If n is a positive integer and r is the remainder when (n - 1)(n + 1)

(1) n must be odd. Try n = 1 and n = 3.
Not sufficient.

(2) Try n = 1 and n = 2.
Not sufficient.

(1) and (2) together: When divided by 6, the remainders can be 0, 1, 2, 3, 4, 5. Since n is not divisible neither by 2, nor by 3, when divide by 6, n can give remainder 1 or 5, which means n is of the form 6k + 1 or 6k - 1, for some positive integer k.
If n = 6k + 1, then (n - 1)(n + 1) = 6k(6k + 2) = 12k(3k + 1) = 12k(k + 2k + 1). Since 2k + 1 is odd, k and 3k + 1 are of different parities, so their product is for sure divisible by 2. Altogether, (n - 1)(n + 1) is divisible by 24.
If n = 6k - 1, then (n - 1)(n + 1) = (6k - 2)6k)= 12k(3k - 1) = 12k(k + 2k - 1). Since 2k - 1 is odd, k and 3k - 1 are of different parities, so their product is for sure divisible by 2. Altogether, (n - 1)(n + 1) is divisible by 24.
The remainder must be 0.
Sufficient.

Answer C.

n-1,n, n+1 are consecutive +ve intigers, and thus if n is even both n-1,n+1 are odd and vice versa. also in every 3 consecutive numbers we get one that is a multiple of 3

from 1

n is odd thus both n-1, n+1 are even and their product has at least 2^3 as a factor however if n = 3 thus n-1,n+1 are 2,4 and since , 24 = 2^3*3 , thus reminder is 3 but if n = 5 for example thus n-1,n+1 are 4,6 and therofre in this case r = 0.....insuff

from 2

n is a multiple of 3 and thus both n-1,n+1 are either even (e.g: n=3) or odd (n=6) and therfore this is insuff

both together

n is odd and is a multiple of 3 and therfore the reminder of the product (n-1)(n+1) when devided by 24 is always 3..suff

C

A few things:
1. Exactly one of any two consecutive positive integers is even.
2. Exactly one of any three consecutive positive integers must be a multiple of 3
3. Exactly one of any four consecutive positive integers must be a multiple of 4
etc

a) n is not divisible by 2

Since every alternate number is divisible by 2, (n-1) and (n+1) both must be divisible by 2. Since every second multiple of 2 is divisible by 4, one of (n-1) and (n+1) must be divisible by 4. Hence, the product (n-1)*(n+1) must be divisible by 8. But if n is divisible by 3, then neither (n-1) nor (n+1) will be divisible by 3 and hence, when you divide (n-1)(n+1) by 24, you will get some remainder. If n is not divisible by 3, one of (n-1) and (n+1) must be divisible by 3 and hence the product (n-1)(n+1) will be divisible by 24 and the remainder will be 0. Not sufficient.


b) n is not divisible by 3
We don't know whether n is divisible by 2 or not. As discussed above, we need to know that to figure whether the product (n-1)(n+1) is divisible by 8. Hence not sufficient.

Take both together, we know that (n-1)*(n+1) is divisible by 8 and one of (n-1) and (n+1) is divisible by 3. Hence, the product must be divisible by 8*3 = 24. So r must be 0. Sufficient.

Answer (C)

Another approach:

(n-1), n and (n+1) are consecutive integers.

If 2 is not a factor of n (i.e. n is odd), it must be a factor of (n-1) and (n+1) (the numbers around n must be even).
Also, out of any two consecutive even numbers, one has to be divisible by 4 because every alternate multiple of 2 is divisible by 4.
Hence, (n-1)*(n+1) must be divisible by 8.

Out of any 3 consecutive integers, one has to be divisible by 3 because every third integer is a multiple of 3. Out of (n-1), n and (n+1), one has to be divisible by 3. If n is not divisible by 3, one of (n-1) and (n+1) must be divisible by 3.
Hence, (n-1)*(n+1) must be divisible by 3.

Using both statements, (n-1)*(n+1) must be divisible by 8*3 = 24. Remainder must be 0.

Answer C

Given:

(n-1)(n+1) = 24m + r - (1) where m=1,2,3...

Statement 1:

n = 2w + x - (2)

Statement 1 alone is not sufficient

Statement 2:

n= 3y + z - (3)

Statement 2 alone is not sufficient.


Taken together:

1. x has to be 1 and z can be 1 or 2

2. When x and z are 1, the values of n are 7, 13, 19 etc

3. When x=1 and z=2, the values of n are 5, 11, 17 etc

4. Substitute one of the above values, say 5 in (1)

5. For n=5 we have 4*6 = 24m + r or
24m +r = 24
r=0

We will get the same value of r for the other values of n too.

Therefore answer is choice C.

1) n not divisible by 2=> n is odd=> (n-1) and (n+1) must be consective even numbers.

if n=1, 0*2/24 leaves remainder 0
if n=3, 2*4/24 leaves remainder 8
not sufficient


2) n not divisible by 3=> n can be even or 1, 5, 7, 11, 13....

if n=5, 4*6/24 leaves remainder 0
if n=2, 1*3/24 leaves remainder 3
not sufficient

together,
n must be odd and not divisible by 3=> n can be 1, 5, 7, 11, 13...
if n=7, 6*8/24 leaves remainder 0
if n=11, 10*12/24 leaves remainder 0

hence C.

Bunuel

VeritasPrepKarishma

What if n=1?

If so, (n-1) will = 0.

Please advise.

Thank you.

0 is divisible by every positive integer.

By combining both the statement, we know that if a number is a prime number apart from 2, and 3, then it can be written in the form of (6n+1) or (6n-1) - plugging these values, we get that (n-1)(n+1) is always a multiple of 24; hence sufficient. Answer C.

* n-1, n and n+1 are three consecutive integers, so one of them must be a multiple of 3, since every third integer is a multiple of 3. From Statement 2, n is not a multiple of 3, so one of n-1 or n+1 is.

* If, from Statement 1, n is odd, then n-1 and n+1 are even. Since every second even number is a multiple of 4, one of n-1 and n+1 is a multiple of 4 (at least) and the other a multiple of 2. So (n-1)(n+1) is a multiple of 8.

So with both Statements together we know that (n-1)(n+1) is a multiple of 3 and 8, and therefore of 24, and the remainder is zero.

Given , n > 0, r is remainder when (n-1)(n+1) is divided by 24, r = ?

(n-1)(n+1) is the product two consecutive even or odd integers, depending on whether n is odd or even.

Statement 1: n is not divisible by 2

n = odd, then we have, for n = 3, (n-1)(n+1) = (3-1)(3+1) = 8, hence r = 8
for n = 5, (n-1)(n+1) = (5-1)(5+1) = 4*6 = 24, hence r = 0

Statement 1 alone is Not Sufficient.


Statement: n is not divisible by 3.
hence, for n = 5, we have from above r = 0
& for n = 8, we have (n-1)(n+1) = (8-1)(8+1) = 7*9 = 63, hence r = 15

Statement 2 alone is Not Sufficient.

Combining the two statements, we get, n is not divisible by 2 or 3
for n = 7, (n-1)(n+1) = (7-1)(7+1) = 6*8 = 48, hence r = 0
for n = 11, (n-1)(n+1) = (11-1)(11+1) = 10*12 = 120, hence r = 0


Both Statements together are Sufficient.

Answer C.

Thanks,
GyM

Statement 1)
When n is not divisible by 2, then n can be 1,3,5,7,9
For n=1,3 & 5 , Different answers. Hence insufficient.

statement 2)
When n is not divisible by 3, then n can be 1,2,4,etc
Here also different remainders.
Insufficient.

On combining these two statements, n is 1,5,7
For such numbers, the remainder is 0.

C is Correct

Statement 1: n is not divisible by 2
Options for n:
1, 3, 5, 7, 9, 11, 13...

If n=1, then dividing (n+1)(n-1) by 24 yields the following:
\(\frac{(1+1)(1-1)}{24} = \frac{0}{24} =\) 0 R0
If n=3, then dividing (n+1)(n-1) by 24 yields the following:
\(\frac{(3+1)(3-1)}{24} = \frac{8}{24} =\) 0 R8

Since R can be different values, INSUFFICIENT.

Statement 2: n is not divisible by 3
Options for n:
1, 2, 4, 5, 7, 8...

If n=1, then dividing (n+1)(n-1) by 24 yields the following:
\(\frac{(1+1)(1-1)}{24} = \frac{0}{24} =\) 0 R0
If n=2, then dividing (n+1)(n-1) by 24 yields the following:
\(\frac{(2+1)(2-1)}{24} = \frac{3}{24} =\) 0 R3

Since R can be different values, INSUFFICIENT.

Statements combined:
Options for n:
1, 5, 7, 11...

If n=1, then dividing (n+1)(n-1) by 24 yields the following:
\(\frac{(1+1)(1-1)}{24} = \frac{0}{24} =\) 0 R0
If n=5, then dividing (n+1)(n-1) by 24 yields the following:
\(\frac{(5+1)(5-1)}{24} = \frac{24}{24} =\) 1 R0
If n=7, then dividing (n+1)(n-1) by 24 yields the following:
\(\frac{(7+1)(7-1)}{24} = \frac{48}{24} =\) 2 R0
If n=11, then dividing (n+1)(n-1) by 24 yields the following:
\(\frac{(11+1)(11-1)}{24} = \frac{120}{24} =\) 5 R0

In every case, R=0.
SUFFICIENT.

.

Alternate approach:

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.

.

Considering (n-1), n and (n+1) as 3 consecutive integers, many students can end up spending a lot of time analyzing the expression, especially when dealing with the individual statements.
Instead, trying simple values is the best approach to proving/disproving statements. Analysis works better when working with the combination of statements

From statement I alone, n is not divisible by 2. This means n is an odd positive integer.

If n = 1, (n-1) (n+1) = 0 * 2 = 0. The remainder when 0 is divided by 24 is 0; therefore, r = 0.
If n = 3, (n-1) (n+1) = 2 * 4 = 8. The remainder when 8 is divided by 24 is 8; therefore, r = 8
Statement I alone is insufficient to find a unique value for r. Answer options A and D can be eliminated.

From statement II alone, n is not divisible by 3.

If n = 1, (n-1) (n+1) = 0 * 2 = 0. The remainder when 0 is divided by 24 is 0; therefore, r = 0.
If n = 2, (n-1) (n+1) = 1 * 3 = 3. The remainder when 3 is divided by 24 is 3; therefore, r = 3
Statement II alone is insufficient to find a unique value for r. Answer option B can be eliminated.

Combining statements I and II, we have the following:
From statement I, n is not an even number; from statement II, n is not a multiple of 3.

Since n is odd, (n-1) and (n+1) are consecutive even numbers. If (n-1) = 2k, (n+1) = 2k +2.

(n-1) (n+1) = 2k (2k + 2) = 2k * 2(k+1) = 4k(k+1); k(k+1) itself is an even number since it represents the product of 2 consecutive integers. Therefore, 4k(k+1) must be divisible by 8.
Since n is not a multiple of 3, one of (n-1) or (n+1) should be divisible by 3.

Therefore, the expression (n-1) (n+1) is divisible by 8 and 3; hence it is divisible by 24 and so the value of r will always be ZERO.
The combination of statements is sufficient to answer the question. Answer option E can be eliminated

The correct answer option is C.

Hope that helps!
Aravind BT

Posting this using numbers because that really helped me with this one and simplify this process -

When i see (n-1) (n+1) I always think of patterns involving consecutive numbers. In this case it is alternate numbers, meaning a number less than n and 1 greater than n. So to give an example,
If n=2, (1,3)
If n=3 (2,4)

You get the hang...
Now its good to know the rule that with 3 consecutive numbers, the number is divisible with the number of terms. Eg: 1,2,3 ; 2,3,4 ; 3,4,5 ; and on we go.. you'll notice are divisible by 3.
BUT, notice we're looking at consecutive numbers so in some cases the numbers may not be divisible by 3 and in some it could be.
1. n=2 (1,3) = Div by 3
2. n=3 (2,4) = NOT div by 3
3. n=4 (3,5) = Div by 3
4. n=5 (4,6,) = Div by 3
5. n=6 (5,7) = NOT div by 3

24 = 2 x 2 x 2 x 3
Okay lets go

St 1. n is not divisible by 2
Means? n is not even and thus the first and third number are odd. Now this means anything from (1,3) - 3,5 - 5,7 - etc and it the remainders are clearly ranging and also clearly not divisible by 24. Not Suff

St 2. n is not divisible by 3 -> Ok so could be divisible by 2,4,5,7.... but not 3 ... same problem -

By combining,

pattern 1,(2),3 - eliminated - n div by 2
2,(3),4 - div by 3
3,(4),5 - eliminated - div by 2
4,(5),6 - KEEP because not div by 2 or 3 => 4*6= 2x2x2x3 = 24 => R =0
6,(7),8 - KEEP because not div by 2 or 3 => 6x8= 2x2x2x2x3 = 24 x 2 => R=0 and div by 24

Thats the trend

Hi Bunuel, for the Algebraic approach, what if n=1 => (n-1)x(n+1)=0x2 not divisible by 4?

0 is divisible by every integers (except 0 itself). Divisible means divisible without a remainder, so integer x is divisible by integer y, means that x/y = integer. Since 0/4 = 0 = integer, then 0 is divisible by 4.

Hope it's clear.

Hi Bunuel, With statement 1, if we assume n = 1, then, why can we still assume that n is a multiple of 8, when we solve together? My guess is, we cant, but if we were to assume n = 1, we still get r = 0 (which is the same case as when we assume n is a multiple of 8 and 3), and hence, we still consider this sufficient together? Would just like to validate my thinking on this.

Bunuel.