1) = 144

Odd factors of 17! means powers of all the prime numbers except 2.
Power of 3 in 17! = 5+1 = 6
Power of 5 in 17! = 3
Power of 7 in 17! = 2
Power of 11 in 17! = 1
Power of 13 in 17! = 1
Power of 17 in 17! = 1
Odd factors = (6+1) (3+1) * (2+1) * (1 +1) * (1+1) * (1+1) = 672

Odd factors of 19! will be more than 1000.

Hence there are only two values that p can take 13 & 17.

If p is a prime number and p! has more than 100 but less than 1,000 positive odd factors, how many different values can p take ?
the number of odd factors in n! is product of (powers+1) of all prime factors except 2
if P = 13
13! has odd powers = 3^5 * 5^2 * 7^1*11^1
number of odd factors = (5+1)*(2+1)*(1+1)*(1+1) = 18*4 = 72
next prime number is 17
17! has odd factors = 336
next prime number is 19
19! has odd factors = 1728

therefore P can take 1 value for the given condition
Answer:A

Given that p is a prime number. And p! has more than 100 but less than 1000 positive odd factors.
We know that any number can be expressed in terms of prime factors e.g
N = p1^a * p2^b * p3^c...
Where p1, p2, p3 are prime numbers.
But we need to omit any powers of 2 because we need to have only odd factors.
Now if consider 11! then we see no. of odd factors are 60. Hence it doesn't fit the criteria.
For 13 we see 13! we find no. of odd factors are 144.
For 17 we see 17! has 672 no. of odd factors.
But for 19! the no. of odd factors are 1724. Hence it doesn't fit the criteria.
So p can have to values. 13 and 17.
Hence answer choice (B)

The answer to this question has to be Option D ie 5