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单词 AFewExamplesOfSolutionsToZnamsProblem
释义

a few examples of solutions to Znám’s problem


For k=5 in Znám’s problem, two sets of integers can be constructed. To check each set involves calculating k different deleted productsPlanetmathPlanetmath.

To check that 2, 3, 7, 47, 395 is a solution, we verify that

3747395+1=389866, and 2|389866

2747395+1=259911, and 3|259911

2347395+1=111391, and 7|111391

237395+1=16591, and 47|16591

23747+1=1975, and 395|1975.

To check that 2, 3, 11, 23, 31 is a solution, we verify that

3112331+1=23530, and 2|23530

2112331+1=15687, and 3|15687

232331+1=4279, and 11|4279

231131+1=2047, and 23|2047

231123+1=1519, and 31|1519.

If we chose to ignore the requirement of proper divisors, and allow all divisorsMathworldPlanetmathPlanetmath, then for k=4 there would be the solution 2, 3, 7, 43. The verification would show that

3743+1=904, and 2|904

2743+1=603, and 3|603

2343+1=259, and 7|259

237+1=43, and obviously 43|43 because 43=43.

The vast majority of the sets in the known solutions include 2. A set is known for k=13 with its smallest two elements being 3 and 4. It’s not known if there are any sets consisting entirely of odd numbersMathworldPlanetmathPlanetmath. But since 2 is a prime numberMathworldPlanetmath, sets consisting entirely of prime numbers are possible, such as the second one given above, and for k=7 there is 2, 3, 11, 17, 101, 149, 3109.

For a completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath listing of solutions up to k=8, see http://www.geocities.com/primefan/ZnamProbSols.htmlPrimeFan’s listing

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