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?

Number plugging method:
(n-1)(n+1)=n^2-1
(1) n is not divisible by 2 --> pick two odd numbers: let's say 1 and 3 --> if n=1, then n^2-1=0 and as zero is divisible by 24 (zero is divisible by any integer except zero itself) so remainder is 0 but if n=3, then n^2-1=8 and 8 divided by 24 yields remainder of 8. Two different answers, hence not sufficient.

(2) n is not divisible by 3 --> pick two numbers which are not divisible by 3: let's say 1 and 2 --> if n=1, then n^2-1=0, so remainder is 0 but if n=2, then n^2-1=3 and 3 divided by 24 yields remainder of 3. Two different answers, hence not sufficient.

(1)+(2) Let's check for several numbers which are not divisible by 2 or 3:
n=1 --> n^2-1=0 --> remainder 0;
n=5 --> n^2-1=24 --> remainder 0;
n=7 --> n^2-1=48 --> remainder 0;
n=11 --> n^2-1=120 --> remainder 0.
Well it seems that all appropriate numbers will give remainder of 0. Sufficient.

Algebraic approach:
(1) n is not divisible by 2. Insufficient on its own, but this statement says that n=odd --> n-1 and n+1 are consecutive even integers --> (n-1)(n+1) must be divisible by 8 (as both multiples are even and one of them will be divisible by 4. From consecutive even integers one is divisible by 4: (2, 4); (4, 6); (6, 8); (8, 10); (10, 12), ...).

(2) n is not divisible by 3. Insufficient on its own, but form this statement either n-1 or n+1 must be divisible by 3 (as n-1, n, and n+1 are consecutive integers, so one of them must be divisible by 3, we are told that it's not n, hence either n-1 or n+1).

(1)+(2) From (1) (n-1)(n+1) is divisible by 8, from (2) it's also divisible by 3, therefore it must be divisible by 8*3=24, which means that remainder upon division (n-1)(n+1) by 24 will be 0. Sufficient.

Answer: C.

Hope it helps.
_________________

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


What are GMAT Club Tests?
25 extra-hard Quant Tests

Find out what's new at GMAT Club - latest features and updates

Question 2:

To be divisible by 24, a number must be divisible by 8 and by 3.

Statement 1: n is not divisible by 2. This basically says that n is an odd number. So if n is an odd number, both n-1 and n+1 are even numbers, which means that (n-1)(n+1) must be divisible by 4.

n-1 = 2k
n+1 = 2p

Between any two consecutive even integers, one has to be divisible by 4, so we know that the given number is divisible by 8.

Statement 2: n is not divisible by 3. However, between any three consecutive numbers one has to be divisible by 3. Considering the consecutive numbers (n-1)(n)(n+1) we know that n isn't divisible by 3, so either (n-1) or (n+1) should be divisible by 3.

Combining both together, we know that (n-1)(n+1) is divisible by 8 and 3 and hence 24. So reminder = 0. Hence answer is C.

@bunuel... 3 divided by 24 yields remainder of 3???

n is not divisble by 2 implies n-1 and n+1 are even , this implies the product of these two numbers i divisible by 4 and not 8.
both the statements taken together say that the number is not even and not divisible by 3 , meaning all odd numbers not divisible by 3 .
these are 1,5,7,11,13,17--- , if u (square of any of these minus 1 )/24 = 1 and not 0