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21 Mar 2023, 10:29
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
60% (02:44) correct
40%
(02:37)
wrong
based on 125
sessions
History
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Not Attempted Yet
If x is a positive integer greater than 1, what is the remainder when (x – 1) (x + 1) is divided by 24?
(A) 6 + x is not divisible by 3
(B) 5x is not a multiple of 2
(A) 6 + x is not divisible by 3
(B) 5x is not a multiple of 2
21 Mar 2023, 13:54
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ChandlerBong
Given x is a positive integer and x > 1
Statement 1
(A) 6 + x is not divisible by 3
Therefore x is not divisible by 3. Hence, the remainder of \(\frac{x }{ 3}\) is either 1 or 2.
We can write x as -
x = 3q + 1 or x = 3q + 2
Assume x = 3q + 1; x = 4 | \(_{q = 1}\)
(x – 1) (x + 1) = 3 * 5 = 15
Remainder of 15/24 = 15
Assume x = 3q + 2; x = 2 | \(_{q = 0}\)
(x – 1) (x + 1) = 3
Remainder of \(\frac{3}{24}\) = 3.
As we are getting two different values of the remainder the statement is not sufficient. We can eliminate A and D.
Statement 2
(B) 5x is not a multiple of 2
This statement tells us that x is odd.
However, the statement alone is not sufficient, as we can have multiple possible values of the remainder
Ex. x = 3; Remainder (\(\frac{8}{24}\)) = 8
x = 5; Remainder (\(\frac{6*4}{24}\)) = 0
Hence we can eliminate B
Combined
We know that x is odd, hence if
x = 3q + 1 ; q = even
x = 3q + 2 ; q = odd
As x is odd, x + 1 and x- 1 are even. So the expression (x – 1) (x + 1) will always be divisible by 2.
As the remainder of x when divided by 3 is either 2 or 1, one of the expressions (x – 1) (x + 1) will always be divisible by 3.
Hence, the statements combined are sufficient.
Option C
15 May 2023, 21:40
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ChandlerBong
Statement 1
This tells us that x is not a multiple of 3
so x can have values such as 4/6/....
if x=4 then remainder when (x-1)(x+1) = 15 is divided by 24 is 15
if x=6 then remainder when (x-1)(x+1) = 35 is divided by 24 is 11
Not sufficient
Statement 2
This tells us that x is an odd number
so x can have values such as 3/5/7/9/...
if x=3 then remainder when (x-1)(x+1) = 8 is divided by 24 is 8
if x=5 then remainder when (x-1)(x+1) = 24 is divided by 24 is 0
if x=7 then remainder when (x-1)(x+1) = 48 is divided by 24 is 0
if x=9 then remainder when (x-1)(x+1) = 80 is divided by 24 is 8
Not sufficient.
Combined together
we know x can't be a multiple of 3 so x has to be those-multiples of 5 or 7 which are odd and don't have 3 in them.
I plugged in some values and found out that for all these values remainder was 0 when (x+1)(x-1) was divided by 24.
and thus selected C
Is there any fool-proof (easier) way of arriving at C ?
