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I wonder if a couple of you with good mathematical intellects could do me a favor?
There was a mathematical question which appeared in the wordplay category a couple of weeks back. I was the only who solved it(there were only 5 answers)and it was a good solution, but as u know many people take the answers and run. Anyways one guy said it couldn't be solved, which was not true. I believe he voted for his answer out of spite and I voted for mine, and these are the only 2 votes cast so far.
Anyways if you type in "I have a question please solve the same" into the yahoo questions search engine, it will come up. Just read the problem, and the only 2 serious answers, and cast a vote for what you truly think is the best answer. I don't care about the points since if I receive 10 points for that question, or not, I'll be awarding 10 points for an answer here. When you've done that then answer this and render your opinion. Thanks for doing this if you do.
Thanks for looking at the problem and rendering an opinion. However, of course I disagree. Even though not specifically stated It is very clear that each letter was meant to represent a unique digit from 0 to 9, since there are exactly 10 distinct letters and that's also precisely why the letter A was used twice in different positions. That's why this condition is implicit in the problem. As soon you realize the C can't be 0 for the reason I stated then the argument immedately follows and there is only one possible solution. Of course there are an infinite number of solutions if you do not recognize that which was obviously implicit. The fact that without this assumption the question doesn't make sense and yet with it you get exactly one solution proves that that was what was intended. Anyways, we agree to disagree. Actually, it didn't take long to solve. My credentials: a math background and an IQ of 160. Anyways, thanks and take care.
8 Answers
- Anonymous2 decades agoFavorite Answer
Jimbo, believe in yourself. You have given enough right answers to know what you are capable of. Thanks for an answer you gave me, that I needed, so I could help my wife with her college math class. Keep up the GREAT WORK!
- 2 decades ago
Your solution looks ok, of course based on the assumption that different letters represent different digits.
The argument used by the other party, that there are not enough equations, is really short sighted. It is true for linear equations on the real or complex field, but you cannot extrapolate from there. A typical example would be Fermat's last Theorem, which says for example that x^3+y^3-z^3=0 has NO solution: one equation, three variables, no solution. Or, another example, (x/3)^2+(y/5)^2=2 has a unique solution over the positive integers.
In the case we are talking about, the assumption that all letters represent different digits is enough to force a unique solution. And although it is not stated in the problem, the fact that there are precisely 10 different letters involved in the problem should be a good hint.
- TychaBraheLv 72 decades ago
I'm sorry, StillNoCouch, but you're wrong. These puzzles are common, and it is the teacher's fault that the critical "each letter corresponds to one digit 0-9" was left out.
Jimbo solved this in exactly the same way these puzzles are solved every month by readers of such magazines as Dell Math Puzzles & Logic Problems. Personally, I prefer Word Arithmatic, which is long division problems using ten letters which, when solved and placed in order representing 0-9, spell out a word or phrase.
- memetraderLv 62 decades ago
Your application of logical analysis is correct.
Those people who say in relation to this problem such things as "The question cannot be solved since only two equations are given whereas three variables are unknown" or "The fundamental properties of the commonly accepted mathematical logic dictates that two unknowns require a minimum of two relationships" quite clearly do not understand what they are saying.
In fact they do not even understand the difference between logical inference and arithmetic but that is their problem not yours.
Well done for solving it!
You will find a lot more amusing alphametic and cryptarithm problems here http://www.mathematik.uni-bielefeld.de/~sillke/PUZ...
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- TheOnlyBeldinLv 72 decades ago
I voted for it. Logic seemed good to me. (And I'll get 3 points out of the deal.)
- Anonymous2 decades ago
You could have posted the link, but I'll give it a whirl and update in a few ...
Source(s): UPDATE: Sorry, you loose. The question is not solvable. The fundamental properties of the commonly accepted mathematical logic dictates that two unknowns requrire a minimum of two relationships (see Newton). Additionally, your assumptions are not based within the confines of the question ... you assume too much. There are an infinate number of answers to that question in the base 10 system alone. I'm sorry that you wasted so much time on the answer ... it was very well written, well articulated and well thought out ... unfortunately, it wasn't correct. Hey, that happens ! I've made at least two mistakes in my life ! ;-) Sorry, the problem is not solvable ... the first rule in mathematics is ALWAYS to look at the problem first and have a clue whether or not it's workable. This one wasn't and can't be made so. It was a good try though. P.S. -- Yes, I do this for a living. UPDATE II: I looked at your previous posts and gave you 5 Thumbs-Up's. Your a bright kid. Kudo's to you. - Anonymous2 decades ago
don't be peeved, well atleast you are the brainy one.
