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Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]
 04 Dec 2008, 13:31
Question Stats:
43% (02:10) correct
56% (01:09) wrong based on 2 sessions
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.
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Re: Remainders Question from MGMAT [#permalink]
04 Dec 2008, 13:46
its a good question, I would go with B..
1) range tells me that maybe i got 34/7 etc...maybe the rest all multiples of 7 who knows..
2) tells us we got consecutive us , we got a series that repeats itself every 7 numbers..so thats suff
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Re: Remainders Question from MGMAT [#permalink]
05 Dec 2008, 10:46
My answer will be B. Here is the explanation.
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.
Any OA?
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Re: Remainders Question from MGMAT [#permalink]
05 Dec 2008, 23:20
y nt D.
if the range is 6.
then the nos must be consecutive..
7 - 1 = 6
27 - 21 = 6,etc
now v knw range , v knw nos r consecutive...
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Re: Remainders Question from MGMAT [#permalink]
06 Dec 2008, 02:48
mrsmarthi wrote: My answer will be B. Here is the explanation.
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.
Any OA? Yes, once an integer is divided by 7, the remainder will be either (0,1,2,3,4,5,6) - couldn't be greater than 6.
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Re: Remainders Question from MGMAT [#permalink]
24 Jan 2009, 11:48
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.
I'm not sure where to begin with this question. Stmt1: At least one 0; and at least one 6. Insuff. Stmt2: 0+1+2+3+4+5+6=21 Suff. So,B.
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Re: Remainders Question from MGMAT [#permalink]
24 Jan 2009, 16:16
bindrakaran001 wrote: y nt D.
if the range is 6.
then the nos must be consecutive..
7 - 1 = 6
27 - 21 = 6,etc
now v knw range , v knw nos r consecutive... Range will give you difference between highest and lowest numbers for e.g. 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 multiple solutions : satement1 is not suffcieint B is suffcieint.
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Re: Remainders Question from MGMAT [#permalink]
24 Jan 2009, 17:07
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.
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]
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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Re: Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]
31 Mar 2013, 08:14
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LalaB wrote: 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. Answer: B. Similar questions to practice: seven-different-numbers-are-selected-from-the-integers-1-to-99943.htmln-consecutive-integers-are-selected-from-the-integers-131349.htmlHope it helps.
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Re: Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked
[#permalink]
31 Mar 2013, 08:14
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