N is an integer, r is the remainder when ((N-1)*(N+1)) is divided by 24. What is the value of r ?
1) N is not a multiple of 2.
2) N is not a multiple of 3.
Remainder
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1)N not multiple of 2
=>N=1, 0*3/24 remainder 0
=>N=3, 2*4/24 remainder 8
not sufficient
2) N not multiple of 3
=>N=2, 1*3/24 remainder 3
=>N=5, 4*6/24 remainder 0
not sufficient
together, N not a multiple of 2 and 3
=>N=1,5, 7, 11.. leave a remainder of 0
N=1, (N-1)(N+1)/24=0*2/24 remainder 0
N=5, (N-1)(N+1)/24= 4*6/24 remainder 0
sufficient, hence C
=>N=1, 0*3/24 remainder 0
=>N=3, 2*4/24 remainder 8
not sufficient
2) N not multiple of 3
=>N=2, 1*3/24 remainder 3
=>N=5, 4*6/24 remainder 0
not sufficient
together, N not a multiple of 2 and 3
=>N=1,5, 7, 11.. leave a remainder of 0
N=1, (N-1)(N+1)/24=0*2/24 remainder 0
N=5, (N-1)(N+1)/24= 4*6/24 remainder 0
sufficient, hence C
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Thanks SD.....scoobydooby wrote:1)N not multiple of 2
=>N=1, 0*3/24 remainder 0
=>N=3, 2*4/24 remainder 8
not sufficient
2) N not multiple of 3
=>N=2, 1*3/24 remainder 3
=>N=5, 4*6/24 remainder 0
not sufficient
together, N not a multiple of 2 and 3
=>N=1,5, 7, 11.. leave a remainder of 0
N=1, (N-1)(N+1)/24=0*2/24 remainder 0
N=5, (N-1)(N+1)/24= 4*6/24 remainder 0
sufficient, hence C
Is there a way to solve without using numbers ? Beacuse here calculation is simple, so we did the same....but what if calculation involves very complex numbers.....
Is there a general solution for the same ?
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For the sake of brevity, let's assume (N-1)(N+1) = x.
1. If N is not a multiple of 2, then both N-1 and N+1 are going to be even, so x will be the product of two even numbers, making it a multiple of 4. However, this is not enough to establish the remainder of x/24. Consider x = 16 and x = 20: we get different results for the two values.
2. If N is not a multiple of 3, then either N-1 or N+1 will be a multiple of 3 (there's that age-old rule that out of three consecutive numbers, one of them is a multiple of 3). This doesn't help with 24, however. Take x = 6 and x = 9: they are both divisible by 3, but yield different remainders when divided by 24.
If you put both statements together you get that N is not even and not a multiple of 3. The LCM f 2 and 3 is 6 and all numbers can be expressed according to the remainder they yield when divided by 6 like so:
6k - can't be N, since N is not div by 2
6k + 1
6k + 2 - can't be N, since N is not even
6k + 3 - can't be N, since N is not div by 3
6k + 4 - can't be N, since N is not even
6k + 5
So, out of all the numbers out there, N can be either 6k + 1 or 6k + 5. But watch what happens to x:
a. N = 6k + 1
N - 1 = 6k
N + 1 = 6k + 2
x = 6k(6k + 2) = 12k(3k + 1) - x is obviously divisible by 12, but it will also be divisible by 24. This is because no matter what k is, 3k + 1 will be the opposite of it: if k is even, then 3k + 1 will be odd.
You may wonder how could this be: well, 3k + 1 = (k + 1) + 2k. Adding an even number (2k) to another number (k + 1) will always make the sum even or odd depending on k + 1.
So k(3k + 1) = 2p (where p is an integer) will always be even, since one of its components will be even and the other will be odd.
Finally, x can be rewritten as x = 12 * 2 * p = 24p. x is divisible by 24.
b. N = 6k + 5
N - 1 = 6k + 4
N + 1 = 6(k + 1)
x = (6k + 4)*6*(k + 1) = 12(3k + 2)(k + 1). The demonstration why (3k + 2)(k + 1) will always be even is similar to the one above, since in this case we have (k + 1) and (3k + 1) + 1 instead of k and 3k + 1 respectively.
So again x = 12 * 2 * r = 24r. x is divisible by 24.
IMHO, really, really hard demonstration, or at least the way it look a it. I'm really curious whether someone could come up with a faster way of doing this except number picking.
1. If N is not a multiple of 2, then both N-1 and N+1 are going to be even, so x will be the product of two even numbers, making it a multiple of 4. However, this is not enough to establish the remainder of x/24. Consider x = 16 and x = 20: we get different results for the two values.
2. If N is not a multiple of 3, then either N-1 or N+1 will be a multiple of 3 (there's that age-old rule that out of three consecutive numbers, one of them is a multiple of 3). This doesn't help with 24, however. Take x = 6 and x = 9: they are both divisible by 3, but yield different remainders when divided by 24.
If you put both statements together you get that N is not even and not a multiple of 3. The LCM f 2 and 3 is 6 and all numbers can be expressed according to the remainder they yield when divided by 6 like so:
6k - can't be N, since N is not div by 2
6k + 1
6k + 2 - can't be N, since N is not even
6k + 3 - can't be N, since N is not div by 3
6k + 4 - can't be N, since N is not even
6k + 5
So, out of all the numbers out there, N can be either 6k + 1 or 6k + 5. But watch what happens to x:
a. N = 6k + 1
N - 1 = 6k
N + 1 = 6k + 2
x = 6k(6k + 2) = 12k(3k + 1) - x is obviously divisible by 12, but it will also be divisible by 24. This is because no matter what k is, 3k + 1 will be the opposite of it: if k is even, then 3k + 1 will be odd.
You may wonder how could this be: well, 3k + 1 = (k + 1) + 2k. Adding an even number (2k) to another number (k + 1) will always make the sum even or odd depending on k + 1.
So k(3k + 1) = 2p (where p is an integer) will always be even, since one of its components will be even and the other will be odd.
Finally, x can be rewritten as x = 12 * 2 * p = 24p. x is divisible by 24.
b. N = 6k + 5
N - 1 = 6k + 4
N + 1 = 6(k + 1)
x = (6k + 4)*6*(k + 1) = 12(3k + 2)(k + 1). The demonstration why (3k + 2)(k + 1) will always be even is similar to the one above, since in this case we have (k + 1) and (3k + 1) + 1 instead of k and 3k + 1 respectively.
So again x = 12 * 2 * r = 24r. x is divisible by 24.
IMHO, really, really hard demonstration, or at least the way it look a it. I'm really curious whether someone could come up with a faster way of doing this except number picking.
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Thanks Ian.
Can you please tell....how to solve the same in case statement 2 would have been "4 is not a factor of N"...
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Awesome DanaJ....real awesome explanation!!!!DanaJ wrote:For the sake of brevity, let's assume (N-1)(N+1) = x.
1. If N is not a multiple of 2, then both N-1 and N+1 are going to be even, so x will be the product of two even numbers, making it a multiple of 4. However, this is not enough to establish the remainder of x/24. Consider x = 16 and x = 20: we get different results for the two values.
2. If N is not a multiple of 3, then either N-1 or N+1 will be a multiple of 3 (there's that age-old rule that out of three consecutive numbers, one of them is a multiple of 3). This doesn't help with 24, however. Take x = 6 and x = 9: they are both divisible by 3, but yield different remainders when divided by 24.
If you put both statements together you get that N is not even and not a multiple of 3. The LCM f 2 and 3 is 6 and all numbers can be expressed according to the remainder they yield when divided by 6 like so:
6k - can't be N, since N is not div by 2
6k + 1
6k + 2 - can't be N, since N is not even
6k + 3 - can't be N, since N is not div by 3
6k + 4 - can't be N, since N is not even
6k + 5
So, out of all the numbers out there, N can be either 6k + 1 or 6k + 5. But watch what happens to x:
a. N = 6k + 1
N - 1 = 6k
N + 1 = 6k + 2
x = 6k(6k + 2) = 12k(3k + 1) - x is obviously divisible by 12, but it will also be divisible by 24. This is because no matter what k is, 3k + 1 will be the opposite of it: if k is even, then 3k + 1 will be odd.
You may wonder how could this be: well, 3k + 1 = (k + 1) + 2k. Adding an even number (2k) to another number (k + 1) will always make the sum even or odd depending on k + 1.
So k(3k + 1) = 2p (where p is an integer) will always be even, since one of its components will be even and the other will be odd.
Finally, x can be rewritten as x = 12 * 2 * p = 24p. x is divisible by 24.
b. N = 6k + 5
N - 1 = 6k + 4
N + 1 = 6(k + 1)
x = (6k + 4)*6*(k + 1) = 12(3k + 2)(k + 1). The demonstration why (3k + 2)(k + 1) will always be even is similar to the one above, since in this case we have (k + 1) and (3k + 1) + 1 instead of k and 3k + 1 respectively.
So again x = 12 * 2 * r = 24r. x is divisible by 24.
IMHO, really, really hard demonstration, or at least the way it look a it. I'm really curious whether someone could come up with a faster way of doing this except number picking.
Can someone tell if there is someother easier general approach as compared to the general approach suggested by DanaJ ?
Thanks
Mohit
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Yes, but what if statement #2 would have been ""4 is not a factor of N"...then how to solve.....DanaJ wrote:I think Ian's approach was a much better/faster one...
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Well, the fact that 4 is not a factor of N would not add much value if we consider both statements, since if a number is not even (statement 1: 2 is not a factor of N), then it will surely NOT BE a multiple of 4.
You could use the same approach as above: the LCM of 2 and 4 is 4, so every number can be written as:
4k - out since it's even
4k + 1
4k + 2 - out since it's even
4k + 3
So N can only be 4k + 1 or 4k + 3. Split it up into cases as above:
a. N = 4k + 1
N -1 = 4k
N + 1 = 4k + 2
x = 4k(4k+2)= 8k(2k +1) - here, you can't go any further, since 2k + 1 will always be odd, no matter what k is.
It's largely the same thing with b. N = 4k + 3.
Note that you won't probably see such a question on the real thing. They'd pick numbers like 2 and 3 or 2 and 5, that have GCD 1.
You could use the same approach as above: the LCM of 2 and 4 is 4, so every number can be written as:
4k - out since it's even
4k + 1
4k + 2 - out since it's even
4k + 3
So N can only be 4k + 1 or 4k + 3. Split it up into cases as above:
a. N = 4k + 1
N -1 = 4k
N + 1 = 4k + 2
x = 4k(4k+2)= 8k(2k +1) - here, you can't go any further, since 2k + 1 will always be odd, no matter what k is.
It's largely the same thing with b. N = 4k + 3.
Note that you won't probably see such a question on the real thing. They'd pick numbers like 2 and 3 or 2 and 5, that have GCD 1.
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In Ian's explanation, I have a question.Ian Stewart wrote:(1) says that n is odd. Thus n-1 and n+1 are both even. In fact, one of them must be divisible by 4, because every second even number is divisible by 4. So from (1) we know that (n-1)(n+1) is divisible by 8.unc42 wrote: If "n" is a positive integer and "r" is the 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) says that n is not divisible by 3. Notice that n-1, n, and n+1 are three consecutive integers. If you have any three consecutive integers, exactly one of them must be divisible by 3. So if n is not divisible by 3, either n-1 or n+1 is.
Using both (1) and (2) together, we know (n-1)(n+1) is divisible by both 3 and 8, and therefore by 24.
If we consider n to be 1, then n-1 = 0 and n+1 = 2. None of these 2 consecutive even integers are divisible by 4. How can we assume that for statement 1?
Am I missing something?
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0 is divisile by every integer including 4...since it leaves remainder 0.....I guess this might help.enniguy wrote:
In Ian's explanation, I have a question.
If we consider n to be 1, then n-1 = 0 and n+1 = 2. None of these 2 consecutive even integers are divisible by 4. How can we assume that for statement 1?
Am I missing something?
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Will be C.goelmohit2002 wrote:N is an integer, r is the remainder when ((N-1)*(N+1)) is divided by 24. What is the value of r ?
1) N is not a multiple of 2.
2) N is not a multiple of 3.
For any odd Integer greater than 3 the rem is 0.
Now if N is not a multiple of 3 then N is any odd integer greater than 3 which gives 0 as remainder.