City-Data Forum Mathematical Conundrum (power, display, system, function)
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09-07-2010, 11:01 PM
 Location: Victoria TX 42,668 posts, read 71,820,212 times Reputation: 35910

Here's a mathematical paradox I just thought of, as I was watching Brewers' pitcher Trevor Hoffman go for his 600th save. The fans had a three-card display, in which they could progressively change a card to raise his number by increments of one, each time he earned a save. Let's say you want to prepare a set of cards on flip rings, to line up to display any number, as needed, from 000 to 999. It will require 30 cards---0 through 9 printed on ten cards placed on the flip rings in each of the three positions.

But 1000 is not divisible by 30, so if you sequentially display each number from 000 to 999, some cards will have to be displayed more times than others. Which ones?

If you want to make all the 2-digit numbers, 20 cards are needed to make 100, and for 4-digit numbers, 40 cards will be needed to make 10,000 numbers. 50 cards for 100,000, etc. Those are divisible, so each card is displayed the same number of times. Why doesn't it work for 30?

Last edited by jtur88; 09-07-2010 at 11:09 PM..

09-07-2010, 11:06 PM
 Location: Las Flores, Orange County, CA 26,346 posts, read 80,996,080 times Reputation: 17414
I don't get the problem. The cards for the units digit will be displayed the most. The cards for the highest power of ten will be displayed once each.

What does the even divisibility have to do with it?

09-08-2010, 12:28 AM
 Location: Victoria TX 42,668 posts, read 71,820,212 times Reputation: 35910
No, in each of the three positions (units, tens, hundreds), each card is displayed the same number of times. One card from each position is displayed every time. For example, in the hundreds position, first we display the 0 card 100 times, then the 1 card 100 times, etc. until we exhaust the possible combinations of the tens and units. Every display will have one card from each group: units, tens, hundreds, so each will be displayed an equal number of times.

But we can't display each card an equal number of times, because there are 30 cards, which do not divide equally into the 1,000 different possible displays.

09-08-2010, 05:00 AM
 Location: Westwood, MA 3,514 posts, read 4,389,843 times Reputation: 4522
There are 1000 different numbers, but each number has 3 digits, so there are 3000 different places. 3000/30 = 100, so each digit will be used 100 times, exactly like you would expect (i.e. the 0 in the hundreds digit would be used for the numbers 000 - 099, etc.)

Another interesting question is which base would, in general, require the fewest number of cards for a given number.

09-08-2010, 10:57 AM
 Location: Victoria TX 42,668 posts, read 71,820,212 times Reputation: 35910
Quote:
 Originally Posted by jayrandom There are 1000 different numbers, but each number has 3 digits, so there are 3000 different places. 3000/30 = 100, so each digit will be used 100 times, exactly like you would expect (i.e. the 0 in the hundreds digit would be used for the numbers 000 - 099, etc.) .
But there are only 1,000 permutations, each of which uses specific cards, of which there are 30, and each card is used the same number of times.

I should have put this one over in the POS Forum, and let the Libertarians and Muslim haters apply their logic to it.

09-08-2010, 03:19 PM
 Location: Sherwood, OR 663 posts, read 1,611,236 times Reputation: 664
Quote:
 Originally Posted by jtur88 If you want to make all the 2-digit numbers, 20 cards are needed to make 100, and for 4-digit numbers, 40 cards will be needed to make 10,000 numbers. 50 cards for 100,000, etc. Those are divisible, so each card is displayed the same number of times. Why doesn't it work for 30?
Because its arbitrary that it worked for your other examples. If you wanted to display 1,000,000 different numbers you would need 60 cards. Is 1,000,000 evenly divisible by 60? What about 10,000,000?

09-08-2010, 04:20 PM
 Location: Victoria TX 42,668 posts, read 71,820,212 times Reputation: 35910
Quote:
 Originally Posted by LeftCoastee Because its arbitrary that it worked for your other examples. If you wanted to display 1,000,000 different numbers you would need 60 cards. Is 1,000,000 evenly divisible by 60? What about 10,000,000?
That may be true, but what is the reason why the number of permutations need not be divisible by the number of cards, if each card is used with equal frequency to form the permutations?

09-08-2010, 08:28 PM
 Location: Sherwood, OR 663 posts, read 1,611,236 times Reputation: 664
Quote:
 Originally Posted by jtur88 That may be true, but what is the reason why the number of permutations need not be divisible by the number of cards, if each card is used with equal frequency to form the permutations?
In effect, you are asking the same question as:

Why isn't a number evenly divisible by the number of digits in that number? You will get the same results, just 10 times larger

Ex. 00-99; 100/20 cards=5 or 100/2 digits=50

In your example of 1,000 and 30 cards, they are not 30 unique entities. They are 3 sets of unique numbers. There is one set for every digit in the number. By and large, the 30 required cards is a function of the number of digits combined with our 10-based system.

Try it in a different number system. Lets say we want to be able to represent 0 through 31 (32 unique values) with your same flip ring system, but in binary. You would need 5 positions. Each position would need two cards, a 0 and a 1. There would be 10 total cards and each would get used with equal frequency. Why isn't 32 evenly divisible by 10? Because its not a relevant question.

09-08-2010, 09:32 PM
 Location: Westwood, MA 3,514 posts, read 4,389,843 times Reputation: 4522
Quote:
 Originally Posted by jtur88 But there are only 1,000 permutations, each of which uses specific cards, of which there are 30, and each card is used the same number of times. I should have put this one over in the POS Forum, and let the Libertarians and Muslim haters apply their logic to it.
Yes, but each entry uses 3 cards (digits), so there are 3000 total times that a card will be used.

000 001 010 011
100 101 110 111

There are 8 numbers (2^3) and 6 cards (2 x 3), but since each number has three digits, the total number of spaces filled by cards is 24 (3 x 8), which is the same as the total number of digits above. 24 spaces filled by 6 cards means that each card is used 4 times (24/6), which is half of the time.

In an arbitrary base b with n digits, you get similar relations. b^n numbers filled by b x n different unique cards (assuming each is a digit in a fixed place). The total number of digits used in the numbers is n * b^n, so each digit is used n*b^n/b*n = b^(n-1) times. The fraction of time any one digit is in place is given by the number of uses of a single digit divided by the total number of digits, or b^(n-1)/b^n = 1/b. This is exactly what one would expect for a uniformly distributed variable with b possible values.

---

It's not clear what your comment about Libertarians or Muslim haters has to do with anything--it's seems inflammatory and is out of place in this forum.

09-09-2010, 11:37 AM
 Location: Victoria TX 42,668 posts, read 71,820,212 times Reputation: 35910
Quote:
 Originally Posted by jayrandom It's not clear what your comment about Libertarians or Muslim haters has to do with anything--it's seems inflammatory and is out of place in this forum.
They do, in their forum, what I just did in this one: Keep on insisting that a fallacious conclusion drawn from a deceptive observation proves a point, even when proven wrong.

I did it here as an exercise. But it was interesting to see if anyone would actually come up with a mathematical explanation.
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