BdMO 2017 junior/10

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Thamim Zahin
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BdMO 2017 junior/10

Unread post by Thamim Zahin » Fri Feb 17, 2017 1:34 am

Whenever Avik gets a sequence, he multiplies every two distinct terms of that sequence, and then sums up these products to get the Hocus-pocus sum of the sequence. For example, the Hocus-pocus sum for the sequence $a, b, c, d$ is $ab + bc + ac + ad + bd + cd$. If Avik gets a sequence of $100$ terms, where each term is either $2$ or $-1$, what is the minimum Hocus-pocus sum of that sequence?
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Thanic Nur Samin
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Re: BdMO 2017 junior/10

Unread post by Thanic Nur Samin » Fri Feb 17, 2017 10:49 am

Notice that the hocus pocus sum is essentially square of the sum of all numbers minus the sum of the squares of the numbers divided by two.

So, if there are $x$ $(-1)$'s and $y$ $2$'s, then the sum is $\dfrac{(2y-x)^2-(4y+x)}{2}=\dfrac{(3y-100)^2-3y-100}{2}=\dfrac{(3y-100.5)^2-200.25}{2}$. It clearly achieves its integral minimum on $y=33$ and $y=34$, and so the desired minumum sum is $-66$.
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aritra barua
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Re: BdMO 2017 junior/10

Unread post by aritra barua » Fri Feb 17, 2017 9:14 pm

Let,the maximum number of combinations be x....which follows the form ab+ab+....+ab...where a=2,b=-1;Now,when there are 100 terms in a sequence,the numbers of combinations will be maximum when number of a=number of b.....because here distinct terms will be applicable and 2 terms are followed....In that case,we may find all the combinations considering that there are no distinct terms,then we apply our condition and deduce that number of combinations....Since,every combination follows the value -2,the total minimum sum of hocus pocus series will be -2x..

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