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Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]
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04 Dec 2008, 13:31
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Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked at random from the set of all integers between 10 and 110, inclusive. If each of these integers is divided by 7 and the 7 remainders are all added together, what would be the sum of the 7 remainders?
(1) The range of the remainders is 6. (2) The seven integers are consecutive.
Clue 1 - Range is the difference between the largest and the smallest numbers. Possible remainders when a number is divided by 7 is 0 thru 6. SO range is 6. This can be derived from the question itself. So Clue 1 is the same derivation made is explicit. With this we cannot say what numebers are choosen and what would be ther remainders and in turn the sum. S0 Clue 1 is INSUFFICIENT.
Clue 2 - It says that the numbers are consecutive. This clearly tells that one number is divisible by 7. So remainder is 0. And the rest of the 6 numbers should leave the remainders from 1 thru 6. From this we can get the sum of the remainders. So Clue 2 is sufficient.
Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked at random from the set of all integers between 10 and 110, inclusive. If each of these integers is divided by 7 and the 7 remainders are all added together, what would be the sum of the 7 remainders?
(1) The range of the remainders is 6.
(2) The seven integers are consecutive.
I'm not sure where to begin with this question.
first step: xi = 7*ki+ri where i=1 to 7. sum of ri=? (1) max-min (r1,....,r6)=6 (2) xi is such that x(i+1) = xi+1 So, x1 = 7k1+r1 x2 = x1+1=7k1+r1+1 x3 = x2+1 = 7k1+r1+2 x7 = 7k1+r1+6. sum of remainders = 7r1+1+2+3+4+5+6 = 7r1+21 one remainder is zero. so r1 can be solved for. hence B.
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Re: Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]
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31 Mar 2013, 02:15
I didnt get why (1) is wrong. I think all 7 integers are different. that is why if the range is 6, then there are consecutive integers that match the criteria. how is it possible to have the following, if 7 integers are different and the range of the set is 6? could you please write down any set that match this criteria?
0,1,2,3,4,5,6 -- range (6-0)=6 0,0,0,0,0,0,6 -- range (6-0)=6 0,6,6,6,6,6,6 -- range (6-0)=6
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I didnt get why (1) is wrong. I think all 7 integers are different. that is why if the range is 6, then there are consecutive integers that match the criteria. how is it possible to have the following, if 7 integers are different and the range of the set is 6? could you please write down any set that match this criteria?
0,1,2,3,4,5,6 -- range (6-0)=6 0,0,0,0,0,0,6 -- range (6-0)=6 0,6,6,6,6,6,6 -- range (6-0)=6
For example, {14, 21, 28, 35, 42, 49, 55} --> remainders {0, 0, 0, 0, 0, 0, 6} --> range 6 --> the sum of the remainders 6.
Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked at random from the set of all integers between 10 and 110, inclusive. If each of these integers is divided by 7 and the 7 remainders are all added together, what would be the sum of the 7 remainders?
The trick here is to know that remainder is always non-negative integer less than divisor \(0\leq{r}<d\), so in our case \(0\leq{r}<7\).
So, the remainder upon division of any integer by 7 could be: 0, 1, 2, 3, 4, 5, or 6 (7 values).
(1) The range of the remainders is 6 --> if we pick 6 different multiples of 7 (all remainders 0) and the 7th number is 20 (remainder 6) then the range would be 6 and the sum also 6. But if we pick 7 consecutive integers then we'll have all possible remainders: 0, 1, 2, 3, 4, 5, and 6 and their sum will be 21. Not sufficient.
(2) The seven integers are consecutive. ANY 7 consecutive integers will give us all remainders possible: 0, 1, 2, 3, 4, 5, and 6. It does not matter what the starting integer will be: if it's say 11 then the remainder of 7 consecutive integers from 11 divided by 7 will be: 4, 5, 6, 0, 1, 2, and 3 and if starting number is say 14 then the remainder of 7 consecutive integers from 14 divided by 7 will be: 0, 1, 2, 3, 4, 5 and 6. So in any case sum=0+1+2+3+4+5+6=21. Sufficient.
Re: Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]
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07 Jun 2013, 01:07
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This post received KUDOS
Mspixel wrote:
Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked at random from the set of all integers between 10 and 110, inclusive. If each of these integers is divided by 7 and the 7 remainders are all added together, what would be the sum of the 7 remainders?
(1) The range of the remainders is 6. (2) The seven integers are consecutive.
(1) seven numbers could be all different and the remainders also could be different. For example, if we have 13, 14, 21, 28, 35, 42, 56 the range of the remainders is 6 (6, 0, 0, 0, 0, 0, 0) the sum is also 6. However if we have 10, 11, 12, 13, 14, 15, 16 the range is still 6 but the sum is greater. The statement is not sufficient.
(2) if we have 10, 11, 12, 13, 14, 15, 16 the remainders will be 3, 4, 5, 6, 0, 1, 2 it is easy to see that there is a pattern of remainders, since the numbers are consecutive their remainders will always have a pattern from 0 to 6. Sufficient, thus the answer is B.
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Re: Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]
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30 Sep 2014, 21:15
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Re: Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]
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13 Oct 2015, 00:00
Hello from the GMAT Club BumpBot!
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked at random from the set of all integers between 10 and 110, inclusive. If each of these integers is divided by 7 and the 7 remainders are all added together, what would be the sum of the 7 remainders?
(1) The range of the remainders is 6. (2) The seven integers are consecutive.
There are 7 variables, so we need 7 equations in order to solve the problem; only 2 equations are given, so there is high chance (E) will be the answer. When we look at the conditions together, the remainders become 1,2,3,4,5,6,0 so the sum becomes 1+2+3+4+5+6=21, a unique answer. The condition is sufficient, so the answer seems to be (C), but this is a question with commonly made mistakes in 4(A) (Common Mistake Type 4 4(A)). If we look at the conditions separately, Condition 1 gives 1,2,3,4,5,6,7==> 1+2+3+4+5+6+0=21, but this also works for 7,14,21,35,42,48,49==>0+0+0+0+0+6+0=6, so this condition is not unique, and insufficient by itself. Condition 2, on the other hand, the sum of the remainders becomes 1+2+3+4+5+6+0=21, so this condition is unique and therefore sufficient. So the answer becomes (B).
This type of question is a common type in today GMAT math
For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]
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14 Nov 2016, 02:25
Hello from the GMAT Club BumpBot!
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Re: Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]
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21 Dec 2016, 17:42
Excellent Question Here is what i did in this one => We need the sum of remainders Statement 1=> Clearly not sufficient. It just tells us that there will be attest one number leaving a remainder 6 ad one number that is leaving a remainder zero.
Hence not sufficient
Statement 2--> 7 consecutive integers. Forget the boundary 10,110 => If we pick any 7 consecutive integers =>WE will get all the remainders-> {0,1,2,3,4,5,6}Hence sufficient
Hence B _________________
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Re: Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked
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21 Dec 2016, 17:42
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