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For all real numbers, [x] = the greatest integer less than or equal to

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Bunuel
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For all real numbers, [x] = the greatest integer less than or equal to [#permalink] New post   10 Jun 2019, 00:28
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For all real numbers, [x] = the greatest integer less than or equal to x.

What is the sum of the 28 terms in the series [1/6] + [2/6] + [3/6] + … + [28/6]?

A. 48
B. 50
C. 52
D. 54
E. 56
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E
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nick1816
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For all real numbers, [x] = the greatest integer less than or equal to [#permalink] New post   Updated on: 10 Jun 2019, 12:34
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Write a general form of the terms =[n/6], where n is an integer
when 1≤n≤5, [n/6]=0
when 6≤n≤11, [n/6]=1
when 12≤n≤17, [n/6]=2
when 18≤n≤23, [n/6]=3
when 24≤n≤28, [n/6]=4

[1/6] + [2/6] + [3/6] + … + [28/6]= 0*5 + 1*6 + 2*6 + 3*6 + 4*5 = 56

Bunuel wrote:
For all real numbers, [x] = the greatest integer less than or equal to x.

What is the sum of the 28 terms in the series [1/6] + [2/6] + [3/6] + … + [28/6]?

A. 48
B. 50
C. 52
D. 54
E. 56

Originally posted by nick1816 on 10 Jun 2019, 00:44.
Last edited by nick1816 on 10 Jun 2019, 12:34, edited 1 time in total.
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Re: For all real numbers, [x] = the greatest integer less than or equal to [#permalink] New post   10 Jun 2019, 08:52
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Bunuel wrote:
For all real numbers, [x] = the greatest integer less than or equal to x.

What is the sum of the 28 terms in the series [1/6] + [2/6] + [3/6] + … + [28/6]?

A. 48
B. 50
C. 52
D. 54
E. 56


need to be careful with the given function [x] = the greatest integer less than or equal to x
so sum of 28 terms ; [1/6] + [2/6] + [3/6] + … + [28/6]
from 1/6 to 5/6; [x]=0
6/6 to 11/6 ; [x]=1
12/6 to 17/6; 2
18/6 to 23/6 ; 3
24/6 to 28/6 ; 4

so total sum ; 0*6+1*6+2*6+3*6+4*5 ; 56
IMO E
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Re: For all real numbers, [x] = the greatest integer less than or equal to [#permalink] New post   11 Jun 2019, 20:19
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Bunuel wrote:
For all real numbers, [x] = the greatest integer less than or equal to x.

What is the sum of the 28 terms in the series [1/6] + [2/6] + [3/6] + … + [28/6]?

A. 48
B. 50
C. 52
D. 54
E. 56


We can group the terms by groups of 6, except the first and last groups will have 5 terms each.

First group (5 terms): [1/6], [2/6], …, [5/6]
Second group (6 terms): [6/6], [7/6], …, [11/6]
Third group (6 terms): [12/6], [13/6], …, [17/6]
Fourth group (6 terms): [18/6], [19/6], …, [23/6]
Fifth group (5 terms): [24/6], [25/6], …, [28/6]

We see that each term in the first group is 0, in the second group 1, in the third group 2, in the fourth group 3, and in the fifth group 4. Therefore, the sum is 5 x 0 + 6 x 1 + 6 x 2 + 6 x 3 + 5 x 4 = 6 + 12 + 18 + 20 = 56.

Answer: E
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Re: For all real numbers, [x] = the greatest integer less than or equal to [#permalink] New post   23 Jun 2019, 23:46
ScottTargetTestPrep wrote:
Bunuel wrote:
For all real numbers, [x] = the greatest integer less than or equal to x.

What is the sum of the 28 terms in the series [1/6] + [2/6] + [3/6] + … + [28/6]?

A. 48
B. 50
C. 52
D. 54
E. 56


We can group the terms by groups of 6, except the first and last groups will have 5 terms each.

First group (5 terms): [1/6], [2/6], …, [5/6]
Second group (6 terms): [6/6], [7/6], …, [11/6]
Third group (6 terms): [12/6], [13/6], …, [17/6]
Fourth group (6 terms): [18/6], [19/6], …, [23/6]
Fifth group (5 terms): [24/6], [25/6], …, [28/6]

We see that each term in the first group is 0, in the second group 1, in the third group 2, in the fourth group 3, and in the fifth group 4. Therefore, the sum is 5 x 0 + 6 x 1 + 6 x 2 + 6 x 3 + 5 x 4 = 6 + 12 + 18 + 20 = 56.

Answer: E


Hi Scott,

Why did we group this ?? Is there any other method I can attempt this ?? I couldn't understand the explanation.

Please help.
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Re: For all real numbers, [x] = the greatest integer less than or equal to [#permalink] New post   24 Jun 2019, 02:54
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0- 5=0
6-11=1*6
12-17=2*6
18-23=3*6
24-28=4*5

total is 56.option E
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Re: For all real numbers, [x] = the greatest integer less than or equal to [#permalink] New post   27 Jun 2019, 18:19
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GMATINSECT wrote:
ScottTargetTestPrep wrote:
Bunuel wrote:
For all real numbers, [x] = the greatest integer less than or equal to x.

What is the sum of the 28 terms in the series [1/6] + [2/6] + [3/6] + … + [28/6]?

A. 48
B. 50
C. 52
D. 54
E. 56


We can group the terms by groups of 6, except the first and last groups will have 5 terms each.

First group (5 terms): [1/6], [2/6], …, [5/6]
Second group (6 terms): [6/6], [7/6], …, [11/6]
Third group (6 terms): [12/6], [13/6], …, [17/6]
Fourth group (6 terms): [18/6], [19/6], …, [23/6]
Fifth group (5 terms): [24/6], [25/6], …, [28/6]

We see that each term in the first group is 0, in the second group 1, in the third group 2, in the fourth group 3, and in the fifth group 4. Therefore, the sum is 5 x 0 + 6 x 1 + 6 x 2 + 6 x 3 + 5 x 4 = 6 + 12 + 18 + 20 = 56.

Answer: E


Hi Scott,

Why did we group this ?? Is there any other method I can attempt this ?? I couldn't understand the explanation.

Please help.


We group them because each term in any one particular group has the same value. The only other alternative I see is to add each term, one by one, meaning carrying out the addition 0 + 0 + … + 0 + 1 + 1 + … + 1 + 2 + … etc. ; but there are 28 terms and doing that would be a huge waste of time. However, if you started this laborious task, you would see the pattern emerging that would result in the groupings shown in my original solution.
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For all real numbers, [x] = the greatest integer less than or equal to [#permalink] New post   03 Jan 2024, 22:59
My logic.
1. Multiple of 7 since its 28
2. Second figure had to be divisible by 4 and given that we're rounding off, half of it would be a (-1) and another half (+1), so 4 x 2 =8
3. 7 x 8 = 56
Not sure if my pattern recognition helped.
I didnt allow the 1/6 to affect me so much cause its a round off situation so I tried to understand the logic behind it.
I could be veryyyy wrong here.
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Re: For all real numbers, [x] = the greatest integer less than or equal to [#permalink] New post   04 Jan 2024, 17:03
when you add up the whole amounts by summing up the greatest and smallest number (1/6+28/6, 2/6+27/6 ..) you will have a summation of 14*(29/6) which will be close to 58 but since we know [x] is greatest number less than or equal to x it will naturally be the closest number to 58, that is, 56.
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Re: For all real numbers, [x] = the greatest integer less than or equal to [#permalink]
04 Jan 2024, 17:03
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