Picking your 2th.... as you grin and bear it!
Here’s the problem. You have seven toothpicks on a table arranged in such a way as to make two adjacent squares one on top of the other like the number 8 in calculator typeface. What you have to do is, by moving only two toothpicks make . . . .200! (You can’t stack, bend, break or remove any toothpick.)
Here’s the solution. Take the upper right hand toothpick and place it flat and horizontally on top of the figure of 8. The figure should now look like a 6 with a bar over it. Now take the toothpick on the lower left of this figure of 6 and place it flat and vertically just on top of the bar on top of the 6. Just go ahead, get some toothpicks and just do it and see. But here’s your metaproblem now: How does this solve the original problem?
THROUGHPUT
(The problem submitted by Ajit Athle, ajitathle@gmail.com was: “If in rectangle ABCD there is a point E (not coincident with any of the vertices) such that AE, BE, CE and DE are all distinct integer lengths, what is the minimum value of AE + BE + CE + DE?”)
The minimum integer lengths that constitute a Pythagorean Triple are 3, 4 and 5. Let the
rectangle be ABCD with the lengths of parallel sides AB and CD be each equal to 4. Let the longer parallel sides AD and BC be each of length equal to 6. Let the middle point of the side AD be E such that the length AE = length ED = 3. Join E to B and E to C. Then, by Pythagoras’s theorem, length EB = length EC = square root of (9 + 16) = 5. The sum of lengths (AE + BE + CE + DE) = 3 + 5 + 5 + 3 = 16, which is the minimum value for that sum. -- Narayana Murty Karri, k_n_murty@yahoo.com
(However, for what it’s worth the submitter’s solution is: “By the British Flag Theorem, AE² + CE² = BE² + DE². The smallest integral solution to the equation above is, 1² + 8² = 7² + 4² which gives the smallest value of the expression as (1 + 8 + 7 + 4) = 20.”)
(The second problem was: “Divide the digits 1 to 9 into three sets. Must the product of the integers in one of the sets always exceed 71? If yes, how close to 71 can you get?”)
The product must exceed 71 in at least one of the sets. This is because 1*2*3*4*5*6*7*8*9 can be rearranged 63*64*90 or 70*72*72. If any of the factors were to be altered, the higher value of the factor will go up. If the product can exceed 71 in only one of the sets, the three sets will be (1, 7, 9), (2, 4, 8) and (3, 5, 6). The closest to 71 will be 90. However, if the product can exceed 71 in two sets, the three sets will be
(2, 5, 7), (3, 4, 6) and (1, 8, 9). The closest to 71 will be 72 , albeit twice. -- -J Vaseekhar Manuel, orcontactme@gmail.com
Our problem is to form three groups of integers from the range of 1 to 9 whose products are to be 70, 72 and 72. With eight integers, it can be groups of 2, 3 and 3 integers. With 2 integers we can get a product of 72, by 8 and 9. With left out integers 2 to 7, by trials two groups (2, 5 and 7) and (3, 4 and 6) with products of 70 and 72 can be formed, which are the closest to 72. -- Shashidhara Sultanipur, shashidharasnpwd@gmail.com
Among the first five who also got it right are: Dhruv Narayan, dhruv510@gmail.com; Hemalatha T, hemalatha1956@gmail.com; Seshagiri Row Karry, srkarry@yahoo.com; Abhay Prakash, abhayprakash@hotmail.com; Gaurav Panth, sgurupa@gmail.com.)
(The third one was something we’re beginning to get pretty sick of by now so we’ll just run a correct answer or something and go to sleep forever on it.)
The orbiting pistol and bullet can be reckoned as satellites of Earth and consequently, will obey Kepler’s laws of planetary motion. If the bullet has to catch up with the pistol and overtake it, the velocity of the bullet should keep increasing and that of the pistol keep decreasing. This leads to a violation of the conservation and Kepler’s laws. Accordingly, after the initial spurt in the velocity of the pistol, the bullet and pistol occupy orbits that obey Kepler’s laws of planetary motion and respective laws of conservation. -- Balagopalan Nair K, balagopalannair@gmail.com
BUT GOOGLE THIS NOW
1. Write “oLr” on the first line. Then write “elteltelt” on the second, third and fourth lines below it. Why is this a human (rights) being?
2. 16435934; 51966; 14600926; 14613198; 13286381; 47838; 11649182; 48813. Which is the odd one out?
Mukul Sharma is a scriptwriter and former editor of Science Today magazine.(mukul.mindsport@gmail.com)