EPTMy set:
+ Show Spoiler +
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 19, 29, 39, 49, 59, 69, 79, 89, 99
Math Problem
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motbob
United States12546 Posts
My set: + Show Spoiler + | ||
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ninjafetus
United States231 Posts
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minus_human
4784 Posts
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motbob
United States12546 Posts
On February 06 2009 23:50 minus_human wrote: I don't understand your problem OK, I tried to clarify it. | ||
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o3.power91
Bahrain5288 Posts
 edit: i just noticed i have the exact same comment as the first reply XD | ||
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Saracen
United States5139 Posts
my set: S{0,1,2,3,4,5,6,7,8,9} and 10n+a from n=1 to n=9 where a is an integer between -1 and 10 | ||
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FaCE_1
Canada6185 Posts
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Saracen
United States5139 Posts
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Divinek
Canada4045 Posts
On February 07 2009 01:28 Saracen wrote: how so... because his set has to be able to add every number to get 1-100, not multiply it afterwards to do so. er right, you need 0 for 1, duh. | ||
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huameng
United States1133 Posts
To the OP: I think you have the correct solution | ||
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Saracen
United States5139 Posts
On February 07 2009 01:30 Divinek wrote: because his set has to be able to add every number to get 1-100, not multiply it afterwards to do so. also op why do you have 0 in your set, is that required? which is what my set does {0,1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90} or {0,1,2,3,4,5,6,7,8,9,11,21,31,41,51,61,71,81,91} or {0,1,2,3,4,5,6,7,8,9,12,22,32,42,52,62,72,82,92} ... and yes, 0 is required | ||
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LTT
Shakuras1095 Posts
0, 1, 2, 3, 4, 5, 6, 7, 8, 17, 26, 35, 44, 53, 62, 71, 80, 89, 98 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 19, 29, 39, 49, 59, 69, 79, 89, 99 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 21, 32, 43, 54, 65, 76, 87, 98 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 23, 35, 47, 59, 71, 83, 95 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 25, 38, 51, 64, 77, 90 There are also variations of those. Take his example using 0-9 as his "base". 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 30, 40, 50, 60, 70, 80, 90 All of the above have 19 terms. The total possible summations for a set S(n) containing n elements is C(n,2) + n. From this we can gather tht the absolute minimum of terms to even get 100 summations is 14. C(14,2) + 14 = 105 You will need at least 14 elements to even have a chance of covering 0-100. Whether any below 19 elements can actually cover...dunno. | ||
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blapsd
England121 Posts
First im going to split the set up into 2 subsets the 1st subset consists of the consecutive numbers 0 - 9 and the second subset consists of the rest. We can change the number of integers in our first set and this directly affects how many integers we need in our second. I'm going to call the number we go up to in our first set x (in your case this is 9) ok now lets count how many numbers we have x + 1 in the first subset (the +1 is because of the 0) Now we need to make 101 numbers in total (101 because of the 0 remember) and we now have 101 - x numbers left to make. Now we need a number in our second subset for every x+1th number from x to 100.....for example you needed a number every 9+1 times in your 2nd subset. Now we add up the total number of integers we have: x+1 (from the first subset) + (100 - x)/(x + 1) (from the second subset) now form the equation : y = x + 1 + (100-x)/(x+1) This is the equation for the number of integers in our set. we need to MINIMISE this To do this we simply differentiate to get (TIP: USE THE QUOTIENT RULE) dy/dx = 1 + (-101)/((x+1)^2) put this equal to zero and you finally get x is around 9.05 rounded up a little. This therefore means for the number of integers in your set to be minimum you need x = 9 NOTE: we are only working in integers and you are allowed to round down due to there being remainders when you divide 101-x by x+1. If you're not convinced just test it for x = 9 and x=10 because i've proved the answer lies between 9 and 10 so you were right x = 9 or putting it back into the equation for y, y = 19.1 (which we are allowed to round down to the nearest integer also) so the minimum number of integers is 19. which means your set is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 19, 29, 39, 49, 59, 69, 79, 89, 99 also note theres other sets that allow 19 numbers but no sets that contain less than 19 Hope that wasnt too hard to follow. | ||
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Saracen
United States5139 Posts
On February 07 2009 01:41 LTT wrote: There is more than 1 solution. You can also get it with 0, 1, 2, 3, 4, 5, 6, 7, 8, 17, 26, 35, 44, 53, 62, 71, 80, 89, 98 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 19, 29, 39, 49, 59, 69, 79, 89, 99 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 21, 32, 43, 54, 65, 76, 87, 98 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 23, 35, 47, 59, 71, 83, 95 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 25, 38, 51, 64, 77, 90 no you can't unless you can repeat numbers, but then you wouldn't need 0 | ||
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motbob
United States12546 Posts
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motbob
United States12546 Posts
On February 07 2009 01:46 Saracen wrote: no you can't unless you can repeat numbers, but then you wouldn't need 0 One, you can repeat numbers (it's in the OP) and two, you need zero for 1 (0+1). | ||
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Saracen
United States5139 Posts
the numbers don't have to be non-negative ! On February 07 2009 01:55 motbob wrote: One, you can repeat numbers (it's in the OP) and two, you need zero for 1 (0+1). missed that sorry :\ | ||
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StRyKeR
United States1756 Posts
On February 07 2009 01:42 blapsd wrote: Ok, look at how many numbers you get in your set: First im going to split the set up into 2 subsets the 1st subset consists of the consecutive numbers 0 - 9 and the second subset consists of the rest. We can change the number of integers in our first set and this directly affects how many integers we need in our second. I'm going to call the number we go up to in our first set x (in your case this is 9) ok now lets count how many numbers we have x + 1 in the first subset (the +1 is because of the 0) Now we need to make 101 numbers in total (101 because of the 0 remember) and we now have 101 - x numbers left to make. This isn't true because the (x+1) numbers in the first subset account for more than x+1 numbers from 0 to 101. For example, having {0, 1, 4, 6} hits the set {0, 1, 2, 4, 5, 6, 7, 8, 10, 12}, since you are allowed to double numbers and also take pairs of numbers. In this case, while you have 4 elements from 0 to 9, you actually need to cover only 101 - 10 = 91 numbers now. There are issues with the rest of your argument because of this fact -- a set of numbers can generate more than its own size. | ||
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StRyKeR
United States1756 Posts
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blapsd
England121 Posts
This isn't true because the (x+1) numbers in the first subset account for more than x+1 numbers from 0 to 101. For example, having {0, 1, 4, 6} hits the set {0, 1, 2, 4, 5, 6, 7, 8, 10, 12}, since you are allowed to double numbers and also take pairs of numbers. Now we need a number in our second subset for every x+1th number from x to 100.....for example you needed a number every 9+1 times in your 2nd subset. There are other issues with the rest of the argument because of this fact -- a set of numbers can generate more than its own size. yes but remember you are only allowed to make one addition. If you doubled 6 to get 12 for example, you could never add 12 to any other numbers since we've used our 1 addition already. The first subset needs to be a list of numbers you can still add to the others in the 2nd subset, and hence cant be made by doubling or adding. | ||
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