Here’s the latest GRE Brain Twister! Be sure to come back this Thursday for the answer and explanation! ๐

13! โ 11! is divisible by n but not by n^2. What is the range of values of n if n is a prime number?

(A) 20

(B) 24

(C) 26

(D) 29

(E) 31

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Here’s the latest GRE Brain Twister! Be sure to come back this Thursday for the answer and explanation! ๐

13! โ 11! is divisible by n but not by n^2. What is the range of values of n if n is a prime number?

(A) 20

(B) 24

(C) 26

(D) 29

(E) 31

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Ans.B.24

Answer: 29

Answer : (Oops not divisible by n squared) 31 – 7 = 24

it was the easiest brain twister of all the time ๐

the answer is 24 and I think Daniel Leonard expound the solution in the most reasonable style .

The answer is 24.The prime number list on solving in ascending order is 7 11 31 and the range would be 31-7=24

Yeah correct answer must be 24 range is max-min i.e 31-7=24

(B) 24.

13! – 11!

= 11!*(12*13-1)

= 11!*155

= 11!*31*5

It is obvious that the maximum value of n is 31 and the minimum value of n is 7.

Hence, the range will be 31-7 = 24

I think if range means the the largest value of which it can take then answer must be 31.

(B) 24.

13! – 11! = (13*12*11*10*…*2*1) – (11*10*…*2*1)

Thus the two terms have 11! as a common factor, and we can use the distributive property:

13! – 11! = (11!)(13*12 – 1) = (11!)(156 – 1) = (11!)(155)

“Being divisible by n” means “having n as a factor.” Not being divisible by n^2 means not having n*n as a factorโin other words, not having n as a factor more than once. We’re told n is a prime number, so we need to find values of n which appear exactly once in the prime factorization. To find the range of values of n, we need to determine the highest and lowest prime factors of (11!)(155) for which that prime occurs as a factor only once.

The prime factorization of 155 is 5 * 31. Every prime factor in 11! is less than 31, so 31 is the upper limit of our range.

On the low end, 11! contains 2 as a factor, but it contains it more than once: 1*2*3*4… = 1*2*3*(2*2)… so 2 is not an acceptable value of n.

11! contains 3 as a factor, but it contains it more than once: 11*10*9*… = 11*10*(3*3)…

The pattern is emerging: we need a prime number which has no multiples less than 11. 5 does have a multiple less than 11 (10 = 5*2); 7 does not (7*2 > 11). So 7 occurs only once in the prime factorization of (11!)(155), and it’s the lower limit of our range.

Our range is 31 – 7, which equals 24, choice B.

13!-11!=(13*)(12!)-11!=(13*12*)(11!)-11!= (13*12-1)(11!)= 155*(11!)

155= 5*31

11!= 11*10*9*8*7*6*5*4*3*2*1=11*5^2*2^7*3^4*7

13!-11!=(5*31)*(11*5^2*2^7*3^4*7)=2^7*3^4*5^3*7*11*31

So Range is =31-7= 24

Answer is B