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Question - A Roulette wheel is divided into 36 sectors. Each sector is assigned a random number from 1 to 36. Show that there are three consecutive sectors such that the sum of their assigned numbers is at least 56.

My approach -> (Proof by Contradiction)

Let $a_1, a_2, \ldots, a_n$ be the numbers assigned to the 1st, 2nd, ..., nth sectors respectively. Assume that the sum of all three consequtive sectors is less than 56.

Thus,

$$ a_1 + a_2 + a_3 < 56 \\ a_2 + a_3 + a_4 < 56 \\ \vdots \\ a_{36} + a_1 + a_2 < 56 $$ Adding them all, $$ \Longrightarrow 3(a_1 + a_2 + \ldots + a_{36}) < 56 \cdot 36 \\ \Longrightarrow 3 \frac{36 \cdot 37}{2} < 56 \cdot 36 \\ \Longrightarrow 111 < 112 $$ (which is true and hence, unable to prove by contradiction)

Where am I going wrong?

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  • $\begingroup$ @vishlgoel If you are ok, you can accept the answer and set as solved. Thanks! $\endgroup$ – gimusi Dec 23 '17 at 18:43
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You were on the right track. It is correct that if the statement

There are three consecutive sectors such that the sum of their assigned numbers is at least 56.

does not hold then $$ a_i + a_{i+1} + a_{i+2} < 56 $$ for $i = 1, 2, 3, \ldots$. However, this estimate is not strong enough to obtain a contradiction. Since all $a_i$ are integers we have in fact the better estimate $$ a_i + a_{i+1} + a_{i+2} \le 55 $$ for all $i$. Adding these inequalities now gives a contradiction, as desired: $$ 3 \frac{36 \cdot 37}{2} \le 55 \cdot 36 \Longrightarrow 1998 \le 1980 \, . $$


Remark: Actually there must be three consecutive sectors on the Roulette wheel whose number add up to at least 57, see

A066385 Smallest maximum of sum of 3 consecutive terms in any arrangement of [1..n] in a circle.

in the On-Line Encyclopedia of Integer Sequences for the general question and more information. In particular it is listed that $a(36) = 57$.

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  • $\begingroup$ But if I use <56 instead of <=55, the answer is different. The former concludes to 1998<2016 and the latter concludes to 1998<=1980. Where am I thinking wrong? My brain is so not working! $\endgroup$ – vishalgoel Dec 22 '17 at 18:05
  • $\begingroup$ @vishalgoel: $1998 \le 1980$ is a false statement, this implies that the assumption is wrong. $1998 < 2016$ is a true statement, that does not imply anything. $\endgroup$ – Martin R Dec 22 '17 at 18:10
  • $\begingroup$ Thanks a lot Martin. I feel so stupid right now. $\endgroup$ – vishalgoel Dec 22 '17 at 18:13
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Since the average value of $a_i=18.5$ the average value of the sum of triples is 55.5 thus exists at least one triple such that the sum is at least 56.

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