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09 Mar 2019, 09:36
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
26% (02:35) correct
74%
(02:51)
wrong
based on 72
sessions
History
Date
Time
Result
Not Attempted Yet
How many ways are there to paint each of the integers 2, 3, ..., 9 either red, green, or blue so that each number has a different color from each of its proper divisors?
(A) 144
(B) 216
(C) 256
(D) 384
(E) 432
(A) 144
(B) 216
(C) 256
(D) 384
(E) 432
11 Mar 2019, 08:10
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Noshad
The question is not likely to be form a GMAT source, more of Indian CAT exam.
Having said that the solution could be..
(1) All prime number can take any of three colors, except one of 2 or 3 as both are connected via 6..
so 2, 5, 7 can have 3 ways each - 3*3*3 = 27 ways
(2) Let us take the multiples of 2..
4 can take any of the remaining 2 colors, while 8 would take the remaining colour. As like 4, 6 can also take any of 2 colors other than the colour of integer 2.
Thus 2*2*1=4
(3) Now let us take a look at 3 and its multiple.
3 is restricted because of 6, so it cannot have the same color as 6 and therefore 3 can have any of the remaining 2 colors.
We could also have taken 3 colors for 6, and then 2 and 3 would have 2 colors each and overall 2*2*3.. We are still getting the same
Only integer left is 9, so it will take any of the 2 colors other than that of 3, so 2 ways..
Total 2*2=4
Combined way of all integers = \(27*4*4=432\) ways
General Discussion
11 Mar 2019, 00:26
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2,4,8 require different colors each, so there are 6 ways to color these.
5 and 7 can be any color, so there are 3 x 3 ways to color these.
6 can have 2 colors once 2 is colored, and thus 3 also has 2 colors following 6, which leaves another 2 for 9.
All together: 6 x 3 x 3 x 2 x 2 x 2 = 432
5 and 7 can be any color, so there are 3 x 3 ways to color these.
6 can have 2 colors once 2 is colored, and thus 3 also has 2 colors following 6, which leaves another 2 for 9.
All together: 6 x 3 x 3 x 2 x 2 x 2 = 432
