The first few powers of 5 share a curious property — their digits can be rearranged to express their value:Found in the Futility Closet.
25 = 52
125 = 51 + 2
625 = 56 – 2
3125 = (3 + (1 × 2))5
15625 = 56 × 125
78125 = 57 × 182
It’s conjectured that all powers of 5 have this property. But no one’s proved it yet.
04 December 2010
Mathematical curiosity
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If you're allowed to use *1^something then isn't it always going to be possible, there is always a 5 as the least sig digit.
ReplyDeleteso 5^8 (which is 390625)
the 5 is accounted for.
we can make 8 from the 6+2
then the remaining 390 can be expressed as x 1^390 (which is the same as saying times 1 which is saying nothing.
So, all the conjecture really states is that for all powers of five, the power can be expressed as a calculation of some of the remaining digits of the results.
5^23 = 11920928955078125
= 5^(1+1+9+2+9+1) x 1 ^ 2089550782
or whatever.
^@.\\axxx: You forgot to count the "1" in your expression.
ReplyDeleteSo you can't say that
390 625 = 5^(6+2) x 1^390
fulfills the conjecture.
But your method works whenever there is a "1" in the power of 5 (well, I think it does anyway).
Actually I spoke too soon it doesn't seem to work for 5^11 either..
ReplyDeleteI don't know who conjectured that but I doubt they actually tried it above 5^7 ;-)
5^8 = 390625
ReplyDelete5^(9-((6/3)/2)) + 0
5^11 = 48828125
5^(88/8) * (1^422)
5^23 = 11920928955078125
5^(5+9+9) + (0*218950782211)
5^(5+9+9) * (1^208950782211)
It seems quite trivial if you can simply eliminate extra numbers with tricks like *(1^X) or +(0*X) but i've no idea how you'd go about proving it for all values.
yea, using one to the power of a number is a cop out
ReplyDelete