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given,[100],[100/2],[100/3],[100/4][100/5][100/6][100/7][100/8][100/9][100/10][100/11][100/12][100/13][100/14][100/15][100/16][100/17][100/18][100/19][100/20]
ie simplifying ,100,50,33,25,20,16,14,12,11.10,9,8,7,7,6,6,5,5,5,5
Hence distinct integers=15
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The number of distinct integers in the collection [100/1], [(100/2], [100/3],…,[100/20] , where [x] denotes the largest integer not exceeding x, is
[100], [50], [33.33],[25],[16.66], [14.28], [12.5], [11.11], [10], [9.09], [8.25], [7.69], [7.14], [6.66], [6.25], [5.88], [5.55], [5.26], [5]
[100], [50], [25], [10], [5]
i'd guess ans. A as odd one so out.
Out of B,C,D and E, i'd go with D as it multiple of 5. :lol:
Ans. D
(a) 20
(b) 18
(c) 17
(d) 15
(e) 12
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Answer:D

[x] denotes the largest not exceeding x. meaning the decimal is truncated and the resulting integer is taken into consideration.

[100/1] =100

[(100/2]=50

[100/3]=33
....
..
[100/19]=5
[100/20]=5

100,50,33,25,20,16,14,12,11,10,9,8, 7,7,6,6,5,5,5,5.
Distinct integer values = 15
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for the given collection of integers we get following integer values
[100/1]=1
[100/2]=50
.
.
until [100/12] it would be unique
there after at [100/13] , [100/14] = 7
[100/15], [100/16]=6
[100/17] , [100/18] , [100/19], [100/20]=5

total distinct digits ; 15
IMO D

The number of distinct integers in the collection [100/1], [(100/2], [100/3],…,[100/20] , where [x] denotes the largest integer not exceeding x, is
(a) 20
(b) 18
(c) 17
(d) 15
(e) 12
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Answer is 15. But I solved all the 20 terms which is time consuming. Would like to know if there is any easier way to do this.
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1/13=1/14
1/15=1/16
1/17=1/18=1/19=1/20
So 15 different integers are there.
Option D is the answer.

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The number of distinct integers in the collection [100/1], [(100/2], [100/3],…,[100/20] , where [x] denotes the largest integer not exceeding x, is
(a) 20
(b) 18
(c) 17
(d) 15
(e) 12

[100/1] = 100
[100/2] = 50
[100/3] = [33.33] = 33
.....

[100/18] = [5.x] = 5
[100/19] = [5.x] = 5
[100/20] = [5] = 5

Basically we need to take the quotient only when divided by 100. i.e. the combinations of x*y <= 100; where y is the value we are seeking

Thus, we will have all the combinations as :-
1*100, 2*50, 3*33, 4*25, 5*20, 6*16, 7*14, 8*12, 9*11, 10*10, 11*9, 12*8, 13*7, 14*7, 15*6, 16*6, 17*5, 18*5, 19*5, 20*5

- There are 15 distinct values of y

Answer - D
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Quote:
The number of distinct integers in the collection [100/1], [(100/2], [100/3],…,[100/20] , where [x] denotes the largest integer not exceeding x, is
(a) 20
(b) 18
(c) 17
(d) 15
(e) 12

[100/1] = 100
[100/2]= 50
[100/3] = 33
[100/4] = 25
[100/5] = 20
[100/6] = 16
[100/7] = 14
[100/8] = 12
[100/9] = 11
[100/10] = 10
[100/11] = 9
[100/12] = 8
[100/13] = 7
[100/14] = 7

[100/15] = 6
[100/16] = 6
[100/17] = 6

[100/18] = 5
[100/19] = 5
[100/20] = 5


i.e. 15 different integers

Answer: Option D
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Answer E

Solution
[x] - denotes the greatest integer not exceeding x, so all the numbers in the given will be converted to their respective decimal forms.
when we do an approx calculation
[100/1] = 100,
[100/2] = 50,
[100/3] = 33,
[100/4] = 25,
[100/5] = 20,
[100/6] = 16,
[100/7] = 14,
[100/8] = 12,
[100/9] = 11,
[100/10] = 10,

[100/11] = 9,
[100/12] = 8,
[100/13] = 7,
[100/14] = 7,
[100/15] = 6,
[100/16] = 6,
[100/17] = 5,
[100/18] = 5,
[100/19] = 5,
[100/20] = 5

So we can see that starting from [100/13], the greatest integer of [x] becomes constant and not distinct.

So as per me, the number of distinct integers in the collection: 12.
Answer E
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The number of distinct integers in the collection [100/1], [(100/2], [100/3],…,[100/20] , where [x] denotes the largest integer not exceeding x, is
(a) 20
(b) 18
(c) 17
(d) 15
(e) 12

[100/1] = 100, [100/2]=50,[100/3]=33,[100/4]=25,[100/5]=20...[100/10]=10..[100/20]=5
concentrating on highlighted part, between [100/10] & [100/20] (inclusive) only 5 integer values can be distinct.
this takes the total straight down to 15.
Ans: D
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