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If n is a positive integer and r is the remainder when (n  1)(n + 1)
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21 Jan 2012, 17:57
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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) n is not divisible by 2 (2) n is not divisible by 3
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Re: If n is a positive integer and r is the remainder when (n  1)(n + 1)
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21 Jan 2012, 18:12
If n is a positive integer and r is the remainder when (n1)(n+1) is divided by 24, what is the value of r?Plugin method:\((n1)(n+1)=n^21\) (1) n is not divisible by 2 > pick two odd numbers: let's say 1 and 3 > if \(n=1\), then \(n^21=0\) and as zero is divisible by 24 (zero is divisible by any integer except zero itself) so remainder is 0 but if \(n=3\), then \(n^21=8\) and 8 divided by 24 yields remainder of 8. Two different answers, hence not sufficient. (2) n is not divisible by 3 > pick two numbers which are not divisible by 3: let's say 1 and 2 > if \(n=1\), then \(n^21=0\), so remainder is 0 but if \(n=2\), then \(n^21=3\) and 3 divided by 24 yields remainder of 3. Two different answers, hence not sufficient. (1)+(2) Let's check for several numbers which are not divisible by 2 or 3: \(n=1\) > \(n^21=0\) > remainder 0; \(n=5\) > \(n^21=24\) > remainder 0; \(n=7\) > \(n^21=48\) > remainder 0; \(n=11\) > \(n^21=120\) > remainder 0. Well it seems that all appropriate numbers will give remainder of 0. Sufficient. Algebraic approach:(1) n is not divisible by 2. Insufficient on its own, but this statement says that \(n=odd\) > \(n1\) and \(n+1\) are consecutive even integers > \((n1)(n+1)\) must be divisible by 8 (as both multiples are even and one of them will be divisible by 4. From consecutive even integers one is divisible by 4: (2, 4); (4, 6); (6, 8); (8, 10); (10, 12), ...). (2) n is not divisible by 3. Insufficient on its own, but form this statement either \(n1\) or \(n+1\) must be divisible by 3 (as \(n1\), \(n\), and \(n+1\) are consecutive integers, so one of them must be divisible by 3, we are told that it's not \(n\), hence either \(n1\) or \(n+1\)). (1)+(2) From (1) \((n1)(n+1)\) is divisible by 8, from (2) it's also divisible by 3, therefore it must be divisible by \(8*3=24\), which means that remainder upon division \((n1)(n+1)\) by 24 will be 0. Sufficient. Answer: C. Hope it's clear.
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Re: If n is a positive integer and r is the remainder when (n  1)(n + 1)
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31 Dec 2012, 21:18
Nadezda wrote: If n is positive integer and r is the remainder when (n1)(n+1) is divided by 24, what is the value of r?
(1) is not divisible by 2 (2) is not divisible by 3 Statement 1) When n is not divisible by 2, then n can be \(1, 3, 5, 7, 9 etc\) For n=1, the remainder is 0 For n=3, the remainder is 16. For n=5, the remainder is 0. Different answers. Hence insufficient. statement 2) When n is not divisible by 3, then n can be \(1,2, 4, 6 etc\) Here also different remainders. Insufficient. On combining these two statements, n is \(1,5, 7 etc\) For such numbers, the remainder is 0. Sufficient. +1C
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Re: If n is a positive integer and r is the remainder when (n  1)(n + 1)
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26 Sep 2012, 13:44
ksharma12 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) n is not divisible by 2.
(2) n is not divisible by 3. (1) n must be odd. Try n = 1 and n = 3. Not sufficient. (2) Try n = 1 and n = 2. Not sufficient. (1) and (2) together: When divided by 6, the remainders can be 0, 1, 2, 3, 4, 5. Since n is not divisible neither by 2, nor by 3, when divide by 6, n can give remainder 1 or 5, which means n is of the form 6k + 1 or 6k  1, for some positive integer k. If n = 6k + 1, then (n  1)(n + 1) = 6k(6k + 2) = 12k(3k + 1) = 12k(k + 2k + 1). Since 2k + 1 is odd, k and 3k + 1 are of different parities, so their product is for sure divisible by 2. Altogether, (n  1)(n + 1) is divisible by 24. If n = 6k  1, then (n  1)(n + 1) = (6k  2)6k)= 12k(3k  1) = 12k(k + 2k  1). Since 2k  1 is odd, k and 3k  1 are of different parities, so their product is for sure divisible by 2. Altogether, (n  1)(n + 1) is divisible by 24. The remainder must be 0. Sufficient. Answer C.
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Re: If n is a positive integer and r is the remainder when (n  1)(n + 1)
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12 Jan 2013, 03:34
kiyo0610 wrote: If n is a positive integer and r is the remainder when (n1)(n+1)is divided by 24, what is the value of r?
(1) n is not divisible by 2 (2) n is not divisible by 3 n1,n, n+1 are consecutive +ve intigers, and thus if n is even both n1,n+1 are odd and vice versa. also in every 3 consecutive numbers we get one that is a multiple of 3 from 1 n is odd thus both n1, n+1 are even and their product has at least 2^3 as a factor however if n = 3 thus n1,n+1 are 2,4 and since , 24 = 2^3*3 , thus reminder is 3 but if n = 5 for example thus n1,n+1 are 4,6 and therofre in this case r = 0.....insuff from 2 n is a multiple of 3 and thus both n1,n+1 are either even (e.g: n=3) or odd (n=6) and therfore this is insuff both together n is odd and is a multiple of 3 and therfore the reminder of the product (n1)(n+1) when devided by 24 is always 3..suff C



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Re: If n is a positive integer and r is the remainder when (n  1)(n + 1)
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25 Mar 2013, 21:02
vbodduluri wrote: If n is a positive integer and r is the remainder when (n1)(n+1) is divided by 24, what is the value of r? a)n is not divisible by 2 b) n is not divisible by 3 A few things: 1. Exactly one of any two consecutive positive integers is even. 2. Exactly one of any three consecutive positive integers must be a multiple of 3 3. Exactly one of any four consecutive positive integers must be a multiple of 4 etc Check this post for the explanation: http://www.veritasprep.com/blog/2011/09 ... cormath/a) n is not divisible by 2 Since every alternate number is divisible by 2, (n1) and (n+1) both must be divisible by 2. Since every second multiple of 2 is divisible by 4, one of (n1) and (n+1) must be divisible by 4. Hence, the product (n1)*(n+1) must be divisible by 8. But if n is divisible by 3, then neither (n1) nor (n+1) will be divisible by 3 and hence, when you divide (n1)(n+1) by 24, you will get some remainder. If n is not divisible by 3, one of (n1) and (n+1) must be divisible by 3 and hence the product (n1)(n+1) will be divisible by 24 and the remainder will be 0. Not sufficient. b) n is not divisible by 3 We don't know whether n is divisible by 2 or not. As discussed above, we need to know that to figure whether the product (n1)(n+1) is divisible by 8. Hence not sufficient. Take both together, we know that (n1)*(n+1) is divisible by 8 and one of (n1) and (n+1) is divisible by 3. Hence, the product must be divisible by 8*3 = 24. So r must be 0. Sufficient. Answer (C) Another approach:(n1), n and (n+1) are consecutive integers. If 2 is not a factor of n (i.e. n is odd), it must be a factor of (n1) and (n+1) (the numbers around n must be even). Also, out of any two consecutive even numbers, one has to be divisible by 4 because every alternate multiple of 2 is divisible by 4. Hence, (n1)*(n+1) must be divisible by 8. Out of any 3 consecutive integers, one has to be divisible by 3 because every third integer is a multiple of 3. Out of (n1), n and (n+1), one has to be divisible by 3. If n is not divisible by 3, one of (n1) and (n+1) must be divisible by 3. Hence, (n1)*(n+1) must be divisible by 3. Using both statements, (n1)*(n+1) must be divisible by 8*3 = 24. Remainder must be 0. Answer C
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Re: If n is a positive integer and r is the remainder when (n  1)(n + 1)
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Updated on: 26 Mar 2013, 02:15
vbodduluri wrote: If n is a positive integer and r is the remainder when (n1)(n+1) is divided by 24, what is the value of r? a)n is not divisible by 2 b) n is not divisible by 3 Given: (n1)(n+1) = 24m + r  (1) where m=1,2,3... Statement 1: n = 2w + x  (2) Statement 1 alone is not sufficient Statement 2: n= 3y + z  (3) Statement 2 alone is not sufficient. Taken together: 1. x has to be 1 and z can be 1 or 2 2. When x and z are 1, the values of n are 7, 13, 19 etc 3. When x=1 and z=2, the values of n are 5, 11, 17 etc 4. Substitute one of the above values, say 5 in (1) 5. For n=5 we have 4*6 = 24m + r or 24m +r = 24 r=0 We will get the same value of r for the other values of n too. Therefore answer is choice C.
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Re: If n is a positive integer and r is the remainder when (n  1)(n + 1)
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31 Jan 2017, 17:12
1) n not divisible by 2=> n is odd=> (n1) and (n+1) must be consective even numbers. if n=1, 0*2/24 leaves remainder 0 if n=3, 2*4/24 leaves remainder 8 not sufficient 2) n not divisible by 3=> n can be even or 1, 5, 7, 11, 13.... if n=5, 4*6/24 leaves remainder 0 if n=2, 1*3/24 leaves remainder 3 not sufficient together, n must be odd and not divisible by 3=> n can be 1, 5, 7, 11, 13... if n=7, 6*8/24 leaves remainder 0 if n=11, 10*12/24 leaves remainder 0 hence C.
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Re: If n is a positive integer and r is the remainder when (n  1)(n + 1)
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11 Sep 2017, 06:28
BunuelVeritasPrepKarishmaWhat if n=1? If so, (n1) will = 0. Please advise. Thank you.



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Re: If n is a positive integer and r is the remainder when (n  1)(n + 1)
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11 Sep 2017, 06:32
Eff wrote: BunuellakshmiWhat if n=1? If so, (n1) will = 0. Please advise. Thank you. 0 is divisible by every positive integer.
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Re: If n is a positive integer and r is the remainder when (n  1)(n + 1)
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14 Sep 2017, 03:01
By combining both the statement, we know that if a number is a prime number apart from 2, and 3, then it can be written in the form of (6n+1) or (6n1)  plugging these values, we get that (n1)(n+1) is always a multiple of 24; hence sufficient. Answer C.
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If n is a positive integer and r is the remainder when (n  1)(n + 1)
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02 Jun 2018, 18:09
enigma123 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) n is not divisible by 2 (2) n is not divisible by 3 * n1, 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 n1 or n+1 is. * If, from Statement 1, n is odd, then n1 and n+1 are even. Since every second even number is a multiple of 4, one of n1 and n+1 is a multiple of 4 (at least) and the other a multiple of 2. So (n1)(n+1) is a multiple of 8. So with both Statements together we know that (n1)(n+1) is a multiple of 3 and 8, and therefore of 24, and the remainder is zero.
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Re: If n is a positive integer and r is the remainder when (n  1)(n + 1)
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15 Jun 2018, 08:30
enigma123 wrote: If n is a positive integer and r is the remainder when (n1)(n+1) is divided by 24, what is the value of r?
(1) n is not divisible by 2 (2) n is not divisible by 3
Given , n > 0, r is remainder when (n1)(n+1) is divided by 24, r = ? (n1)(n+1) is the product two consecutive even or odd integers, depending on whether n is odd or even. Statement 1: n is not divisible by 2 n = odd, then we have, for n = 3, (n1)(n+1) = (31)(3+1) = 8, hence r = 8 for n = 5, (n1)(n+1) = (51)(5+1) = 4*6 = 24, hence r = 0 Statement 1 alone is Not Sufficient. Statement: n is not divisible by 3. hence, for n = 5, we have from above r = 0 & for n = 8, we have (n1)(n+1) = (81)(8+1) = 7*9 = 63, hence r = 15 Statement 2 alone is Not Sufficient. Combining the two statements, we get, n is not divisible by 2 or 3 for n = 7, (n1)(n+1) = (71)(7+1) = 6*8 = 48, hence r = 0 for n = 11, (n1)(n+1) = (111)(11+1) = 10*12 = 120, hence r = 0 Both Statements together are Sufficient. Answer C. Thanks, GyM
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Re: If n is a positive integer and r is the remainder when (n  1)(n + 1)
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25 Jun 2019, 09:57
Statement 1) When n is not divisible by 2, then n can be 1,3,5,7,9 For n=1,3 & 5 , Different answers. Hence insufficient.
statement 2) When n is not divisible by 3, then n can be 1,2,4,etc Here also different remainders. Insufficient.
On combining these two statements, n is 1,5,7 For such numbers, the remainder is 0.
C is Correct



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