Armless AliensMath brain teasers require computations to solve.
On planet ?*!@, the inhabitants are robots with various numbers of arms. The robots are grouped into tribes according to the numbers of arms they have so that in any tribe, no member has a different number of arms to the other members. Each tribe follows the same fertility ritual before bed each night, a ritual that also reveals the number of members in a tribe. In the ritual, each arm of a male links with an arm of a female, no male-female pair of robots linking more than once. The minimum number of robots required to link all arms once only is the number belonging to that tribe. Now one night, the Creaky tribe cornered half of the Rusty tribe and despite being outnumbered two to one, unscrewed one arm from each, there being an equal number of male and female victims. However, the Creaky tribe were extremely poor at arithmetic and had failed to realise that they needed 24 times their actual haul of spare arms in order that the number of links their members could then make, each with the same number of extra arms, was half the maximum number of links the Rusty tribe could then make. How many Creakies and Rusties were there?
AnswerARMLESS ALIENS. There were 14 Creakies and 56 Rusties. A robot with n arms requires n females and vice-versa, and the number of members is 2n with n squared links. Let the Creaky tribe have x arms each and the Rusty tribe have y. Then the maximum number of links that the Rusty tribe can make after the mutilation is y(2y-1)/2. The number of extra arms each Creaky needs is 24y/2x allowing x(x + 24y/2x) links. Forming the links equation we have 2(y squared) - 49y - 4(x squared) = 0. We must have y=4x so that y - 4x is a factor. A long division shows that the condition for zero remainder is x=7 so y=28. The numbers of members are twice these.
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