2020-02-09, 13:20 | #1 |
Feb 2020
11_{10} Posts |
Raising groups of digits to powers, equalling itself
My first attempt at sharing some recreational mathematics seems to have backfired somewhat, being accused of not being "maths". Maybe this will do better, in a recommended subforum?
I couldn't think of a snazzy title, but essentially the concept here is raising groups of digits to powers that result in the answer equalling the whole number. Might be better to give an example: 1 * 7 * 5^{2} = 175. This is an infinite series in which you can add arbitrarily many 6s before the 7 as a group and the result will be true - i.e 1 * 6667 * 5^{2} = 166675. There are a number of similar ideas out there, but this particular thing I haven't found any results for, on OEIS or otherwise. Anyway, all of these I have found with pen and paper, plus the assistance of a calculator. Integer powers only: 1 * 2^{4} * 8 = 128 1 * [6]7 * 5^{2} = 1[6]75, arbitrarily many 6s 4 * 3^{3} * 2^{2} = 432 1 * 02^{2} * 4^{4} = 1024 1 * 59 * 3^{3} = 1593 1 * 53 * 17^{2} = 15317 -of the form 2^{n} * 3: 3 * 8 * 4^{2} = 384 6 * 1 * 4^{2} * 4^{3} = 6144 12 * 2^{4} * 8 * 8 = 12288 something that is just divisible by 2 and 3 is great for this as you have a lot of single digits to use, and a group if you are lucky. I have also done fractional powers, though I think it is a bit less impressive. Fractional powers: 3^{3} * 7 * 8^{1/3} = 378 -of the form 2^{n} * 3: 1 * 9^{1/2} * 2^{6} = 192 9 * ^{1/4} * 8^{3} * 3^{1/2} * 04^{3} = 98304 3 * 9^{-1/2} * 3 * 2 * 16^{4} = 393216 1 * 006^{1/3} * 6^{1/3} * 3 * 2^{24} * 9^{-1/2} * 6^{1/3} = 100663296 3^{-3/2} * 2^{7} * 2^{7} * 1 * 2^{7} * 2^{7} * 54^{1/2} * 72^{1/2} = 3221225472 I haven't done other combinations with 2^{n} * 3^{m}, but I suspect 3221225472 to be the biggest of its type. As for 15317, there may be bigger ones, they are just hard to find (I literally found it playing with powers on a calculator). |
2020-02-09, 15:23 | #2 |
Feb 2017
Nowhere
61×107 Posts |
I'll stick to integer powers.
It occurred to me to look for positive integers which are divisible by the product of their decimal digits (none of which can be 0). There are of course trivial examples like repunits, or repunits plus 1 or 4 (or 2, if the repunit has 1, 4, 7, etc digits). I then looked at the cofactors to see if anything interesting appeared. One amusing case: 17136 = 1*7*1*3*6*136 |
2020-02-09, 18:35 | #3 |
Feb 2020
13_{8} Posts |
After some more calculations, it seems that I overlooked something: negative integer powers can be to to create integers of seemingly arbitrary length. Case in point: 2^{60}*3^{2}
10376293541461622784 = 1 * 03 * 7^{-1} * 6 * 2^{5} * 9 * 3 * 54^{-1} * 1 * 4^{6} * 6^{-1} * 1 * 6 * 2^{6} * 2^{6} * 7 * 8^{6} * 4^{6} The only requirement is that you can cancel out "wrong" digits by having an even number of them. Ignoring negative integers for now, I have a new largest: 2^{29} * 3^{2}: 4831838208 = 4^{2} * 8^{2} * 3 * 1 * 8^{2} * 3 * 8^{2} * 2 * 08^{2}. Looks like with a few more powers of three in the factorisation, should be possible to get a few digits bigger. The largest non trivial grouping seems to be 17 in 15317 so far. Last fiddled with by Boltzmann brain on 2020-02-09 at 18:40 |
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