The problem hanging off the end of this first paragraph can be solved in less than a yocto second by a Cray XK7 supercomp but that’s not the pomp. The real point is can you do it using nothing but the O2 guzzling stuff that weighs about 1400 gms and dreams the world with your eyes?
I have a ten-digit number which contains each of the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 and is exactly divisible by 10. If you remove the digit in the units place, the remaining nine-digit number is exactly divisible by 9. Again, if you remove the digit in the units place, the remaining eight-digit number is exactly divisible by 8. This process can be continued as described above till all the numbers have been knocked off one by one. What’s the number?
(The older problem was to discover in which, and where, in a Shakespearian play do the words “Want My Baby” appear in an acrostic as if the bard was hiding a message to someone?)
I was trying to unravel your giving away of the hint. After wracking my brain and going through the names of all plays of the bard, I decided to look at The Comedy of Errors. There it is in act 1/scene 1. This drama opens in the palace of the Duke of Ephesus as he is giving audience to a Syracusan merchant Aegon. -- Saishankar Swaminathan, email@example.com
It is from The Comedy of Errors. It occurs in the second to tenth lines in the fifth dialogue of Duke Solinus in Act 1 Scene 1 with heading. Since you had given away a huge opening by mentioning Act 1 Scene 1, the search online was quite easy with focus on ‘my’ and either side. -- Abhay Prakash, firstname.lastname@example.org
(The other problem was: “At sunrise a monk climbs a hill to meditate. He walks at varying speeds, even stopping frequently. After some days, he comes down -- starting again at sunrise -- with varying speed along the same path. Can it be proven that there is a spot along the path that the monk will occupy on both trips at precisely the same time of day?”)
When the monk starts climbing down the hill at sunrise, assume that another person starts climbing up the hill from the bottom. The monk and the other person cross each other at one point only on their respective ways. If we assume that second person travelled at same speed as the monk on his onward journey (Actually speed doesn’t matter. -- MS), it follows that there is one spot the monk occupies at the same time of the day on both trips. -- Ravi Nidugondi, email@example.com
Instead of one monk going up and down a hill before and after meditation, let us meditate on, sorry, imagine two monks, one going up the hill and the other coming down the hill starting at sunrise. They traverse the path in similar manner: hurrying, slowing and halting at same places. They meet at a point at a particular time of the day but continue their walk. The meeting point is the spot where a monk going up and down the hill occupies at that time of the day. -- Ramakrishna Bhogadi, firstname.lastname@example.org
(Among the first five who also got it correct are: Guruchandran P S, email@example.com; Narayana Murty Karri, firstname.lastname@example.org; Dhruv Narayan, email@example.com; Ajit Athle, firstname.lastname@example.org; Balagopalan Nair, email@example.com.)
(The third one was: “You’re given a large bucket filled with water, a wooden ruler and a soccer ball. Determine – approximately -- the diameter of the ball using only these items.”)
Let the volume of bucket be (A). When the ball is completely immersed, water will flow out. After taking the ball out, the volume of bucket will reduce as water level has reduced. Let this volume be (B). The volume of the soccer ball will be A - B. Let this volume be (C). The volume of sphere is 4/3*22/7*r^3 = volume of ball (C) From this the radius can be found out and by multiplying by 2 we get the diameter. -- Advaith Ram Ravichandran, firstname.lastname@example.org
BUT GOOGLE THIS NOW
1. Although the Celsius and centigrade scales are the same today, originally there were two differences between them. (One of them is easy to Google, the other isn’t.)
2. Bowler bowls, batsman misses, ball misses wicket, keeper can’t collect, ball keeps going. Batsmen go for a run, fielder fields ball, throws it to keeper who, before non-striker can make it to crease, whips the bails off. Who’s out: striker stumped or non-striker run out?
— Sharma is a scriptwriter and former editor of Science Today magazine.(email@example.com)