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

Question

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

ThatDudeKnows · Accepted Answer

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.

gmatophobia · Answer

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

akash2703 · 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.

av1901 · Answer

If a number on prime factorization = a^p * b^q * c^r, then total number of factors = (p+1)(q+1)(r+1)
But if you want to find the number of odd factors, then only take the odd bases into consideration
For example
30 = 3*5*2
Total factors = 2*2*2 = 8
Odd factors = 2*2 = 4 (Did not take the 2 into consideration because it is even and we want only odd factors)
So now, coming to the question at hand, best would be to consider some values > 7 because 7! only has two 3(s), one 7 and one 5 which will be less than 100:

11! = 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 = 11 x (2x5) x (3x3) x (2x2x2) x (7) x (3x2) x (5) x (2x2) x (3) x (2)
Taking out all the odd bases and their powers = 11 x 7 x 5^2 x 3^4 
Odd factors = 2 x 2 x 3 x 5 = 60 Less than 100

13! = 13 x 12 x 11! = 13 x (2^2 x 3) x 11!: So we add a 2 for 13 and add 1 to the exponent of 3 in 11! = 2 x 2 x 3 x 6 x 2 = 144
YES (1)

17! = 17 x 16 x 15 x 14 = 17 x (2^4) x (3x5) x (2x7) = So we add a 2 for 17, increase the exponent of 5 by 1, increase the exponent of 7 by 1 and increase the exponent of 3 by 1 in 13! = 2 x 2 x 2 x 4 x 7 x 3 = 672 YES (2)

Now 19! will at the very least have a another 2 because of 19 which will take the total beyond 1000 and not in the scope of the question

So: two numbers p can take : 13 and 17

Answer - B

bumpbot · Answer

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