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number prop [#permalink] New post 15 Sep 2009, 00:41
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If N is positive integer & r is remainder when (n-1)(n+1) is divided by 24, what is the value of r
(1) 2 is not a factor of n
(2) 3 is not a factor of n



2-min solution?
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Re: number prop [#permalink] New post 15 Sep 2009, 03:49
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thailandvc wrote:
If N is positive integer & r is remainder when (n-1)(n+1) is divided by 24, what is the value of r
(1) 2 is not a factor of n
(2) 3 is not a factor of n



2-min solution?


* n-1, n and n+1 are three consecutive integers, so one of them must be a multiple of 3, since every third integer is a multiple of 3. From Statement 2, n is not a multiple of 3, so one of n-1 or n+1 is.

* If, from Statement 1, n is odd, then n-1 and n+1 are even. Since every second even number is a multiple of 4, one of n-1 and n+1 is a multiple of 4 (at least) and the other a multiple of 2. So (n-1)(n+1) is a multiple of 8.

So with both Statements together we know that (n-1)(n+1) is a multiple of 3 and 8, and therefore of 24, and the remainder is zero.
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Re: number prop [#permalink] New post 15 Sep 2009, 02:36
thailandvc wrote:
If N is positive integer & r is remainder when (n-1)(n+1) is divided by 24, what is the value of r
(1) 2 is not a factor of n
(2) 3 is not a factor of n



2-min solution?


(n-1).(n+1)=n^2-1
if divided by 24, remainder is r.
1)insuff take any number that is not divisible by 2 and you get a lot of differents r values.
for 3, r=8
for 5, r=0
for 7, r=0
for 9, r=8

2)insuff take any number that is not divisible by 3 and you get a lot of different r values
for 4, r=15
for 5, r=0
for 7, r=0
for 8, r=15

Together
sufficient. By 2 ways.
1-As you see, numbers that are not divisible by 2 or 3 gives the remainder 0.
2-One must work on modul 6 (6=2x3).
numbers equal to 0 in modul 6 can be divided both by 2 and 3
numbers equal to 1 in modul 6 can not be divided either by 2 or 3
numbers equal to 2 in modul 6 can be divided by 2 but not by 3
numbers equal to 3 in modul 6 can be divided by 3 but not by 2
numbers equal to 4 in modul 6 can be divided by 2 but not by 3
numbers equal to 5 in modul 6 can not be divided either by 2 or 3

So the numbers that are equal to 1 or 5 in modul 6 can not be divided by 2 or 3.
These numbers also can be 1, 5, 7, 11,13,17,19,23 in modul 24.
So
n^2-1=0 in modul 24 as you see for all of those numbers. (1,5,7,11,13,17,19,23).
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Re: number prop [#permalink] New post 15 Sep 2009, 06:42
IanStewart wrote:
thailandvc wrote:
If N is positive integer & r is remainder when (n-1)(n+1) is divided by 24, what is the value of r
(1) 2 is not a factor of n
(2) 3 is not a factor of n



2-min solution?


* n-1, n and n+1 are three consecutive integers, so one of them must be a multiple of 3, since every third integer is a multiple of 3. From Statement 2, n is not a multiple of 3, so one of n-1 or n+1 is.

* If, from Statement 1, n is odd, then n-1 and n+1 are even. Since every second even number is a multiple of 4, one of n-1 and n+1 is a multiple of 4 (at least) and the other a multiple of 2. So (n-1)(n+1) is a multiple of 8.

So with both Statements together we know that (n-1)(n+1) is a multiple of 3 and 8, and therefore of 24, and the remainder is zero.

Very very good solution. Impressed!
Re: number prop   [#permalink] 15 Sep 2009, 06:42
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