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Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]
04 Dec 2008, 12:31

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

68% (01:56) correct
32% (01:09) wrong based on 298 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.

Re: Remainders Question from MGMAT [#permalink]
05 Dec 2008, 09:46

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.

Re: Remainders Question from MGMAT [#permalink]
24 Jan 2009, 16:07

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. _________________

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

Re: Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]
31 Mar 2013, 07:14

2

This post received KUDOS

Expert's post

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.

Re: Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked [#permalink]
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]
30 Sep 2014, 20:15

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