yvonne0923 wrote:A string of holiday lights has a repeating colot pattern of green, white, red, blue, orange, yellow. If the first light of the string is green and the last light is red, then the number of lights could be:

A. 23
B. 30
C. 37
D. 45
E. 55




















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[spoiler]O.A: D[/spoiler]
For such questions, do we have to follow the sequence the question gives us? Because I reorganized the sequence of the lights, and I treated it as a sequential problem first. How to attack such problems?


Thanks,

repeating pattern of colors given to us is green, white, red, blue, orange, yellow; which is 6 in no.
now suppose we have 7 holiday then seventh light would be green; {green, white, red, blue, orange, yellow, green}; because when we divide 7 by 6 it gives 1 as a remainder, and 1 here is the first light which is green,, now lets start working with options;
23/6 gives 5 as a option which means last light would be orange;
30/6 remainder is 0 therefore last light would be yellow;
37/6 remainder is 1 therefore last light would be green;
45/6 remainder 3 therefore last would be red;*bingo*
55/6 remainder is 1 therefore last light would be green,

hence D is an answer..!!!

manpsingh87 wrote:

yvonne0923 wrote:A string of holiday lights has a repeating colot pattern of green, white, red, blue, orange, yellow. If the first light of the string is green and the last light is red, then the number of lights could be:

A. 23
B. 30
C. 37
D. 45
E. 55




















_____________________________________________________________________________________________________
[spoiler]O.A: D[/spoiler]
For such questions, do we have to follow the sequence the question gives us? Because I reorganized the sequence of the lights, and I treated it as a sequential problem first. How to attack such problems?


Thanks,

repeating pattern of colors given to us is green, white, red, blue, orange, yellow; which is 6 in no.
now suppose we have 7 holiday then seventh light would be green; {green, white, red, blue, orange, yellow, green}; because when we divide 7 by 6 it gives 1 as a remainder, and 1 here is the first light which is green,, now lets start working with options;
23/6 gives 5 as a option which means last light would be orange;
30/6 remainder is 0 therefore last light would be yellow;
37/6 remainder is 1 therefore last light would be green;
45/6 remainder 3 therefore last would be red;*bingo*
55/6 remainder is 1 therefore last light would be green,

hence D is an answer..!!!

Thanks for the explanation, I'm feeling very clear now!