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Seven different numbers are selected from the integers 1 to [#permalink]
 28 Aug 2010, 02:58
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50% (01:57) correct
50% (00:31) wrong based on 0 sessions
Seven different numbers are selected from the integers 1 to 100, and each number is divided by 7. What is the sum of the remainders? (1) The range of the seven remainders is 6. (2) The seven numbers selected are consecutive integers.
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Re: Sum of Remainders [#permalink]
28 Aug 2010, 06:31
(1) The range of the seven remainders is 6.
Plugging numbers:
1,2,3,4,5,6,7 range - 6
remainder = 1,2,3,4,5,6,0 => sum of remainders= 21
7, 13, 14, 21, 28, 35, 42
remainder = 0,6,0,0,0,0,0. = range 6 sum of remainders = 6
Statement 1 not sufficient.
(2) The seven numbers selected are consecutive integers.
Plugging numbers:
1,2,3,4,5,6,7 range - 6
remainder = 1,2,3,4,5,6,0 => sum of remainders= 21
10,11,12,13,14,15,16 => remainder = 3,4,5,6,0,1,2 => sum of remainders = 21.
For any set of 7 consecutive integers divided by 7 the sum of remainders is always 21.
Statement 2 is sufficient.
answer is "B"
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Re: Sum of Remainders [#permalink]
28 Aug 2010, 07:07
udaymathapati wrote: Seven different numbers are selected from the integers 1 to 100, and each number is divided by 7. What is the sum of the remainders?
(1) The range of the seven remainders is 6.
(2) The seven numbers selected are consecutive integers. 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 remainder upon division of any integer by 7 can be: 0, 1, 2, 3, 4, 5, or 6 (7 values). (1) The range of the seven remainders is 6 --> if we pick 6 different multiples of 7 (all remainders 0) and the 7th number 6 (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 numbers selected are consecutive integers --> 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. Hope it helps.
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DS - Divisibility & Primes (600-700) [#permalink]
02 Nov 2010, 19:57
Seven different numbers are selected from the integers 1 to 100, and each number is divided by 7. What is the sum of the remainders? 1. The range of the seven remainders is 6. 2. The seven numbers selected are consecutive integers I've read the solutions both within the OG and within Manhattan GMAT's Official Guide Companion. I disagree with the answer they are presenting. If anyone can please explain the concept I'm missing I would greatly appreciate it! FYI, when figuring out this problem, I chose the following seven consecutive integers 3, 4, 5, 6, 7, 8, 9. Based on my knowledge the sum of the remainders when divided by 7 = 3 (0 + 0 + 0 + 0 + 0 + 1 + 2 = 3). It's getting late so perhaps I'm missing something simple here, but choosing these numbers, from my perspective, negates the explanation/answer given.
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Re: DS - Divisibility & Primes (600-700) [#permalink]
02 Nov 2010, 20:18
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cgl7780 wrote: Seven different numbers are selected from the integers 1 to 100, and each number is divided by 7. What is the sum of the remainders? 1. The range of the seven remainders is 6. 2. The seven numbers selected are consecutive integers I've read the solutions both within the OG and within Manhattan GMAT's Official Guide Companion. I disagree with the answer they are presenting. If anyone can please explain the concept I'm missing I would greatly appreciate it! FYI, when figuring out this problem, I chose the following seven consecutive integers 3, 4, 5, 6, 7, 8, 9. Based on my knowledge the sum of the remainders when divided by 7 = 3 (0 + 0 + 0 + 0 + 0 + 1 + 2 = 3). It's getting late so perhaps I'm missing something simple here, but choosing these numbers, from my perspective, negates the explanation/answer given. Red part in your solution is not correct. Positive integer a divided by positive integer d yields a reminder of r can always be expressed as a=qd+r, where q is called a quotient and r is called a remainder, note here that 0\leq{r}<d (remainder is non-negative integer and always less than divisor).So according to above, when positive integer a is less than divisor d then remainder upon division a by d is always equals to a, for example 5 divided by 10 yields reminder of 5. So when 1, 2, 3, 4, 5, or 6 is divided by 7 remainder is 1, 2, 3, 4, 5, and 6 respectively. Or algebraically: 3 divided by 7 can be expressed as 3=0*7+3, so r=3. For complete solution of this problem above posts. Hope it's clear.
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Re: Sum of Remainders [#permalink]
03 Nov 2010, 07:54
Thank you for the explanation, yes I believe my exhaustion was getting to me. I completely understand why I was calculating incorrect remainders. When we divide 3 by 7, 7 goes into 3, 0 times but there is three left over still, so r = 3. Very basic mistake on my part. So taking my original set (referencing statement 2): {3, 4, 5, 6, 7, 8, 9} Remainders would be; 3, 4, 5, 6, 0, 1, 2 Therefore yeilding; 3 + 4 + 5 + 6 + 0 + 1 + 2 = 21 Thank you very much for the explanation and for the reference to other explanations. + 1 Kudos!
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Re: Sum of Remainders
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03 Nov 2010, 07:54
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