Seven different numbers are selected from the integers 1 to

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 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.

Similar question to practice: n-consecutive-integers-are-selected-from-the-integers-131349.html

Hope it helps.
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(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"

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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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!

The first thing to note here is that when a number is divided by 7, it can give one of the following numbers as remainder 0/1/2/3/4/5/6. There can be no other remainder.
Stmnt 1: This implies that one remainder has to be 0 and another has to be 6. That is how we will get the range (greatest - smallest) = 6-0 = 6. But we do not know anything about the other remainders. They could be 1,1,1,1,1 or 1,2,2,6,6 etc. Hence the sum of the remainders is not known.

Stmnt 2: Note that when I divide 10 by 7, I get 3 as remainder. When I divide 11 by 7, I get 4 as remainder. When I divide 12 by 7, I get 5 as remainder. So when I divide consecutive integers by the same number, the remainders will also be consecutive (I will get 3, 4, 5, 6 and then back to 0, 1, 2 as remainders). If the seven numbers selected are consecutive, the remainders will also be consecutive i.e. they will be 0, 1, 2, 3, 4, 5, 6 (or 3, 4, 5, 6, 0, 1, 2 ) or any such sequence. In any case, the sum of the remainders will be 0+1+2+3+4+5+6 = 21. Sufficient.
Answer (B)
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How is the range = 6?

The range of a set is the difference between the largest and smallest elements of a set. So, the range of {0, 6, 0, 0, 0, 0, 0} is 6 - 0 = 0.
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stat 1 : range = highest - lowest
there is a possibilty that one of the remainders is 0 and others are 6..
and lot of other cases..

stat 2: in consecutive 7 numbers.. one number will be divisible by 7,, rest will not be,,
hence the remainders will be 0,1,2,3,4,5,6

ans B

Hello everyone..
I have just started preparing for GMAT. Pardon me for the silly doubt i am about to ask..
My doubt is not particular to this question..
Can anyone please guide me as to how to approach data sufficiency questions. Should i try and solve the answer choices first and then fit them into the question or try to find the answer to the question first and then look at the options.. i am a little confused about my approach...
I will be glad if anyone can guide me on this..
Thanks

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Hi Bunuel,

According to question stem, seven numbers selected have two properties:
1) They are more than 1 but less than 100 AND
2) Each are divided by 7.

So, if we select seven numbers which are divisible by 7, in each case the remainder is 0. So the range or sum is.

I will appreciate if you clarify my missing points here. Thanks.
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What is your question?

I think you misunderstood what is given in the stem. We are NOT told that numbers selected are DIVISIBLE by 7. We are told that "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?"
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I thought each selected number is divisible by 7. But now its clear.

Thank you.

Posted from my mobile device
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Hi avigutman,

Will appreciate your response for this sum.
Any simpler/easy method to solve this sum ?


Thanks

PriyamRathor, no need to solve this sum. Just to confirm that we could solve it if we wanted to.
(1) isn't sufficient on its own because, to take an extreme example, you could have six numbers whose remainder is zero (e.g. 7, 14, 21, 28, 35, 42) and one number whose remainder is 6 (e.g. 6). This set of numbers is consistent with the free info and statement (1), and the sum of the seven remainders is 6 - but hopefully you can see that many other examples with many different sums can be easily constructed.
Statement (2) locks us in. In the world of divisibility by 7, there are exactly seven different "types" of numbers (remainder 0, 1, 2, ..., 5, 6). By analogy, the world of divisibility by 2 has exactly two different "types" of numbers: remainder 0 and remainder 1 (and, since there are only two types, we invented names for them: even and odd).
So, given that the seven numbers are consecutive (so their remainders will cycle through 0 - 6), we know exactly what the seven remainders are and we could find their sum if we wanted to.
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