Mathematical Induction: Showing Sums of 3 Consecutive Integers

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The discussion centers on proving that in any arrangement of the integers from 1 to 30 on a circular disk, there must be three consecutive integers whose sum is at least 45. By defining sums of three consecutive integers and assuming all such sums are less than 45, a contradiction arises when calculating the total sum of these sums, which exceeds 1350. This leads to the conclusion that at least one sum must be 45 or greater. Additionally, it is noted that it can be shown that there are three adjacent integers whose sum is at least 47. The general principle established indicates that for integers arranged in a circle, a specific number of adjacent integers will always meet a defined sum threshold.
dwyane wade
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On the outside rim of a circular disk the integers from 1 through 30 are painted in
random order. Show that no matter what this order is, there must be three successive
integers whose sum is at least 45.
 
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Hi dwayne wade.

Let the integers be $a_1,a_2,\ldots,a_{30}$ as arranged around the the disc, and let
$$s_1\ =\ a_1+a_2+a_3 \\ s_2\ =\ a_2+a_3+a_4 \\ s_3\ =\ a_3+a_4+a_5 \\ \vdots \\ s_{28}\ =\ a_{28}+a_{29}+a_{30} \\ s_{29}\ =\ a_{29}+a_{30}+a_1 \\ s_{30}\ =\ a_{30}+a_1+a_2.$$
Suppose to the contrary that all the sums are less than 45, i.e. $s_i<45$ for all $i=1,\ldots,30$. Then
$$\sum_{i=1}^{30}s_i\ <\ 30\cdot45\ =\ 1350.$$
But
$$\sum_{i=1}^{30}s_i\ =3(a_1+\cdots+a_{30})\ =\ 3(1+\cdots30)\ =\ 3\cdot\frac{30\cdot31}2\ =\ 1395\ >\ 1350.$$PS: It can be shown that there must be three adjacent numbers on the disc whose sum is at least $47$. In general, if the integers $1,\ldots,n$ are arrange in a circle, in any order, there must be $r$ adjacent ones whose sum is at least $\frac12r(n+1)$.
 
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