1. 2 is not a factor of n

If n = 3, then (n-1)(n+1) = 8, remainder = 8
If n = 5, then (n-1)(n+1) = 24, remainder = 0
More than one answer, Hence insufficient!

2. 3 is not a factor of n

If n = 2, then(n-1)(n+1) = 3, remainder = 3
If n = 5, then(n-1)(n+1) = 24, remainder = 0
More than one answer, Hence insufficient!

From 1 and 2

n = 1 or 5 or 7 or 11...the remainder is 0 every time.
sufficient

Option C

If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r?

For the benefit of fellow BTGers, here is another approach, The Algebra Way !

1. 2 is not a factor of n

Then n can be written in the form 2x+1 or 2x-1

If n = 2x+1, then(n-1)(n+1) = (2x+1-1)*(2x+1+1)= 2x*(2x+2) = 4*x*(x+1).
(n-1)*(n+1) = 4*x*(x+1), is divisible by 24 if x*(x+1) is a multiple of 6 (x=2) and is not divisible by 24 if x*(x+1) is NOT a multiple of 6 (x=4).

If n = 2x-1, then(n-1)(n+1) = (2x-1-1)*(2x-1+1)= 2x*(2x-2) = 4*x*(x-1).
(n-1)*(n+1) = 4*x*(x-1), is divisible by 24 if x*(x-1) is a multiple of 6 (x=3) and is not divisible by 24 if x*(x-1) is NOT a multiple of 6 (x=2).

Hence Insufficient!

2. 3 is not a factor of n

Then n can be written in the form 3x+1, 3x-1, 3x+2, or 3x-2

If n = 3x+1, then(n-1)(n+1) = (3x+1-1)*(3x+1+1)= 3x*(3x+2)
(n-1)*(n+1) = 3x*(3x+2), is divisible by 24 if x*(3x+2) is a multiple of 8 (x=2) and is not divisible by 24 if x*(3x+2) is NOT a multiple of 8 (x=3).

If n = 3x-1, then(n-1)(n+1) = (3x-1-1)*(3x-1+1)= 3x*(3x-2)
(n-1)*(n+1) = 3x*(3x-2), is divisible by 24 if x*(3x-2) is a multiple of 8 (x=2) and is not divisible by 24 if x*(3x-2) is NOT a multiple of 8 (x=3).

If n = 3x+2, then(n-1)(n+1) = (3x+2-1)*(3x+2+1)= (3x+1)*(3x+3) = 3*(x+1)*(3x+1)
(n-1)*(n+1) = 3*(x+1)*(3x+1), is divisible by 24 if (x+1)*(3x+1) is a multiple of 8 (x=3) and is not divisible by 24 if (x+1)*(3x+1) is NOT a multiple of 8 (x=4).

If n = 3x-2, then(n-1)(n+1) = (3x-2-1)*(3x-2+1)= (3x-1)*(3x-3) = 3*(x-1)*(3x-1)
(n-1)*(n+1) = 3*(x-1)*(3x-1), is divisible by 24 if (x-1)*(3x-1) is a multiple of 8 (x=3) and is not divisible by 24 if (x-1)*(3x-1) is NOT a multiple of 8 (x=4).

From 1 and 2

n can be written in the form 6x+1(n>=0) or 6x-1(n>0)

If n = 6x+1, then(n-1)(n+1) = (6x+1-1)*(6x+1+1)= 6x*(6x+2) = 12*x*(3x+1).
If x is even, 3x+1 = odd and 12*x*(3x+1) = 12*even*odd = 24*An Integer. Hence the value of r = 0
If x is odd, 3x+1 = even and 12*x*(3x+1) = 12*odd*even = 24*An Integer. Hence the value of r = 0

If n = 6x-1, then(n-1)(n+1) = (6x-1-1)*(6x-1+1)= 6x*(6x-2) = 12*x*(3x-1).
If x is even, 3x-1 = odd and 12*x*(3x-1) = 12*even*odd = 24*An Integer. Hence the value of r = 0
If x is odd, 3x-1 = even and 12*x*(3x-1) = 12*odd*even = 24*An Integer. Hence the value of r = 0

Hence Sufficient, Answer C
Please correct me if I am wrong !

Hey guys I like your approaches but was just wondering once we start testing whether both statements together are sufficient, doesn't the remainder change if we use prime numbers greater than 24?

If that is the case then the answer would be E.

Experts would love to hear your thoughts.

The quest becomes: what is remainder of (n^2-1)/(2^3*3)

statement 1: 2 is not a factor of n. i. e. n is odd number >2, and n^2-1= even. Not Enough.
statement 2: 3 is not a factor of n. i. e. n is odd number >3, and n^2-1= even. Not Enough.
Statement 1+ 2: n is prime number >3: now try for 5, 7, and 11. The remainder for (n^2-1)/(2^3*3) is same.

briology wrote:Would love some help with this problem. I struggle with remainder questions .

If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r?

1. 2 is not a factor of n

2. 3 is not a factor of n

If someone can show me the best strategy for tackling this sort of problem, i'd greatly appreciate it!

Statement 1: 2 is not a factor of n.
Thus, n = odd.
Thus, (n-1)(n+1) = the product of two consecutive even integers.
Of every two consecutive even integers, exactly one is a multiple of 4.
Thus, the product of 2 consecutive even integers = the product of an even integer and a multiple of 4 = a multiple of 8.
Since a multiple of 8 can be a multiple of 24 (in which case r=0) or not be a multiple of 24 (in which case r≠0), INSUFFICIENT.

Statement 2: 3 is not a factor of n
Since one of every 3 consecutive integers is a multiple of 3, and n is not a multiple of 3, either (n-1) or (n+1) must be a multiple of 3.
Thus, (n-1)(n+1) = a multiple of 3.
If (n-1)(n+1) is also a multiple of 8, then (n-1)(n+1) = a multiple of 24, in which case r=0.
If (n-1)(n+1) is not a multiple of 8, then (n-1)(n+1) ≠ a multiple of 24, in which case r≠0.
INSUFFICIENT.

Statements 1 and 2 combined:
Since (n-1)(n+1) = a multiple of 8, and either n-1 or n+1 must be a multiple of 3, (n-1)(n+1) = a multiple of 24.
When a multiple of 24 is divided by 24, r=0.
SUFFICIENT.

The correct answer is C.

n = 1,5,7
all cases r=0
hence C

neelgandham wrote: For the benefit of fellow BTGers, here is another approach, The Algebra Way !

I sure have learned a lot from you neelgandham!