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Xyzzy 2017-01-02 15:18

January 2017
 
[url]https://www.research.ibm.com/haifa/ponderthis/challenges/January2017.html[/url]

Xyzzy 2017-02-15 16:00

[url]https://www.research.ibm.com/haifa/ponderthis/solutions/January2017.html[/url]

R. Gerbicz 2017-02-15 16:32

My sent solution with code(!):

x=(1+1/80)^80
y=(1+1/80)^81
the first and last ten digits of x: 2701484940, 6337890625
the first and last ten digits of y: 2735253502, 1142578125

For positive integer n value it is known that x=(1+1/n)^n and y=(1+1/n)^(n+1)
gives a rational solution of the equation, because:
x^y=((1+1/n)^n)^((1+1/n)^(n+1))=(1+1/n)^((n+1)^(n+1)/n^n)
y^x=((1+1/n)^(n+1))^((1+1/n)^n)=(1+1/n)^((n+1)^(n+1)/n^n)
hence x^y=y^x. If we want x, where we can see the last eight (or some) digits
then n=2^a*5^b for some non-negative a,b integers.
Restricting the search for a,b<10 we have found 3 solutions,
the simple PARI-GP solution, using only integer arithmetic:

for(a=0,9,for(b=0,9,n=2^a*5^b;e=max(a,b);M=10^8;\
r=Mod(n+1,M)^n*Mod(2,M)^(n*(e-a))*Mod(5,M)^(n*(e-b));\
r=lift(r);test=1;v=vector(10,i,0);for(i=1,8,\
res=1+(r%10);r\=10;v[res]+=1;if(v[res]>1,test=0));\
if(test,X=[a,b,n];print([a,b,n]))))

what has found n=80;3125;1953125 as a solution,
this gives for example that x=(1+1/80)^80 and y=(1+1/80)^81 a solution.

A "cut" function used to get the first and last
L digits of a positive rational number:

cut(r,L)={tmp=r;d=10^L;while(type(tmp)!="t_INT",tmp*=10);\
t=tmp;while(tmp>=d,tmp\=10);return([tmp,t%d])}

In the second email:

Just for small correction obvioulsy
x starts as 2.701484940
and y with 2.735253502
if you needed also the first (ten) digits with a decimal point.

ps. use cut() only if r>0 has a finite representation in base 10.

uau 2017-02-17 01:20

It's pretty easy to directly show starting from the question rationals of that form are answers and are the only rational answer:

x^y = y^x
Let 'a' be such that x^a = y, and replace y by that in above
x ^ x ^ a = x ^ (a*x)
Take x-based logarithm
x ^ a = a*x
Since x^a = y, this means a = y/x, thus rational too
Let 'b' be such that a = x^b
x ^ x ^ b = x^b * x
Take x-based logarithm
x^b = b + 1
x^b equals a, thus b = a-1 and rational
Let x = p/q, b = n/m where p,q,n,m integers
(p/q)^(n/m) = (n+m)/m
If the value on the left is rational (as per above it equals a and must be rational for a valid solution), it must be n:th power of one. Thus the value on the right must be too. So (n+m) and m are n:th powers. But it's not possible for two integers n apart (m and n+m) to both be n:th powers unless n=1. Thus:
(p/q)^(1/m) = (1+m)/m
p/q = ((1+m)/m)^m
And this gives a solution for any integer m.

R. Gerbicz 2017-02-22 21:34

A longer article on this topic: [url]http://www.komal.hu/cikkek/loczy/powers/commpower.e.shtml[/url] in English, but the original was in Hungarian from 2000 in K├Âmal ([url]http://db.komal.hu/KomalHU/showpdf.phtml?tabla=Cikk&id=200047[/url]).


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