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If n is an integer, what is the remainder when n is divided [#permalink]
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31 May 2009, 10:07
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If n is an integer, what is the remainder when n is divided by 7? (1) n+1 is divisible by 7 (2) n+8 is divisible by 7
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Last edited by Bunuel on 12 Oct 2013, 09:19, edited 1 time in total.
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Re: remainder [#permalink]
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31 May 2009, 10:16
Question: what is the remainder when n is divided by 7 ?
(1) if n+1 has a remainder of 0 when divided by 7, then, if positive, n has a remainder of six, but, if negative, it has a remainder of one. insufficient
(2) if n+8 has a remainder of 0 when divided by 7, the same discrepancy between negative and positive numbers occurs. insufficient.
(1+2) if n+8 and n+1 have remainders of 0, then if n is positive, n still has a remainder of six, and, negative, n still has a remainder of one. Either is possible. insufficient.
The answer, therefore, is E.



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Re: remainder [#permalink]
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31 May 2009, 10:33
Agree,
If N = 13, Remainder = 6
If N = 15, Remainder = 1
Therefore, Insufficient.



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Re: remainder [#permalink]
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31 May 2009, 18:02
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vcbabu wrote: If n is an integer, what is the remainder when n is divided by 7?
1) n+1 is divisible by 7 2) n+8 is divisible by 7 D. Reminder(s) can never be negative but is(are) always: 0 <= r =< 7. Lets say n = ax + r, where a = 7, x is quotient, and r is reminder. Or, n = 7x + r Then in each case above, r = 6. 1) If n = 7x + r, n+1 = (7x+r) + 1. If so, r must be 6. Suff............ If "n= 7x + r" is ve, x has to be ve. Then n +1 = (7x + r) + 1 If suppose x = 1, n+1 = 7(1) + r + 1 = 6+r. What has to be r to have (n+1) divisible by 7? r = +6. Somebody might say 1 but remember r can never be ve. So what is the minimum r can be 6 because r must be >0 but smaller than 7. 2) If n = 7x + r, n+8 = (7x+r) + 8. Or, n+8 = 7(x+1) + r +1 Now the equation is similar to the eq. in 1. Therefore r = 6 again. Suff................ So d.
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Re: remainder [#permalink]
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31 May 2009, 19:22
yes but even if the remainder is always positive, but either expression could make the remainder either 1 or 6. n is not necessarily always positive, and if n+1 is divisible by 7, n could easily be 50 or 48, giving you two different remainders, whether they are 1<= r <= 7 .
Still say E.



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Re: remainder [#permalink]
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31 May 2009, 20:24
dk94588 wrote: yes but even if the remainder is always positive, but either expression could make the remainder either 1 or 6. n is not necessarily always positive, and if n+1 is divisible by 7, n could easily be 50 or 48, giving you two different remainders, whether they are 1<= r <= 7 .
Still say E. Still D even if n = 50 or 48 because r = 6 in either case. n +1 = 7(8) + (r + 1) where (r+1) = 7 or r = 6. n +1 = 56 + 7 n = 491 = 50 Simple logic, if n+1 is divided by 7 leaves reminder 0, then n/7 must have 6 reminder no matter n is +ve or ve..
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Re: remainder [#permalink]
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31 May 2009, 21:32
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D Question: (N/7 remainder)? (1) (N+1)/7 = X, where X is any integer N+1=7X N = 7X1 = 7(X1)+71 = 7(X1) + 6 Remainder of 6. (2) Same thing as 2, (N+8)/7 = X, where X is any integer N+8=7X N = 7X8 = 7X  7  1 = 7(X1)  1 = 7X  1 (same as A) Final Answer, \(D\).
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Re: remainder [#permalink]
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31 May 2009, 21:36
Although there is a better method, my Finding The Pattern method when in doubt, FTP!!!!!!!! Because there is always a pattern. Question: (N/7 remainder?) (1)N+1 is divisible by 7 N=6,13,20,27,34,... Question=6,6,6,6,6,...... Sufficient (2)N+8 is divisible by 7 N=6,13,20,27,34,... Question=6,6,6,6,6... Sufficient General rule of thumb generate the possible values and check the remainder really fast... if it changes insufficient, otherwise sufficient.
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Re: remainder [#permalink]
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01 Jun 2009, 13:15
I obviously need to brush up on division skills I guess.
pls explain how 50/7 has a remainder 6.
wouldn't it be 7 Remainder 1, because it divides seven times and has a remainder of 1, or would it be 8 Remainder 6, since it divides 8 times (56) and has a positive 6 remainder?



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Re: remainder [#permalink]
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01 Jun 2009, 14:20
Generally a remainder has to be between 0 & what you're dividing byit can be negative, but you can always convert it to a positive integer. So as a rule of thumb on the GMAT, work with positive remainders. If you do get a negative remainder you can always convert it to a positive one. For example, what's the remainder when 83 is divided by 3? 83 = 3*(27)  2 or 83 = 3*(28) + 1 The remainder is 2 or 1.
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Re: remainder [#permalink]
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01 Jun 2009, 14:31
dk94588 wrote: I obviously need to brush up on division skills I guess.
pls explain how 50/7 has a remainder 6.
wouldn't it be 7 Remainder 1, because it divides seven times and has a remainder of 1, or would it be 8 Remainder 6, since it divides 8 times (56) and has a positive 6 remainder? pls explain how 50/7 has a remainder 6. Q & R are integers 50/7 = 7*(6)  8 50/7 = 7*(7)  1 50/7 = 7*(8) + 6 50/7 = 7*(9) + 13 But generally we want the remainder to be \(0<=R<7\), so we'd go with remainder of 6.
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Re: remainder [#permalink]
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01 Jun 2009, 18:25
so in data sufficiency problems, we do assume that the remainder is the lowest possible positive integer?
because if it could be 1 then it would be insufficient



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Re: remainder [#permalink]
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01 Jun 2009, 19:49
goldeneagle94 wrote: Hades wrote: Generally a remainder has to be between 0 & what you're dividing byit can be negative, but you can always convert it to a positive integer. So as a rule of thumb on the GMAT, work with positive remainders.
If you do get a negative remainder you can always convert it to a positive one.
For example, what's the remainder when 83 is divided by 3?
83 = 3*(27)  2
or
83 = 3*(28) + 1
The remainder is 2 or 1. Does the OG has this rule in it ? I just checked, it doesn't look like it's in the Quant review. I've never seen a remainder question where you're given negative numbers...
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Re: remainder [#permalink]
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03 Jun 2009, 00:22
Good guestion..
Agree the answer is D..
n+1 and n+8 will leave the same remainder when divided by 7..
and if n+1 is evenly divided by 7, then the n will defnitely remain a remainder of 6 i.e falling short of 1.



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Re: remainder [#permalink]
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03 Jun 2009, 05:28
Wikipedia has an interesting article on remainder. http://en.wikipedia.org/wiki/RemainderIt says Remainder can have negative values as well, unless specified by the condition x < r ≤ x+d (or x ≤ r < x+d), where x is a constant, d is the divisor, and r is the remainder.



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Re: remainder [#permalink]
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18 Jun 2009, 09:19
Hades wrote: dk94588 wrote: I obviously need to brush up on division skills I guess.
pls explain how 50/7 has a remainder 6.
wouldn't it be 7 Remainder 1, because it divides seven times and has a remainder of 1, or would it be 8 Remainder 6, since it divides 8 times (56) and has a positive 6 remainder? pls explain how 50/7 has a remainder 6. Q & R are integers 50/7 = 7*(6)  8 50/7 = 7*(7)  1 50/7 = 7*(8) + 6 50/7 = 7*(9) + 13 But generally we want the remainder to be \(0<=R<7\), so we'd go with remainder of 6. As you have just shown, ve numbers can be written in two ways (actually you have shown more than two) Given that the Q does not tell whether n is positive or negative or whether the remainder is exclusively positive to eliminate some of the candidates. we have to consider both cases. All that we need with DS is one case to make it E. I believe that there exists such a case.



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Re: Remainder of n [#permalink]
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18 Jun 2009, 12:25
It's the same same remainder, whether it's 6 or +1 as I explained above, and to make it simpler the standard convention is to express it between 0 & the divisor. If your logic were correct then every single remainder question would be E on the GMAT as there is a billion different ways of rewriting it.
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Re: Remainder of n [#permalink]
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18 Jun 2009, 13:03
Hades wrote: It's the same same remainder, whether it's 6 or +1 as I explained above, and to make it simpler the standard convention is to express it between 0 & the divisor.
If your logic were correct then every single remainder question would be E on the GMAT as there is a billion different ways of rewriting it. K, I guess you got me wrong. When ever we divide a number with x the remainder is between 0 and x1. This is the general rule of thumb, Which I am not debating. How ever, the concepts get dicey if n is just an integer and not positive integer and happens to be a ve integer as it is here. To be clear, I don t consider all re writes of a ve integer. I only consider two of them the one with ve remainder and +ve remainder. With out much fuss, I will agree that we need to know what GMAT thinks if an integer is not explicitly stated as positive.



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Re: remainder [#permalink]
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21 Jun 2009, 10:40
goldeneagle94 wrote: Wikipedia has an interesting article on remainder. http://en.wikipedia.org/wiki/RemainderIt says Remainder can have negative values as well, unless specified by the condition x < r ≤ x+d (or x ≤ r < x+d), where x is a constant, d is the divisor, and r is the remainder. Several points here: First, as that wikipedia article makes clear, mathematicians do not allow remainders to be negative (see where they say "as is usual for mathematicians", when giving the positive solution). ; So, when you divide an integer by 7, there are only seven possible remainders: 0, 1, 2, 3, 4, 5 and 6. Remainders can't be negative, they can't be decimals, and they can't be greater than or equal to what you're dividing by; I have never seen a real GMAT question that asks about remainders when dividing by a negative number. Nor have I seen questions which ask for the remainder when dividing a negative by a positive number; Still, we can see how to find the remainder when dividing, say, 15 by 7. First, we find the remainder when dividing, for example, 20 by 7 as follows: we find the nearest multiple of 7 which is *smaller* than 20. That's 14; we subtract: 20  14 = 6. That is, 20 is 6 larger than the nearest smaller multiple of 7, so the remainder is 6 when we divide 20 by 7; Doing the same for 15: the nearest multiple of 7 which is *smaller* than 15 is 21; now, 15  (21) = 6. That is, 15 is 6 greater than the nearest multiple of 7 which is smaller than 15, and the remainder is 6 when 15 is divided by 7. So, going back to the original question in this post, the remainder is the same regardless of whether n is positive or negative, and the answer is D.
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Re: Remainder of n [#permalink]
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23 Oct 2009, 00:56
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D for me
n = 7a + r (1) n+1 = 7a + r + 1. Because n + 1 and 7a are divisible by 7, r+1 must be divisible by 7. Thus, r = 6
(2) n+8 = 7a + r + 8. Because n + 8 and 7a are divisible by 7, r+8 must be divisible by 7. r + 8 have the same remainder as r + 1 and thus r = 6




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