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| For The Mathematically Inclined |
Anil Saigal
07/21/2008
FOR STUDENTS PROBLEM # 1 FOR GROWNUPS PROBLEM # 2 Please send your solutions to anil@lokvani.com. ------------------------------------------------------------------------------------------------------------- Solution: Solution: Congratulations to Siddharth Annaldasula, Somnath Chattopadhyay and Shailesh Goregaokar, who were winners of the last set of puzzles.
What is the height of a pentagon with side length of 1?
You have a circular piece of paper with radius 1 and wish to form a cone shaped cup. You may
cut out any wedge you like from the paper, call the angle of the wedge x. The
point of the wedge must be at the center of the circle. After
cutting out the wedge you then attach the two straight edges remaining to form a
cone. Assuming the paper cone can hold water, what should x be to maximize the water
holding capacity of the cone?
Use Problem Solutions M-072408 as the subject line. Please include your full name in the text of the main message. Everyone with the right answer will be acknowledged in the next issue of Lokvani.
Please do not post your solution in Post Comments. No credit will be given for solutions not sent to anil@lokvani.com.
If you need clarification on any problem, please contact anil@lokvani.com.
PROBLEM # 1
What is the approximate ratio of the area of the shaded square compared to the area of the large square? (See Fig. 1)
1/8.
FOR GROWNUPS
Show that 10^2008 − 10^8 is divisible by 2008.
10^2008 − 10^8 = 10^8 (10^251 -1)
10^8 is divisible by 8
(10^251 -1) is divisible by 251 from Fermat’s Little Theorem (1640) which states for any integer ‘a’ (in this case, 10) and a prime number ‘p’ (in this case 251), a^p -1 is divisible by p.
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Fig. 1