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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] New post 06 Nov 2012, 18:05
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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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Re: Eight consecutive integers are selected from the integers 1 [#permalink] New post 07 Nov 2012, 04:10
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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.

Answer: C.

Hope it's clear.
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Re: Eight consecutive integers are selected from the integers 1 [#permalink] New post 06 Nov 2012, 19:01
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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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Re: Eight consecutive integers are selected from the integers 1 [#permalink] New post 11 Nov 2013, 00:29
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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] New post 06 Nov 2012, 20:48
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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Re: Eight consecutive integers are selected from the integers 1 [#permalink] New post 14 Aug 2013, 01:23
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Re: Eight consecutive integers are selected from the integers 1 [#permalink] New post 09 Nov 2013, 18:13
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.

Answer: C.

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] New post 09 Nov 2013, 22:33
(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?????
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Re: Eight consecutive integers are selected from the integers 1 [#permalink] New post 11 Nov 2013, 12:27
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.



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Re: Eight consecutive integers are selected from the integers 1 [#permalink] New post 16 Jun 2015, 02:32
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Re: Eight consecutive integers are selected from the integers 1   [#permalink] 16 Jun 2015, 02:32
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