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Post by Deleted on Sept 28, 2009 18:21:55 GMT -5
I've been out of school way too long to figure this out, other than by just trying combinations. Is there an easy solution? I know one answer and I think there are at least 2 more.
x/y = 1/5
x and y are some combination of 1 to 9 with no digit used twice.
I've figured out:
1) x must be 4 digits and y is 5 digits 2) first digit of y can't be greater than 4 (@) 3) last digit of y has to be 5 4) last digit of x must be an odd number (*)
_ _ _ * _______
@_ _ _ 5
But where to from there? Is there a simple way to figure this out? (3297/16485 is one answer.)
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Post by Phaedrus on Sept 28, 2009 19:54:50 GMT -5
I've been out of school way too long to figure this out, other than by just trying combinations. Is there an easy solution? I know one answer and I think there are at least 2 more. x/y = 1/5 x and y are some combination of 1 to 9 with no digit used twice. I've figured out: 1) x must be 4 digits and y is 5 digits 2) first digit of y can't be greater than 4 (@) 3) last digit of y has to be 5 4) last digit of x must be an odd number (*) _ _ _ * _______ @_ _ _ 5 But where to from there? Is there a simple way to figure this out? (3297/16485 is one answer.) Rules 3 and 4 mean the same thing. Rule 1 just means that the first digit of the numerator can not be 1. Rule 2 means that the largest possible first digit for the numerator is 9, which would result in 45000 plus what ever. Since there can be no repeated digits, it can't be 99XX so it saves you from getting close to the limit. Otherwise, just use your calculator.
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Post by Deleted on Sept 28, 2009 20:30:15 GMT -5
Use your calculator? Really? Aren't there an awful lot of possible combinations?
6/30 only uses 2 of the digits and uses one not permitted (0).
No, you don't have to follow those rules. It's just what I've hammered out so far. I was hoping for more.
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Post by mikegarrison on Sept 28, 2009 21:03:10 GMT -5
These kind of number games are not my strength. I can follow how you derived your rules, and can point out that means there are only four possible digits that can go in position * and only four possible digits that can go in position @, which cuts down on your trial and error. But I don't quickly see any more rules, though I wouldn't be surprised if there are some.
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Post by Phaedrus on Sept 28, 2009 21:12:03 GMT -5
Well, the problem is kind of ill formed, the rules limit the numerator to be between 2XXX and 9XXX So, you can start with 2013 and go until 9876 and multiply by 5 until you get a number that does not have repeated digits while skipping the numbers that have repeated digits in the numerator. Thats the scorched earth method.
Another method is to work on the denominator, starting with every five digit number ending in 5 without repeated digits and then pick the ones that end up following the rule.
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Post by The Bofa on the Sofa on Sept 28, 2009 22:48:21 GMT -5
I get 12 of them
2697 13485 2769 13845 2937 14685 2967 14835 2973 14865 3297 16485 3729 18645 6297 31485 7629 38145 9237 46185 9627 48135 9723 48615
admittedly, I did it brute force with excel
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Post by XAsstCoach on Sept 29, 2009 6:46:53 GMT -5
I see...so they must use all 9 digits. Was thinking "what's wrong with 7/35, or 9/45, or 3/15?"
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Post by The Bofa on the Sofa on Sept 29, 2009 7:30:00 GMT -5
Well, the problem is kind of ill formed, the rules limit the numerator to be between 2XXX and 9XXX So, you can start with 2013 and go until 9876 and multiply by 5 until you get a number that does not have repeated digits while skipping the numbers that have repeated digits in the numerator. Thats the scorched earth method. Actually, there are a lot of broad cancelations you can do without too much thinking. For example, it can't be 2013 (no zeros). In fact you can rule out Everything up to 2137 (no 20XX, no 210x, no 211x, no 212x) Heck, everything below 2200 can be ruled out (*5 gives 10XXX) Everything below 2400 can be ruled out (gives 11xxx) Everything below 2600 can be ruled out (gives 12xxx - we already have a 2) It doesn't take much work to show that the only possibilities in the 2600s are 2679 2689 2697 2700s get a little more complicated, but start with 2769 (an answer) 2789 That's it. The key is that it must end in an odd, and it can't have 5 or 3 (because *5 is 13xxx). You also can't with an even number and 7, and it can never end in 49 2800s open a few more options, but their is nothing between 2987 and 3247 (I think that is the next option) Also block out everyone between 3300 and 3400 3800 and 4200 4400 and 4600 5000 and 6200 6600 and 6800 7000 and 7200 7700 and 8200 8800 and 9200 everything above 9873 Given that, there actually aren't that many possibilities. Probably not too more than a couple hundred, at most.
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Post by Deleted on Sept 29, 2009 7:39:08 GMT -5
I guess I don't understand why my 12-year-old would get this as a math problem.
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Post by Thrill of the 'ville on Sept 29, 2009 8:09:48 GMT -5
I guess I don't understand why my 12-year-old would get this as a math problem. I worked in a 5th grade classroom last year and an 8th grade class this year....neither teacher would make this HW for a student. Unless it was for extra credit.
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Post by The Bofa on the Sofa on Sept 29, 2009 9:27:01 GMT -5
I guess I don't understand why my 12-year-old would get this as a math problem. Stupid number tricks aren't math. This is no more math than sodoku is. In particular, the requirement that all the digits 1 - 9 are used only once basically takes the math aspect out of it
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Post by mikegarrison on Sept 29, 2009 9:42:21 GMT -5
I guess I don't understand why my 12-year-old would get this as a math problem. Stupid number tricks aren't math. This is no more math than sodoku is. I partially agree, but partially disagree. This isn't math, but it uses math. Sodoku isn't math, nor does it use math.
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Post by The Bofa on the Sofa on Sept 29, 2009 10:47:38 GMT -5
The only math is "Multiply a 4 digit number by 5"
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Post by mikegarrison on Sept 29, 2009 11:33:16 GMT -5
The only math is "Multiply a 4 digit number by 5" Yeah, but that's more than nothing. Sodoku can use any random symbols.
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Post by The Bofa on the Sofa on Sept 29, 2009 12:33:05 GMT -5
The only math is "Multiply a 4 digit number by 5" Yeah, but that's more than nothing. Sodoku can use any random symbols. I have a great game on my Blackberry called "Greater Than Sodoku," which puts inequality signs between the squares within the groups of 9. It's a lot more fun than normal Sodoku (for me). For one thing, it's fun to start with a completely empy sheet and find the unique solution. That version has as much math as this problem
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