(AMS Bumper Sticker)



Web math-frolic.blogspot.com

Monday, May 4, 2015

Get Your Pencil and Paper...



Currently reading a couple of books simultaneously, including Alfred Posamentier's and Stephen Krulik's new "Problem Solving Strategies in Mathematics" (so far very good... but then that hardly needs saying if it's from Posamentier). Anyway, it looks to include several nice puzzles/problems along the way, one of which I'll adapt to my own wording today:

Donald Chump rolls an ordinary (6-sided) die repeatedly, and records the result of each outcome. He is asked to stop as soon as one number has come up 3 separate times. He thus ends up stopping after the 12th roll, at which point the sum of ALL the rolls is 47.
Which number appeared three times?
.
.
.answer below....
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Answer:  6
(I've given explanation in the comments section)

(image via David Vignoni/WikimediaCommons)


2 comments:

"Shecky Riemann" said...

I'll try to explain answer for any needing it:
After 11 rolls no number has come up 3 times (or game would've ended), so 5 numbers MUST have come up twice and one number once. The latter number, we'll call "L" could not be rolled on the 12th turn, or game would continue with a total value of 42 (i.e., 2 x (1+2+3+4+5+6). Thus, at 11 rolls, the total must be 42-L. But the 12th roll ends game with total of 47, thus 42 plus 12th roll, we'll call "T," equals 47.
So 42-L+T = 47, or T-L = 5. The ONLY die numbers allowing such a result are T=6 and N=1.

"Shecky Riemann" said...

arrrrrghhhh..... that last N=1 above, should of course read L=1.