Letter and Word Puzzle: Bag = 2 + 1 + 7 = 10Date: 02/20/2003 at 19:46:41 From: JJ Subject: What words? What words can I use to make 10? Bag = 2+1+7 = 10 Date: 02/21/2003 at 17:34:50 From: Doctor Ian Subject: Re: What words? Hi JJ, If you really have to get the letters to add up to 10, then you can't use any letters after I, right? a b c d e f g h i 1 2 3 4 5 6 7 8 9 That's going to put some pretty severe limits on what words you'll be able to find. So you might be able to do a systematic search. One thing we can do is consider the first two letters of possible words: 2nd a b c d e f g h i a b c d 1st e f g h i We can rule out a lot of words by noting that we can't start with two consonants. For example, 'bc' can't start a word, so we can put an 'x' where it would go in the table. And we can do it for other consonant pairs, like 'fd' and 'cd'. Note that 'ch' can start a word, so we'll leave that open: 2nd a b c d e f g h i a b x x x x x x c x x x x x d x x x x x x 1st e f x x x x x x g x x x x x x h x x x x x x i In each of the other spaces, we can add up the letters we'd get by combining the two letters. Some combinations (like 'fi') will give us more than 10 with just two letters, so we'll put an '-' in for each of those: 2nd a b c d e f g h i a 2 3 4 5 6 7 8 9 10 b 3 x x x 7 x x x - c 4 x x x 8 x x - - d 5 x x x 9 x x x - 1st e 6 7 8 9 10 x x x - f 7 x x x - x x x - g 8 x x x - x x x - h 9 x x x - x x x - i 10 - - - - - - - - Now, 'ai', 'ia', and 'ee' aren't words, even though they add up to 10. So we can eliminate them from the table: 2nd a b c d e f g h i a 2 3 4 5 6 7 8 9 x b 3 x x x 7 x x x - c 4 x x x 8 x x - - d 5 x x x 9 x x x - 1st e 6 7 8 9 x x x x - f 7 x x x - x x x - g 8 x x x - x x x - h 9 x x x - x x x - i x - - - - - - - - So now we have a small set of remaining possibilities. They're probably easier to see if we remove all the x's and -'s: 2nd a b c d e f g h a 2 3 4 5 6 7 8 9 b 3 7 c 4 8 d 5 9 1st e 6 7 8 9 f 7 g 8 h 9 Note that 'i' has been eliminated completely from the table. Now, we can start eliminating what remains. Let's start with the highest numbers first. We have ed, de, ah, ha all of which add up to 9. The only possible thing we could do would be to add an 'a' on the end, to get eda, dea, aha, haa none of which are words. So we can eliminate these from the table: 2nd a b c d e f g a 2 3 4 5 6 7 8 b 3 7 c 4 8 d 5 1st e 6 7 8 f 7 g 8 Note that we've knocked another letter, 'h', right out of the table as well. Now the highest number is 8. We have ag, ce, ec, ga To complete an 8, we need two more, so we can either add two a's, or one b: agaa, agb, ceaa, ceb, ecaa, ecb, gaaa, gab --- And we have a winner: 'gab'! And since that's the only one we can get from an 8, we can remove the 8's from the table. 2nd a b c d e f a 2 3 4 5 6 7 b 3 7 c 4 d 5 1st e 6 7 f 7 The next highest number is 7. We have af, be, eb, fa To complete a 7, we can add 'aaa', 'ab', 'ba', or 'c': afaaa, afab, afba, afc beaaa, beab, beba, bec ebaaa, ebab, ebba, ebc faaaa, faab, faba, fac Now, none of these is a word... although we can see that we'll eventually get 'babe'! But that should happen in the course of things, so there's no need to rush it. (Similarly, having found 'gab', we know we'll eventually get to 'bag'.) Anyway, we can eliminate the 7's: 2nd a b c d e a 2 3 4 5 6 b 3 c 4 d 5 1st e 6 As you can see, you can keep reducing the possibilities this way. And if you do, you'll eventually consider every possible word (while ignoring the ones that are clearly impossible). Note that as you consider ways to complete words, you can ignore the sillier possibilities. For example, in eliminating the 7's, we didn't really have to add 'aaa' to the ends of words, because no word ends in 'aaa'. So for 6, the theoretical completions are 6 + aaaa, aab, aba, baa, ac, ca, d but you can safely ignore 'aaaa', 'aab', and 'baa'. So you just have to add 'aba', 'ac', 'ca', and 'd' to 'ae' and 'ea'. Can you take it from here? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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