|By Fairyofwind (Fairyofwind) on Tuesday, September 30, 2003 - 07:17 pm: Edit|
YOU + PAY + CASH = PLESE.
Each letter represents a unique digit from 0-9. Find all letters using reasoning, no trial and error.
|By Kewkiekid (Kewkiekid) on Tuesday, September 30, 2003 - 08:28 pm: Edit|
you spelled please wrong.
sorry, no clue how to do these problems.
|By Fairyofwind (Fairyofwind) on Tuesday, September 30, 2003 - 08:29 pm: Edit|
I know that's on purpose
|By Billiam2 (Billiam2) on Tuesday, September 30, 2003 - 09:42 pm: Edit|
whats this for?
|By Fairyofwind (Fairyofwind) on Tuesday, September 30, 2003 - 09:53 pm: Edit|
This was a ridiculous math problem on today's varsity math team meet. Not only did I spend 40 minutes/60 minutes of the contest on this problem, I got two other easy problems wrong, ending with a 7/10. Instead of putting the probability, I put the number (120 instead of 120/200). Everyone laughed at me: 120 probability! And then I got an easy circle sector/segment question wrong. All because I didn't have time to check my work because of this stupid problem. The other USAMO participant from our sister school didn't get it either. *sigh*
|By Jimjunior (Jimjunior) on Tuesday, September 30, 2003 - 09:57 pm: Edit|
Solved it, that was tough.
1: P=1, no more than 1 could have carried to a fifth column
2: C=9, no more than 1 could have carried into the fourth column, therefor C must be 9 for a fifth column to exist
3: L=0, if C + 1=10, then the answer will have a value of 0 in the fourth column
4: The sum of U,Y and H must be greater than 9, because E cannot equal 9, and no three remaining integers sum to less than 9. they cannot sum to more than 20, because then E would have a value of 0 or 1
5: The sum of A and O must be 9, because 1 carries over from first column, and A+O+S+1 has S as a units digit
6: A+P+Y+1=A+Y+2=U+Y+H, we know this because the units digit of the sum from the first and third columns are equivalent. It follows that A+2=U+H
7: We can postulate that A+O=U+H=9, this would mean that Y-1=E
8: It follows that A=7, O=2, U=6, H=3 (U and H can be switched). They have values of 6 and 3 because the two of the remaining numbers must have a difference of 1
9: Our three numbers left are 5,4 and 8. So Y=5 and E=4 since 5-4=1 The remaining variable is S, so S=8
We can now check to see if this works.
526+175+9783=10484 It does, IT WORKS
any questions, ask
|By Fairyofwind (Fairyofwind) on Tuesday, September 30, 2003 - 10:04 pm: Edit|
|By Billiam2 (Billiam2) on Tuesday, September 30, 2003 - 10:26 pm: Edit|
is this a local meet? mind posting some of the other questions?
|By Billiam2 (Billiam2) on Tuesday, September 30, 2003 - 10:31 pm: Edit|
how often do you have contests? We havent had one yet, and there usually arent that many throughout the year. I wish there were more.
|By Fairyofwind (Fairyofwind) on Tuesday, September 30, 2003 - 10:46 pm: Edit|
There is a contest once a week. Either Mandelbrot, DVML, CVC, NJML, APML, AMC/AIME/USAMO, AMTNJ, HARVARD/MIT, ARML, ANDRUSHKIW. Today was a DVML 10-question 1 hour travel meet.
|By Billiam2 (Billiam2) on Tuesday, September 30, 2003 - 10:52 pm: Edit|
wow, you're pretty lucky. What team for ARML are you? Central Jersey?
|By Geniusash (Geniusash) on Tuesday, September 30, 2003 - 11:23 pm: Edit|
This problem is A LOT easier if you write it this way.
|By Jimjunior (Jimjunior) on Tuesday, September 30, 2003 - 11:53 pm: Edit|
There are more solutions if P is set to 0.
I have done the Harvard/MIT tests and found them to be very hard, for the wrong reasons. Old AIME is my favorite.
|By Billiam2 (Billiam2) on Wednesday, October 01, 2003 - 12:07 am: Edit|
why are they hard for the wrong reasons? My opinion is that the harvard/mit is more about how many tricks you've seen, since many of them appear tough, but are trivial if you know the tricks. If you dont know the trick, its very difficult to get the right answer. But the AIME is not about how many tricks you know necessarily, i think its more about your ingenuity than anything else.
|By Jimjunior (Jimjunior) on Wednesday, October 01, 2003 - 02:07 am: Edit|
Most of the Harvard/MIT ones use advanced math and are only a few steps. If one hasnt seen the math on it before, then one has little hope in answering the questions. The AIME problems can be solved in a variety of ways, and are really all about seeing through the problem and coming up with a solution
|By Ali_Liu (Ali_Liu) on Wednesday, October 01, 2003 - 10:39 pm: Edit|
I did it!! oh god..i'm so happy!!!! It took me 15 minutes, however=(
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