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# Eight consecutive integers are selected from the integers 1

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Eight consecutive integers are selected from the integers 1 [#permalink]

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06 Nov 2012, 19:05
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Eight consecutive integers are selected from the integers 1 to 50, inclusive. What is the sum of the remainders when each of the integers is divided by x?

(1) The remainder when the largest of the consecutive integers is divided by x is 0.
(2) The remainder when the second largest of the consecutive integers is divided by x is 1.
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Kudos [?]: 115 [1], given: 1

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Re: Eight consecutive integers are selected from the integers 1 [#permalink]

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07 Nov 2012, 05:10
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Eight consecutive integers are selected from the integers 1 to 50, inclusive. What is the sum of the remainders when each of the integers is divided by x?

Note that when a positive integer is divided by positive integer x, then possible remainders are 0, 1, 2, ..., (x-1). For example, when a positive integer is divided by 4, then the possible remainders are 0, 1, 2, or 3.

(1) The remainder when the largest of the consecutive integers is divided by x is 0. This implies that the largest integer is a multiple of x. Now, if eight consecutive integers are {1, 2, 3, 4, 5, 6, 7, 8} and x=2, then the sum of the remainders will be 1+0+1+0+1+0+1+0=4, but if x=1, then the sum of the remainders will be 0+0+0+0+0+0+0+0=0. Not sufficient.

(2) The remainder when the second largest of the consecutive integers is divided be x is 1. This implies that the second largest integer is 1 more than a multiple of x. Now, if eight consecutive integers are {1, 2, 3, 4, 5, 6, 7, 8} and x=2, then the sum of the remainders will be 1+0+1+0+1+0+1+0=4, but if x=6, then the sum of the remainders will be 1+2+3+4+5+0+1+2=18. Not sufficient.

(1)+(2) We have that:
7th largest integer when divided by x yields the remainder of 1;
8th largest integer when divided by x yields the remainder of 0.

Thus, we have that 1 is the largest remainder possible, which implies that x=2 (x-1=1 --> x=2). The sum of the remainders when any eight consecutive integers are divided by 2 is 1+0+1+0+1+0+1+0=4. Sufficient.

Hope it's clear.
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Kudos [?]: 124141 [5], given: 12072

Director
Status: Done with formalities.. and back..
Joined: 15 Sep 2012
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Re: Eight consecutive integers are selected from the integers 1 [#permalink]

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06 Nov 2012, 20:01
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superpus07 wrote:
Eight consecutive integers are selected from the integers 1 to 50, inclusive. What is the sum of the remainders when each of the integers is divided by x?

1. The remainder when the largest of the consecutive integers is divided by x is 0.
2. The remainder when the second largest of the consecutive integers is divided by x is 1.

Thanks.

Statement 1: We know only that largest integer is multiple of x. eg, largest integer is 24 then x could be 8 or 24. But it would not help in determining remainder of any other previous 7 integers. eg if x is 8 then 23 would give a remainder of 7 while it would give a remainder of 23 if x is 24.
Not sufficient.

Statement 2: We know only that second largest is multiple of x +1. Again similar examples as given above could be given. Not sufficient.

Combining these statements we know that remainder when the largest of the consecutive integers is divided by x is 0 and remainder when the second largest of the consecutive integers is divided by x is 1.
This could happen only if x=2. In which case there will be alternate remainders of 1,0,1,0,1,0.....

Hence sum of 8 remainders could be found out (= 4*1+4*0)
Sufficient.

Ans C it is!
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Kudos [?]: 642 [2], given: 23

Math Expert
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Re: Eight consecutive integers are selected from the integers 1 [#permalink]

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11 Nov 2013, 01:29
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Expert's post
dreambig1990 wrote:
(1)+(2) We have that:
7th largest integer when divided by x yields the remainder of 1;
8th largest integer when divided by x yields the remainder of 0.

Thus, we have that 1 is the largest remainder possible,

---WHY?????

When a positive integer is divided by positive integer x, then the pattern of the remainders is 0, 1, 2, ..., (x-1), 0, 1, 2, ..., (x-1), 0, 1, 2, ..., (x-1), .... So, the remainder preceding 0 is the largest.
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Re: Eight consecutive integers are selected from the integers 1 [#permalink]

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06 Nov 2012, 21:48
superpus07 wrote:
Eight consecutive integers are selected from the integers 1 to 50, inclusive. What is the sum of the remainders when each of the integers is divided by x?

1. The remainder when the largest of the consecutive integers is divided by x is 0.
2. The remainder when the second largest of the consecutive integers is divided by x is 1.

8 Consecutive integers...

say 1,2,3,4,5,6,7,8

A) NS because x can be 8,4,2.....reaminders will vary
B) NS same as above

Combine
The 7th leaves remainder 1 and 8th zero...this can only happen if X = 2
Sufficient.
Sum of Remainder = 4
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Kudos [?]: 61 [0], given: 22

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Re: Eight consecutive integers are selected from the integers 1 [#permalink]

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14 Aug 2013, 02:23
Bumping for review and further discussion.
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Re: Eight consecutive integers are selected from the integers 1 [#permalink]

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09 Nov 2013, 19:13
Bunuel wrote:
Eight consecutive integers are selected from the integers 1 to 50, inclusive. What is the sum of the remainders when each of the integers is divided by x?

Note that when a positive integer is divided by positive integer x, then possible remainders are 0, 1, 2, ..., (x-1). For example, when a positive integer is divided by 4, then the possible remainders are 0, 1, 2, or 3.

(1) The remainder when the largest of the consecutive integers is divided by x is 0. This implies that the largest integer is a multiple of x. Now, if eight consecutive integers are {1, 2, 3, 4, 5, 6, 7, 8} and x=2, then the sum of the remainders will be 1+0+1+0+1+0+1+0=4, but if x=1, then the sum of the remainders will be 0+0+0+0+0+0+0+0=0. Not sufficient.

(2) The remainder when the second largest of the consecutive integers is divided be x is 1. This implies that the second largest integer is 1 more than a multiple of x. Now, if eight consecutive integers are {1, 2, 3, 4, 5, 6, 7, 8} and x=2, then the sum of the remainders will be 1+0+1+0+1+0+1+0=4, but if x=6, then the sum of the remainders will be 1+2+3+4+5+0+1+2=18. Not sufficient.

(1)+(2) We have that:
7th largest integer when divided by x yields the remainder of 1;
8th largest integer when divided by x yields the remainder of 0.

Thus, we have that 1 is the largest remainder possible, which implies that x=2 (x-1=1 --> x=2). The sum of the remainders when any eight consecutive integers are divided by 2 is 1+0+1+0+1+0+1+0=4. Sufficient.

Hope it's clear.

Did you just happen to know that the only way two consecutive integers can have this happen is if they're both divided by 2? I'm curious where you know that from, and if there's more rules like that which I could learn.

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Re: Eight consecutive integers are selected from the integers 1 [#permalink]

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09 Nov 2013, 23:33
(1)+(2) We have that:
7th largest integer when divided by x yields the remainder of 1;
8th largest integer when divided by x yields the remainder of 0.

Thus, we have that 1 is the largest remainder possible,

---WHY?????

Kudos [?]: 3 [0], given: 77

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Re: Eight consecutive integers are selected from the integers 1 [#permalink]

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11 Nov 2013, 13:27
Bunuel wrote:
dreambig1990 wrote:
(1)+(2) We have that:
7th largest integer when divided by x yields the remainder of 1;
8th largest integer when divided by x yields the remainder of 0.

Thus, we have that 1 is the largest remainder possible,

---WHY?????

When a positive integer is divided by positive integer x, then the pattern of the remainders is 0, 1, 2, ..., (x-1), 0, 1, 2, ..., (x-1), 0, 1, 2, ..., (x-1), .... So, the remainder preceding 0 is the largest.

wow---Thank you very much!!! I saw ppl nominate you in many posts-- I join them too. Thank you!

Kudos [?]: 3 [0], given: 77

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Re: Eight consecutive integers are selected from the integers 1 [#permalink]

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16 Jun 2015, 03:32
Hello from the GMAT Club BumpBot!

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Re: Eight consecutive integers are selected from the integers 1 [#permalink]

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05 Feb 2017, 00:35
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).

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Re: Eight consecutive integers are selected from the integers 1 [#permalink]

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01 Mar 2017, 22:39
Recall that remainders follow a repeating pattern when x is divided into consecutive integers. For
example, when the integers 1 to 50 are divided by x = 4, the remainders form a [1, 2, 3, 0] repeating
pattern, and the sum of any consecutive 8 of these remainders would be 2 × [1 + 2 + 3 + 0] = 12.
However, when the integers 1 to 50 are divided by x = 3, the remainders are a [1, 2, 0] repeating pattern,
so the sum of any consecutive 8 of these remainders would depend on which term of the pattern was the
starting term.
(1) INSUFFICIENT: Statement (1) simply indicates that the largest of the eight consecutive integers is
divisible by x. It does not indicate the value of x, which determines the remainder pattern and, indirectly,
the sum of the remainders.
(2) INSUFFICIENT: Statement 2 simply indicates that the third largest of the eight consecutive integers is
divisible by x. It does not indicate the value of x, which determines the remainder pattern and, indirectly,
the sum of the remainders.
(1) & (2) SUFFICIENT: Together, the statements indicate that Largest Term and (Largest Term – 2) are
each divisible by x. Alternatively, the statements indicate a remainder pattern of [1,0] repeating. Thus, x
must be 2, and the sum of the remainders is 4 × [1 + 0] = 4.
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Re: Eight consecutive integers are selected from the integers 1   [#permalink] 01 Mar 2017, 22:39
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