PUZZLE: 9
The answer is that the oranges should be kept in the baskets in the following manner: 1,2,4,8,16,32,64,128,256,512,1024,1953. Using some or all of the baskets one can get any number of oranges between 1 and 4000.The principle is the basis of the binary system of notation of numbers in contrast to denary system we use normally every day.
It can be seen that the above is nothing but the series 20,21,22,23,24,25,26,27,28,29,210 and 211 etc and the last number is [4000- the sum of all the previous numbers].
The principle involved in the answer is actually nothing other than the number logic that any positive integer can be expressed as either 2n(if the number is even) or (2n+1) if the number is odd and also the binary system of notation [01=1,10=2,11=3,100=4,101=5,110=6,111=7 etc.,] – by which using a single binary number consisting of zeroes and ones we can express any number.
Here in this basket case, by using a single basket or many baskets (excepting the first basket), which are all multiples of 2, and adding the first basket whenever an odd number is encountered we can express any natural number (positive integer).
The way in which the above series progresses is astonishing.While with two baskets we could have given any numbers up to only 3 and with 3 baskets only up to 7, with 10 baskets we are able to go up to 1023 and with 15 baskets up to 32,767 oranges. Such is the power of geometric progression and the number system.
In the puzzle3, we talked of weighing up to 40 kilograms using only 4 weights and we found that if we divide the weights as per series 1,3,9,27 etc-30,31,32,33 etc- we are able to weigh any weight up to the summation of all the weights.
What is the difference between the above two? The later series can also be used in the binary system of notation to get all the natural numbers we want. How? We can imagine a new system wherein the numbers can be notated as [01=1,10=3,11=4,100=9,101=10,111=13 etc.,]. But then how do we get the intermediate numbers? You remember what we did to get the intermediate weights. We just put the required weight on the other side of the pan. So in this system of notation, we need to ‘add’ or ‘subtract’ or ‘add and subtract’ the binary notated numbers to get all the numbers.
PUZZLE 6
AN UPHILL TASK
The answer is actually simple if one is not carried away by the unnecessary stress made on the fact that the climb and descent were made at inconsistent and totally unpredictable varying speeds.
The answer is that there will be one place in the mountain path where Tom was on both the days at the same time. How and Why? Let us assume a third day where two separate people mimic both the ascent and descent of Tom simultaneously- one starting from the top and another from the bottom at 06.00hrs. They are bound to cross each other at some point in the path. Won’t they? That would be the place where Tom was on both the days at that time.
For this problem it is not necessary that both the start and end time of both the journeys should coincide. What condition of timings would be enough for Tom to be in the same place on both the days? I am sure that by tracing the above logic, you can figure it out.
PUZZLE-7
TO BOAT AGAINST THE CURRENT?
Well, on the face of it, it appears as if Arthur is correct. Isn’t it? The speed of the boat in the river upstream (against the current) will be 15 Km/ hr [25-10] and the speed of the boat in the downstream (along the current) will be 35 Km/ hr [25+10]. They are after all traveling the same distance up and downstream; so the loss and gain should cancel out. Should it?
Appearances can always be deceptive. The main point is, “ Yes! If you take the arithmetic average of both the speeds, they will cancel out. But the time taken in the upstream and downstream journeys are not the same."
Let us take the distance between A and B as ‘x’. Then for the upstream journey the time taken will be x/15hrs and for the downward journey will be x/35 hrs. Hence the total time taken will be {(x/15) + (x/35)}=(2x/21) hrs. In a still water lake for the same up and down journey, the total time taken would have been (2x/25) hrs, which is definitely a lesser time.
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