SIR , there is another shortest path problem of a solid cuboid…
an ant starts at one lower corner.. and wants to reach the diagonally opposite corner of the upper edge of the opposite face…
and it comes out that path is shortest when it passes through midpoint of sides… for not just opposite face , any face!
Girish, the midpoint would only be for a cube. For a random cuboid, it would have to be checked a bit more carefully. And coincidentally, tomorrow’s post will cover precisely that 😀
Thank You Sir
its a base ques quite simple but coudnt get it . What is the number of zeroes at the end of 100! if it is expressed in base 21?
Shalini, the numbers of zeroes in a base is equivalent to the highest power of that base present in the original number. So 21 = 3 * 7. Now the highest power of 3 in 100! is 48 whle that of 7 is 16 (lower than 48 and hence the limiting case) and hence the highest power of 21 is also going to be 16. So there should be 16 zeroes I guess.
Thnk u so much 🙂
LIKE FOR NOS. IN BASE 2.. WE WILL HAVE TO CHCK WD 2 DEN ??
Sir, shouldn’t it be 7 pi, not 14 pi in the last example ?
sir i am having hard time visualizing the second one…..i am thinking if we cut along the PB path and then hammer down the half cylinder, the 28cm should be double ??what part of my logic is wrong?
I am not sure what you mean – but I suspect either you are changing the shape of the figure or else trying to go through the interior. The ant must crawl, not fly.
but sir how the PA length got half,what i was meaning by the above comment is for the ant to crawl what if the cylinder got cut be by half along PB then to visualize how the rectangle is made the cylinder is hammered down to be a rectangle so PA should be greater that 28 in that scenario…
Fold a piece of A4 paper to make a hollow cylinder. Identify the points for the ant and the jam. Open it up. You will get your answer.
Yes….got it now…thanks much….!!
Can you please explain the second problem?
Finding it difficult to visualize the same and arrive at the values.
Instead of my telling you, try to visualise it physically. Take various rectangles of paper (4 x 3, 5 x 3, 6 x 2 etc), draw their diagonals, and fold them to form cylinders. You should be able to understand for yourself then, always a better way than taking someone’s word for it!
In ant question, You said there can be the case where d2 > d1.
However, for triangle APB:
d1 = AP + PB
d2 = AB
AP + PB > AB (sum of 2 sided of a triangle)
d1 > d2
So, as per my understanding there can never be case when d2>d1.
The triangle is formed in the second case, after opening, and AP there is along the circumference of the upper face. Which is 14pi or around 44. So AB is definitely less than AP + PB in that figure i.e. 44 + 33. However, in the first case, AP is the diameter of the top face which is just 28.
Sorry, I got where I was wrong.
Thank you for wrriting this
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