BdMO 2017 junior/10

Thamim Zahin
Posts: 98
Joined: Wed Aug 03, 2016 5:42 pm

BdMO 2017 junior/10

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
Posts: 176
Joined: Sun Dec 01, 2013 11:02 am

Re: BdMO 2017 junior/10

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
Posts: 48
Joined: Sun Dec 11, 2016 2:01 pm

Re: BdMO 2017 junior/10

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..