a necklace is made by stringing n individual beads....

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factor26: Senior | Next Rank: 100 Posts; Posts: 99; Joined: Sun Mar 13, 2011 12:42 pm

by factor26 » Sat Oct 01, 2011 7:25 am

a necklace is made by stringing n individual beads together in the repeating pattern, red bead, green bead, white bead, blue bead and yellow bead. if the necklace design begins with a red bead and ends white bead then n could equal?

a - 16
b - 32
c - 41
d - 54
e - 68

OG ANSWER IS E ... CAN SOMEONE EXPLAIN?
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Source: Beat The GMAT — Problem Solving | Sat Oct 01, 2011 7:25 am

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by GmatMathPro » Sat Oct 01, 2011 7:51 am

There are 5 beads in the sequence. Every time the sequence is completed, you add 5 beads. So the sequence of 5 beads goes on and on and on for a while, and every time it is completed, the total number of beads is a multiple of 5: 5, 10, 15, 20... Eventually, though, we don't complete a full sequence; we stop on white. The white bead is the third bead in the sequence, so the number of beads we have is 3 more than some multiple of 5. This means the total number of beads must end in 3 or 8. The only one that does is E

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sl750: Master | Next Rank: 500 Posts; Posts: 496; Joined: Tue Jun 07, 2011 5:34 amFollowed by: 1 member; Thanked: 38 times

by sl750 » Sun Oct 02, 2011 4:40 am

R G W B Y represent the color of each bead. There are 5 beads, so count multiples of 5. Only in choice E, can be arrive at 68 beads as the 68th bead will be white

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edvhou812: Master | Next Rank: 500 Posts; Posts: 143; Joined: Sat Nov 06, 2010 9:06 pmFollowed by: 1 member; Thanked: 10 times

by edvhou812 » Sun Oct 02, 2011 8:10 am

Write down 'Red, Green, White, Blue, Yellow'

When you examine and count, you will find the Yellow always ends on 5 and 10. So from here, you can count after 10 and see where you land in the list.

A: 10 is Yellow. Count six more, and end on Red (out)
B: 30 is Yellow. Count two more and end on Green (out)

You can go through this process until you see that E works. 60 is yellow, count eight more and land on White.

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factor26: Senior | Next Rank: 100 Posts; Posts: 99; Joined: Sun Mar 13, 2011 12:42 pm

by factor26 » Sun Oct 02, 2011 11:05 am

Thanks for the help all!!

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rooster: Senior | Next Rank: 100 Posts; Posts: 44; Joined: Wed Aug 25, 2010 12:23 am; Thanked: 6 times

by rooster » Sun Oct 02, 2011 8:24 pm

Let me explain my method and maybe you can let me know what you all think.

To add onto sl750, you set up the problem with R, G, W, B, Y thus to fill the rotation, a complete round is 5

BUT you need to end on white, so you are looking for an answer that is divisible by 5 with a remainder of 3

68/5 = 13 R3

E is the only choice with this type of requirement.

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mad2011: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Mon Aug 29, 2011 4:42 am; Thanked: 1 times

by mad2011 » Mon Oct 03, 2011 1:16 am

It simple
5n+3

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