Live Chat with Admission Manager and Current Student of NUS SIngapore - Join Chat Room to Participate.

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]

Show Tags

04 Dec 2008, 12:31

12

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

66% (01:00) correct
34% (01:13) wrong based on 540 sessions

HideShow timer Statistics

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.
_________________

Re: Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]

Show Tags

31 Mar 2013, 01:15

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
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

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]

Show Tags

07 Jun 2013, 00:07

1

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.
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

Re: Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]

Show Tags

30 Sep 2014, 20:15

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]

Show Tags

12 Oct 2015, 23:00

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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.
_________________

Re: Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]

Show Tags

14 Nov 2016, 01:25

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]

Show Tags

21 Dec 2016, 16:42

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

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...