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
Remainder with division by 24
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1. x is not multiple of 2.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
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
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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,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.
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,
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Number not div by 2 can be 2n+1 (n = 0,1.....)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 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.
If the problem is Easy Respect it, if the problem is tough Attack it
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
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
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OK!! I am breaking down one. Try the other one yourselfneilcao 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
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.
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this is a good solution...very nice. i also find the answer is Cshovan85 wrote:1. x is not multiple of 2.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
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
(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.
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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
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