OG12 #87

GMATbeater12 wrote:Could someone please help me with the following problem solving problem? It is problem 87 on page 164 of the Official Guide, 12th edition.

87. 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 with a white bead. then N could equal

(A) 16
(B) 32
(C) 41
(D) 54
(E) 68


I answered A because I figured that you could have RGWBYRGW, but the answer is E. Why?

Check the pattern
i.e. (RGWBY)(RGWBY)......A times (RGW)
=> 5A+3
68 SATISFIES (for A=13)
HENCE E is the answer.

GMATbeater12 wrote:
I answered A because I figured that you could have RGWBYRGW,

using the same logic, u can figure out that B is also ok 32/8=4

see the q.stem. it is said that u have a repeating patern -RGWBY, and need to end with W
so ur pattern is RGWBY + RGW . it means that u have 5x+3
now after checking each answer choice, u will find out that only option E is correct
5x+3=68 ;x=13

Thank you for your help.

I still do not understand it, though. I see how if 16 were be correct, so would 32, but both fulfill the requirement that the last bead is white. Why is that not correct?

GMATbeater12 wrote:Thank you for your help.

I still do not understand it, though. I see how if 16 were be correct, so would 32, but both fulfill the requirement that the last bead is white. Why is that not correct?

87. 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 with a white bead. then N could equal

(A) 16
(B) 32
(C) 41
(D) 54
(E) 68

The part in bold is not correct. In any case 16 is not correct.

you need to follow the pattern (RGWBY) start with red till white.

see:possible arrangements are

(RGWBY)(RGW)--> 8 beads
(RGWBY)(RGWBY)(RGW)--> 13 beads
(RGWBY)(RGWBY)(RGWBY)(RGW)--> 18 beads
etc.

it is basically

(RGWBY)--> 5 beads
suppose you repeat it A times i.e. 5A beads

(RGW)--> 3 beads
hence number of beads will always satisfy 5A+3

put A=1,2,3,4.......infinity
i.e. 8, 13, 18, 23, 28.......

for A=13 its 68 beads

hence E.

HTH

Ok, I see it now. Thank you!