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D
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Difficulty:
95%
(hard)
Question Stats:
45% (02:21) correct
55%
(02:23)
wrong
based on 175
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If P is a positive integer, what is the reminder when P^2-1 is divided by 24?
(1) P is a prime number greater than 19
(2) P is an odd number that is not a multiple of 3
(1) P is a prime number greater than 19
(2) P is an odd number that is not a multiple of 3
22 Apr 2015, 04:09
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Any prime number \(p\) greater than 3 could be expressed as \(p=6n+1\) or \(p=6n+5\) (\(p=6n-1\)), where \(n\) is an integer >1.
That's because any prime number \(p\) greater than 3 when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3).
But:
Note that, not all number which yield a remainder of 1 or 5 upon division by 6 are primes, so vise-versa of above property is not correct. For example 25 (for \(n=4\)) yields a remainder of 1 upon division by 6 and it's not a prime number.
Now, back to the original question:
If P is a positive integer, what is the reminder when P^2-1 is divided by 24?
(1) P is a prime number greater than 19.
According to the above, P^2 - 1 = (6n - 1)^2 - 1 = 12n(3n - 1) or P^2 - 1 = (6n + 1)^2 - 1 = 12n(3n + 1). Since either n or 3n - 1 (3n + 1) is even, then 12n(3n - 1), as well as 12n(3n + 1), is a multiple of 24. Sufficient.
(2) P is an odd number that is not a multiple of 3 --> P could be 1, 5, 7, 11, 13, ... As you can see all of them are of the form of 6x - 1 or 6x + 1 (the same as above). So, the remainder is 0. Sufficient.
Answer: D.
Hope it's clear.
General Discussion
20 Apr 2015, 13:20
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1) remainder will be zero for all primes except 2 and 3, so sufficient
2) remainder will be zero for all numbers including 1, so sufficient
and answer is D.
