22! is divisible by x but not by x^2. If x is a prime number, what is

Question

22! is divisible by x but not by x^2. If x is a prime number, what is the range of possible values of x?

A. 4
B. 6
C. 8
D. 16
E. 17

chetan2u · Accepted Answer

22! Has following prime numbers - 2,3,5,7,11,13,17,19
Now if 2x >22 or x>11..
So min value 13 and largest 19..
And 19-13=6
B

manoj1115 · Answer

We can eliminate the prime numbers below 10 because they will have a double between 10-20 which will make the number 22! divisible by their squares. 
To find the range we need to find the lowest and the highest number. Between 11 and 22 we have the lowest prime as 11 the square of which also being a square prime will not divide 22! The highest prime between 11 and 22 is 19, the square of which also will not divide 22!
Having got our lowest and highest number, we now have the range too, i.e. 19-11=8. Hence, option C

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kunalsinghNS · Answer

All possible prime numbers that is less than 22 ! 
2 3 5 7 11 13 17 19

So 2x is not a factor of 22! hence it should be greater than 22. So, 13, 17 and 19 are the ones

range = largest value - smallest value = 6

hence, answer = B

Luckisnoexcuse · Answer

11 comes twice in 22!..hence lowest would be 13 and highest would be 19 ...range will be 19-13 =6

Boundtohappen · Answer

if we look at the prime factors we will find that till 11 everyone will have atleast 2 as its power.. so clearly the number is divisible by x as well x^2... from 13 to 19 we have the option so range is 6

NamVu1990 · Answer

I think the question should be " 22! is divisible by x but not by x^2." instead of "22! is divisible by x but not by x2."
be cause there is a 2 in 22! so there is no such prime number x that is a factor of 22! but 2x is not.

septwibowo · Answer

First, we can list all of the prime numbers below 22, to make it divisible by X.
- We get a set {2,3,5,7,11,13,17,19}

Second, we must evaluate whether  X^2  is hiding in the 22!.
- 2 :  X^2  hiding all of the even numbers in the 22!
- 3 :  X^2  --> we have more than 2 multiple of 3 in 22!
- 5 :  X^2  --> we have more than 2 multiple of 5 in 22!
- 7 :  X^2  --> we have more than 2 multiple of 7 in 22!
- 11 :  X^2  --> we have exactly 2 multiple of 11 in 22!
- 13 :  X^2  --> we have only 1 multiple of 13 in 22!, hence 22! is not divisible by  13^2 
- 17 :  X^2  --> we have only 1 multiple of 17 in 22!, hence 22! is not divisible by  17^2 
- 19 :  X^2  --> we have only 1 multiple of 19 in 22!, hence 22! is not divisible by  19^2

Difference : 19-13 = 6.

B.

sobby · Answer

22! /11 =remainder = 0 
22!/11*11 = remainder =0 
same result for 2,3,5,7
lets take 13 ..
22!/13*13=remainder will not be 0 
we can go till 19 ..as 23 will not divide 22!
so the answer is 19 -13 = 6

ScottTargetTestPrep · Answer

We know that the largest prime number that divides into 22! is 19 and 19^2 will not divide into 22! because 22! does not have two factors of 19. Now we need to find the smallest prime that divides into 22! but whose square doesn’t. That prime number will be 13, since 22! contains only one factor of 13 but not two factors of 13. Therefore, the range of possible values of x is 19 - 13 = 6.

Answer: B

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22! is divisible by x but not by x^2. If x is a prime number, what is

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22! is divisible by x but not by x^2. If x is a prime number, what is

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