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Nigel Parsons

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Non-Adaptive 12 Identical Balls

I've already given this solution in the original 12 balls problem, but here goes again: There is a solution to this which allows you to know in advance what three weighings/comparisons you will make, so that the choice of balls for the second and third weighing is not dependant on what has gone before. Label/identify each ball with a single letter: (using the letters from) "THE KP FORMULA" Weigh: TAKE against FOUR Weigh: PARK against THEM Weigh: HALF against MORE Record the results of the three weighings (Left pan heavy / left pan light / pans balance) By comparing the results with the pan contents you can identify the odd ball and whether it is heavy or light every time. eg. if the left pan always goes down then 'A' is heavy, if it always goes up 'A' is light. each ball has its own set of three unique weighings.

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Boys and Girls

It does seem to need further consideration, but the basis of 50/50 is true.

For those still unsure, start with 128 couples trying for families. The chances are (allowing for slight deviations from the norm) that 64 couples will have boys, and have to cease reproduction, the other 64 will have girls, and may choose to continue, but at this stage the score is 64 all.

The 64 families who choose to continue can expect to produce 32 boys & 32 girls. Again we have an equal number, and 32 families who can choose to carry on. Continuing this, we can expect (if everyone continues having children for as long as they are permitted) that the 128 families will end up with 128 boys & 128 girls, just some families will be smaller than others.

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Power of 2 (Microsoft Interview)

Mod2 will give zero for any even number, not just powers of 2.

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Good luck at an interview with Google

‚ó¶How much should you charge to wash all the windows in Seattle?

$75 per day (or whatever you think is a reasonable amount of pay). Just make it clear that you will do your best, but can't guarantee that you will be around long enough to finish the job.

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12 Identical Balls

There is a solution to this which allows you to know in advance what three weighings/comparisons you will make, so that the choice of balls for the second and third weighing is not dependant on what has gone before. Label/identify each ball with a single letter: THE KP FORMULA Weigh: TAKE against FOUR PARK against THEM HALF against MORE Record the results of the three weighings (Left pan heavy / left pan light / pans balance) By comparing the results with the pan contents you can identify the odd ball and whether it is heavy or light every time. eg if the left pan always goes down then 'A' is heavy, if it always goes up 'A' is light. each ball has its own set of three unique weighings.

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The Famous Hole in a Sphere

No, (or not necessarily). The puzzle doesn't state that. To drill a hole 6in long through the centre of the sphere, the sphere must have a minimum diameter of 6 inches. In which case, the hole would be of negligible diameter (and the remainder would effectively still be a 6 inch sphere) as otherwise the hole would be less that 6 inches long. If you start with a sphere of 7 inches then for the hole to go through it, and be only 6 inches long the hole would have to be wide enough so that the 'caps' taken off each end have a maximum thickness (at the centre) of 1/2 inch each. This will mean that the larger the sphere, the wider must be the 6 inch hole in order for it to be only 6 inches in length. As such the result can be calculated without knowing the size of the sphere, strange as that may sound.

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