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Sir Col
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Pandigital Products
« on: Jun 13th, 2003, 11:05am » |
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I'm not sure if this is better suited to 'cs' problems, but I'll post it here anyway... We shall call 30576 a 0 to 9 pandigital product, because it can be factored in such a way that is uses all the digits 0 to 9: 30576 = 8 x 42 x 91. Similarly, 364 is a 1 to 7 pandigital product, as 364 = 1 x 7 x 52. 1. What is the smallest pandigital product? 2. Find the smallest n for which you can form a 1 to n pandigital product. 3. How many 1 to 9 pandigital products can you find? 4. What about 0 to 9 pandigital products?
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wowbagger
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Re: Pandigital Products
« Reply #1 on: Jun 13th, 2003, 11:32am » |
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on Jun 13th, 2003, 11:05am, Sir Col wrote:2. Find the smallest n for which you can form a 1 to n pandigital product. |
| How about n=4?
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Sir Col
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Re: Pandigital Products
« Reply #2 on: Jun 13th, 2003, 11:36am » |
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The product being... ? Okay, it was trivial, so I suppose the obvious question to ask is, what is the next smallest n and what is the product?
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ThudnBlunder
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Re: Pandigital Products
« Reply #3 on: Jun 13th, 2003, 2:17pm » |
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Oh, wait till Leonid sees this one. Nitpick: "...it uses all the digits" once only. 3) : 4396 = 28 x 157 5346 = 18 x 297 5346 = 27 x 198 5796 = 42 x 138 5796 = 12 x 483 5986 = 2 x 41 x 73 7254 = 39 x 186 7632 = 48 x 159 : 4) Are leading zeroes allowed? : 28651 = 7 x 4093 :
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« Last Edit: Jun 13th, 2003, 2:18pm by ThudnBlunder » |
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Sir Col
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Re: Pandigital Products
« Reply #4 on: Jun 13th, 2003, 4:01pm » |
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Good point, T&B: yes, "...uses the digits once and only once." Nice work on 3#, but there are more. In fact there are more 1 to 9 pandigital products that can be written as a product of 3 factors than 2. And, as you would expect, there are lots more for #4. Out of interest, how did you find them? Some combination of computer, trial and/or analysis?
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Chronos
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Re: Pandigital Products
« Reply #5 on: Jun 13th, 2003, 11:55pm » |
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Quote:We shall call 30576 a 0 to 9 pandigital product, because it can be factored in such a way that is uses all the digits 0 to 9: 30576 = 8 x 42 x 91. |
| I think I'm totally missing the point here... 8 x 42 x 91 does not use 0, 3, 5, 6, nor 7. Similar problems occur for the second example, though I can see that neither reuses a digit. Further, it seems like n! is a 1 to n pandigital product, for n < 10 (I'm not sure what it would mean for n to be greater than 10). Could anyone care to enlighten me?
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ThudnBlunder
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Re: Pandigital Products
« Reply #6 on: Jun 14th, 2003, 2:35am » |
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Quote:Out of interest, how did you find them? |
| During my 'research'. Quote:I think I'm totally missing the point here... 8 x 42 x 91 does not use 0, 3, 5, 6, nor 7. |
| OK, his wording was not so accurate, but his meaning is discernible by inspection. He means that the digits of the number - together with the digits of its factors - use all the digits 0-9 once, and once only.
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« Last Edit: Aug 18th, 2003, 10:06pm by ThudnBlunder » |
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Sir Col
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Re: Pandigital Products
« Reply #7 on: Jun 14th, 2003, 4:30am » |
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I thought the meaning was clear, but I'll restate it: We shall call A a 1 to n pandigital product if it can be factored in such a way that the equality A=f1 x f2 x ... x fm contains all the digits 1 to n once and only once. I forgot to answer your other question, T&B: no, you cannot include leading zeroes.
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« Last Edit: Jun 14th, 2003, 4:31am by Sir Col » |
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Leo Broukhis
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Re: Pandigital Products
« Reply #8 on: Jun 14th, 2003, 9:44am » |
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I'm not sure if I have time to use my brute force toward this problem, but I'd like to point out that factoring of a number never includes 1 as one of the factors. If you want to include 1, you need to call them divisors.
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Sir Col
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Re: Pandigital Products
« Reply #9 on: Jun 14th, 2003, 10:09am » |
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Good point, Leonid. I'm not sure if brute force alone will help much in solving this problem. Unless you apply an efficient algorithm your computer could spend a long time churning through possibilities. I know with problems like this that standards/benchmarks are difficult to determine, but to give others something to work around, the algorithm I wrote found all the solutions for 0 to 9 pandigital products in less than 3 minutes. For your reference I used VB6 and ran it on a P4 -1.8GHz processor with 512MB RAM.
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Chronos
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Re: Pandigital Products
« Reply #10 on: Jun 17th, 2003, 4:00pm » |
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Can I use the excuse that I was posting that really late at night, and my brain wasn't working properly? I really should have seen that.
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Lightboxes
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Re: Pandigital Products
« Reply #11 on: Aug 6th, 2003, 10:06pm » |
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The only next smallest n I could get is: 162 = 54 * 3
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« Last Edit: Aug 6th, 2003, 10:07pm by Lightboxes » |
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Sir Col
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Re: Pandigital Products
« Reply #12 on: Aug 7th, 2003, 4:40am » |
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For reference: :: 1 to 4 pandigitals... 3 x 4 = 12 1 solution 1-5 pandigitals... 4 x 13 = 52 1 solution 1-6 pandigitals... 3 x 54 = 162 1 solution 1-7 pandigitals... 0 solutions 1-8 pandigitals... 3 x 582 = 1746 6 x 453 = 2718 24 x 57 = 1368 34 x 52 = 1768 37 x 58 = 2146 58 x 64 = 3712 6 solutions 1-9 pandigitals... 4 x 1738 = 6952 4 x 1963 = 7852 12 x 483 = 5796 18 x 297 = 5346 27 x 198 = 5346 28 x 157 = 4396 39 x 186 = 7254 42 x 138 = 5796 48 x 159 = 7632 9 solutions :: Of course, there is nothing to stop you using more than two factors.
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Icarus
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Re: Pandigital Products
« Reply #13 on: Aug 7th, 2003, 6:02pm » |
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on Jun 13th, 2003, 11:05am, Sir Col wrote:Similarly, 364 is a 1 to 7 pandigital product, as 364 = 1 x 7 x 52. |
| on Aug 7th, 2003, 4:40am, Sir Col wrote:1-7 pandigitals... 0 solutions |
| Have the rules of arithmetic changed in the last 2 months, so that 1x7x52 != 364 anymore?
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Sir Col
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Re: Pandigital Products
« Reply #14 on: Aug 7th, 2003, 6:57pm » |
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Not forgetting, 1 x 6 x 57 = 342. Interestingly, the first 1-n pandigial product made up of 3 factors is for n=7, of which there are two examples. There are no pandigital products made up of three factors for n=8, but for n=9 there are are 35! After that there are only two that are made up of 4 factors, one for n=8 and one for n=9: 1 x 3 x 8 x 24 = 576 1 x 6 x 9 x 47 = 2538
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