crackthegmat2011 wrote:If x is an integer > 1, r is remainder when (x-1) (x+1) is divided by 24, what is r ?

1. x is not multiple of 2

2. x is not multiple of 3

1. x is not multiple of 2.

x = 3,5,7,9,....

So, x is odd then x-1 and x+1 both are even. Even * Even = Even. When a Even number is divided by 24 then r can be anything.

For example let x = 3 then (x-1) (x+1) = 8 thus r = 8
For example let x = 5 then (x-1) (x+1) = 24 thus r = 0

Thus, as no concrete value of r is found, this is INSUFFICIENT.

2. x is not multiple of 3.

So x can be 2,4,5,7,8.....

For example let x = 2 then (x-1) (x+1) = 2 thus r = 2
For example let x = 4 then (x-1) (x+1) = 15 thus r = 15

Thus, as no concrete value of r is found, this is INSUFFICIENT.

Combine both (1) and (2),

x can be 5,7,11,13,17,19...

Take the value of x,

For example let x = 5 then (x-1) (x+1) = 24 thus r = 0
For example let x = 7 then (x-1) (x+1) = 48 thus r = 0
For example let x = 11 then (x-1) (x+1) = 120 thus r = 0

etc.

As we can alsways see r = 0 thus sufficient.

IMO C

shovan85 wrote: x can be 5,7,11,13,17,19...

Take the value of x,

For example let x = 5 then (x-1) (x+1) = 24 thus r = 0
For example let x = 7 then (x-1) (x+1) = 48 thus r = 0
For example let x = 11 then (x-1) (x+1) = 120 thus r = 0

As we can alsways see r = 0 thus sufficient.

I Think this is really a very special property of Number 24, That the square of every prime number after 3, Leaves the remainder 1 when divided by 24,

There must be some Hidden Reasoning behind this one, Which i am not to understand , IF anyone can spot why is this happening please share it,

goyalsau wrote: I Think this is really a very special property of Number 24, That the square of every prime number after 3, Leaves the remainder 1 when divided by 24,

There must be some Hidden Reasoning behind this one, Which i am not to understand , IF anyone can spot why is this happening please share it,

Number not div by 2 can be 2n+1 (n = 0,1.....)
Number not div by 3 can be 3n+1 or 3n-1 (n = 1,2.....) -------(1)

Thus combined number will be of the form 6n+1 or 6n-1

Now (x+1)(x-1) is either (6n+2)*(6n) or (6n)*(6n-2)

=> 36n^2+12n OR 36n^2-12n

=> 12n(3n+1) Or 12n(3n-1) --------(2)

Now, from(1) we know 3n+1 or 3n-1 is not divisible 3.

Thus (2) is a even multiple of 12. And all even multiple of 12 is a multiple of 24.

Hi,

Undertand the overall logic now. But not sure how you generated the following:

For example let x = 2 then (x-1) (x+1) = 2 thus r = 2
For example let x = 4 then (x-1) (x+1) = 15 thus r = 15

Why is the remainder r =2 and 15 in these two situations? Can you please break it down into more detailed process?
Thanks

neilcao wrote:Hi,

Undertand the overall logic now. But not sure how you generated the following:

For example let x = 2 then (x-1) (x+1) = 2 thus r = 2
For example let x = 4 then (x-1) (x+1) = 15 thus r = 15

Why is the remainder r =2 and 15 in these two situations? Can you please break it down into more detailed process?
Thanks

OK!! I am breaking down one. Try the other one yourself

For example let x = 4 then (x-1) (x+1) = 15 thus r = 15

Let x = 4 then
x - 1 = 4 - 1 = 3
x + 1 = 4 + 1 = 5

Thus, (x-1) (x+1) = 3 * 5 = 15

Now divide 15 by 24 the quotient is Zero and Remainder is 15. Thus r = 15.

Same when u take x = 2 u ll get r = 3. Earlier have made a mistake it should be 3 not 2.

shovan85 wrote:

crackthegmat2011 wrote:If x is an integer > 1, r is remainder when (x-1) (x+1) is divided by 24, what is r ?

1. x is not multiple of 2

2. x is not multiple of 3

1. x is not multiple of 2.

x = 3,5,7,9,....

So, x is odd then x-1 and x+1 both are even. Even * Even = Even. When a Even number is divided by 24 then r can be anything.

For example let x = 3 then (x-1) (x+1) = 8 thus r = 8
For example let x = 5 then (x-1) (x+1) = 24 thus r = 0

Thus, as no concrete value of r is found, this is INSUFFICIENT.

2. x is not multiple of 3.

So x can be 2,4,5,7,8.....

For example let x = 2 then (x-1) (x+1) = 2 thus r = 2
For example let x = 4 then (x-1) (x+1) = 15 thus r = 15

Thus, as no concrete value of r is found, this is INSUFFICIENT.

Combine both (1) and (2),

x can be 5,7,11,13,17,19...

Take the value of x,

For example let x = 5 then (x-1) (x+1) = 24 thus r = 0
For example let x = 7 then (x-1) (x+1) = 48 thus r = 0
For example let x = 11 then (x-1) (x+1) = 120 thus r = 0

etc.

As we can alsways see r = 0 thus sufficient.

IMO C

this is a good solution...very nice. i also find the answer is C
(x^2-1)= 24a+r , what is r
statement 1: x is not multiple of 2 so x could be odd numbers such as 5,9,15 ...
just try those 3 numbers : x=5 so 25-1/24=1 the remainder is 0
x=9 >> 81/24=3 the remainder is 12
so insufficient
statement 2L x is not multiple of 3 so x could be 5,8
x=5 the remainder is 0
x=8 the remainder is 16
so insufficient
both statement 1+2, x is not multiple of 2 and 3 so it also not multiple of 6,8,9,12,15...
x could be only 7 or 11, 13 or prime numbers
so sufficient.
this data is time-consuming. it takes me 2 mins.

hi guys i was so admired with discuss, that decided to add my small note, despite the problem is elaborated thoughoutly
x^2-1 is always divisible by 24, if x is prime and x>3, as well as x^2-y^2 is always divisible by 24 if x, y are prime >3,
but we need to prove it
(1) x is not divisible by 2. so x can be written in form x=2k+1, where k is integer, insert in given expression
(2k+1-1)(2k+1+1)=2k(2k+2)- here important note we are dealing with product of two consecutive even numbers.
2k and 2k+2. and their product will always divisible by 8. consider pairs (0*2) (2*4) (4*6) and so on...
fron here we know for sure that ( x-1)(x+1) is a multiple of 8 (given that x is not multiple of 2) but it is insufficient
(2) x is not multiple of 3, possible remaiders are 1 and 2, so x can be written in a form
x=3a+1, or x=3a+2. where a is an integer, insert in a given expression
if x=3a+1, then (3a+1-1)(3a+1+1)=3a(3a+2) divisible by 3 for sure
if x=3a+2, then (3a+2-1)(3a+2+1)=(3a+1)(3a+3)- also divisible by 3 but not sufficient to determine remainder
from both we know that (x-1)(x+1) is divisible by 8 and by 3 as 8 and 3 has no common factors it is divisible by LCM(3,8)=24
sufficient