If n is a positive integer, is n^2 - 1 divisible by 24?

bunuel can you please explain the colored part of your answer.
if n is not there, why it n-1 & n+1 must be divisible by 3.
it may be possible that only n is divisible by 3 example; n-1=2,n=3,n+1=4 i.e 2,3,4
if 3 is not there,, why 2x4 must be divisible by 3 ??

For (1)+(2) we have that n is odd and not a multiple of 3. Next, (n-1), n and n+1 represent three consecutive integers. Out of ANY three consecutive integers one is always divisible by 3, we know that it's not n, so it must be either n-1 or n+1.

Hope it's clear.

I have another solution which came to me.
Clearly first and second statement alone each can not be sufficient to find out whether expression is divisible by 24 or not.
Now considering both, we know that n is prime number which is greater than 191.
All prime numbers except 2 and 3 are in the form of 6n+1 or 6n-1.

n^2 -1 = (6n+1)^2 - 1 or n^2 -1 = (6n - 1)^2 - 1

n^2 -1 = 36n^2 + 12n or n^2 -1 = 36n^2 - 12n

= 12(n^2 + n) or 12(n^2 - n)

Now we know that this expression is divisible by 12 and this will be divisible by 24 if (n^2+n) or (n^2-n) is even numbers.
As we know that n is prime number which is greater than 191 so it has to be odd so
n^2 will also be odd and now we can say that n^2+n will be even (odd+odd = even).
Same way n^2 - n will be even.

So this expression will definitely divisible by 24.

Bunuel - Please let me know if I am doing any wrong over here.

Your approach is correct, but when expressing prime number \(n\), you shouldn't use the same variable (\(n\)). Also there are some other little mistakes.

It should be: since any prime number greater than 3 could be expressed as \(6k+1\) or\(6k+5\) (\(6k-1\)), where \(k\) is an integer >1, then \(n=6k+1\) or \(n=6k-1\).

\(n^2-1=36k^2+12k=12k(3k+1)\) or \(n^2-1=36k^2-12k=12k(3k-1)\).

We can see that \(n^2-1\) is divisible by 12. Next, notice that if \(k=odd\) then \(3k+1=even\) (\(3k-1=even\)) and if \(k=even\) then \(3k+1=odd\) (\(3k-1=odd\)). So, \(n^2-1\) is also divisible by 2, which means that \(n^2-1\) is divisible by 12*2=24.

Hope it's clear.

St 1 : If n =2 then n^2-1 is not divisibly by 24
but if n=7 then n^2-1 is divisible by 24

There is a there property that for any prime (p) greater than 6, p^2 -1 is always divisible by 6
Option A,D ruled out
St 2 says n>191 again if n is not prime than n^2-1 may or may not be divisible by 24 but if n is prime and greater than 191 then it surely divisible by 24
since 2 choices are possible so option B is ruled out

Combining we get n is prime and n>191 and hence the expression n^2-1 will be divisible by 24 for any value of n

Ans is C

Any prime number greater than 3 can be represented as \(6k\pm1\) but NOT necessarily vice-versa, where k = 1,2,3..etc

I. If n is prime,\(n^2-1 = (6k+1)^2-1\) --> (6k+2)*6k --> 12*k*(3k+1) . Either k is odd and (3k+1) is even OR k is even and (3k+1) is odd. Either ways, with one even factor in the expression, \(n^2-1\) would always have a factor of 24.

II.If n is prime, \(n^2-1 = (6k-1)^2-1\) --> (6k-2)*6k --> 12*k*(3k-1) . Either k is odd and (3k-1) is even OR k is even and (3k-1) is odd. Either ways, with one even factor in the expression, \(n^2-1\) would always have a factor of 24.

Thus, any prime number n>3,\(n^2-1\) is ALWAYS divisible by 24.

F.S 1- We don't know whether n>3 or not. Insufficient.

F.S 2- We don't know whether n is prime or not. For n= 240 we get a YES, for n = 241, we get a NO.Insufficient.

Taking both togehter, we know that n is prime and n>3. Thus, \(n^2-1\) will always be divisible by 24. Sufficient.
C.

amazing explanation,Bunuel!You rock!

We need to determine whether (n^2 - 1)/24 = integer. Notice that 24 is 2^3 x 3, or 8 x 3.

Since n^2 - 1 = (n + 1)(n - 1), when n is odd and not a multiple of 3, we will have the product of two consecutive even integers, one of which is a multiple of 3, and thus n^2 - 1 is divisible by 24. For instance, when n is 5, we have 4 x 6, which is divisible by 24.

Statement One Alone:

n is a prime number.

When n is 5, we see that 24/24 = integer; however, when n = 2, 3/24 does not equal an integer. Statement one is not sufficient to answer the question.

Statement Two Alone:

n is greater than 191.

If n = 192, then n^2 - 1 will be odd and will not be divisible by 24. If n = 199, then n^2 - 1 = (199 + 1)(199 - 1) = 200 x 198 is divisible by 24, since 200 is divisible by 8 and 198 is divisible by 3. Statement two is not sufficient to answer the question.

Statements One and Two together:

From both statements, we see that n is a prime that is greater than 191, and thus it satisfies the case that n is odd and not a multiple of 3. So, n^2 - 1 will be divisible by 24.

Answer: C

First, look at this post: https://anaprep.com/arithmetic-factors- ... -integers/

\((n^2 - 1) = (n-1)(n+1)\)

(1) n is a prime number.
n could be 2 in which case 2^2 - 1 is not divisible by 24.
n could be 5 in which case 5^2 - 1 is divisible by 24.

(2) n is greater than 191
Clearly insufficient.

Using both, n is a large prime number. It will be odd and will have no factors other than 1 and itself.
So (n-1) and (n+1) would be even and one of them would have to be divisible by 4. Also, since n is not divisible by 3, one of (n-1) and (n+1) will have to be.
Hence (n-1) and (n+1) together will have 3 and 8 as factors and hence will be divisible by 24.

Answer (C)


Note that the only relevance of n > 191 is to tell us that n is neither 2 nor 3. For all other prime numbers, n^2 - 1 will be divisible by 24.
Say n = 11;
(n-1) = 10 (has 2 as factor)
(n+1) = 12 (has 4 and 3 as factors)
Hence the product 10 * 12 will be divisible by 24.

I arrived at the answer by substituting n for different values and found that for any prime number greater than or equal to 5, the condition is met.

St. 1 says the number is prime but n can be 2 or 3 too, which don't satisfy the condition. Not sufficient.

St. 2 says n is greater than 191, but we can't be sure about the exact value of n. Not sufficient.

C is the answer because after combining both we know N will meet the condition.