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19 Jan 2021, 01:15
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C
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Difficulty:
55%
(hard)
Question Stats:
55% (01:39) correct
45%
(01:38)
wrong
based on 273
sessions
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Time
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Not Attempted Yet
Is 729 a member of set S?
(1) The first 20 positive integers are members of set S.
(2) Product of ANY two members of the set S, provided that the value of the product is less than 1000, is a member of the set S.
Are You Up For the Challenge: 700 Level Questions
(1) The first 20 positive integers are members of set S.
(2) Product of ANY two members of the set S, provided that the value of the product is less than 1000, is a member of the set S.
Are You Up For the Challenge: 700 Level Questions
19 Jan 2021, 02:32
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Bunuel
(1) The first 20 positive integers are members of set S.
So 1 to 20 are surely there. But nothing more
(2) Product of ANY two members of the set S, provided that the value of the product is less than 1000, is a member of the set S.
Nothing about the numerical values.
Combined
729=9*81=27*27.
Now the set surely has 3,9, and their product 27 too.
If 3, 9 and 27 are there, then 27*3 or 81 will be there.
If 3,9,27,81 are there, then 81*9 or 729 will also be there.
C
rocky620
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19 Jan 2021, 06:48
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(1) The first 20 positive integers are members of set S.
There might be other numbers also, which may include 729.
Insufficient.
(2) Product of ANY two members of the set S, provided that the value of the product is less than 1000, is a member of the set S.
There might be factors of 729 or not. Yes and No both are possible.
Insufficient
1+2
729 = \(3^6\)
since 3 & 9 both are there, so will be 27 (3*9), 81 (27*3), 243 (81*3), 729 (243*3)
So, 729 will be there in the set.
Sufficient
Option C
There might be other numbers also, which may include 729.
Insufficient.
(2) Product of ANY two members of the set S, provided that the value of the product is less than 1000, is a member of the set S.
There might be factors of 729 or not. Yes and No both are possible.
Insufficient
1+2
729 = \(3^6\)
since 3 & 9 both are there, so will be 27 (3*9), 81 (27*3), 243 (81*3), 729 (243*3)
So, 729 will be there in the set.
Sufficient
Option C
