And The Winners Are… Contest #2 (Four Twos)
Hopefully, a lot of you DID “Try this one at home?” For those who didn’t and would still like to give it a go, the rules and a hint are still posted in our blogs.
But again, the numbers of entries was disappointing - only 4, but they all qualified by having at least 15 or more correct expressions. Congratulations go to Robert Madonna, Joe Scocas, John Haynes, and David Levine. David actually got all 20!!
Like Contest #1, I’m trying to encourage more to participate in these and have decided to award a prize to all 4 contestants. I didn’t have the heart to disappoint 3 of them for the hard work they did by drawing only one name. Robert, Joe, and John have each won a $25 gift certificate, good toward any 3/4 ” green-tagged items in the shop. David for actually getting all 20, something those Mensans back in 1976 couldn’t do, wins a $50 gift certificate for any 3/4″ green-tagged items. Prizes must be claimed and redeemed by October 26, 2007!
Now, on to some of those interesting solutions. The two numbers that eluded everyone back then were 17 & 19. Until David came up with a solution to 17, I had never seen one except for my own, different, solution. 19 wasn’t terribly difficult, once you realized that (.2), two-tenths, was acceptible. 19 = (2 + 2 - (.2)) / (.2)
David used the “double factorial” function. Normal (single) factorial, introduced in basic algebra and involving simple probability is indicated by an exclamation mark and is used as such: 4! = 4 times 3 times 2 times 1 = 24. 5! = 5X4X3X2X1 = 120. The double factorial (!!) is not as common and is used as such: 6!! = 6 times 4 times 2 = 48 (note 6 is even). However, 5!! or (2 / .2 / 2)!! = 5 times 3 times 1 = 15. Notice that David only used three twos here. Add one more two and ya got yer 17!!! Sweet!
The solution I offered so many years ago involved the use of the radical. Not all radical signs indicate finding the square root. Without an index number, it is assumed the index number is “2″ and one is to find the square root. The number under the radical sign is being raised to the one over (whatever the index number is) power. A square root is the 1/2 power. Should you see a “3″ in the index number position, you are finding the cube root, the 1/3 power. What if the index number is (.2)? Then the number under the radical would be raised to the 1/(.2) power; but 1/(.2) =5. 2 raised to the 5th power is 32. Add two, then divide the 34 by two and you get…
Joe Scocas had an interesting attempt: He used a summation symbol, setting the range of “i” from (-x) to (+x). “x” was defined as 2+2+2+2. The expression was (i - i)! The introduction of the extra variables was a little fuzzy, but I liked the effort!
I’m not going to go over all of the different solutions. Some have many possibilities and perhaps what you have read here will inspire you to try again. There will be at least one more contest. If we don’t have a better response, I will give up. Check back!
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