Define the flipflop function, applied to a positive integer, as
the result of having the 10^2i and 10^(2i+1) digits switch places.
Moreover, if the integer has an odd number of digits, append
a leading zero to the left side of the number so that
it can flipflop with the first nonzero digit.
For example, flipflop(9876) = 8967 and flipflop(1234567) is 10325476.
warm-up:
What is the smallest positive integer such that flipflop(m) = m*4?
octuple the number:
What is the smallest positive integer such that flipflop(n) = n*8?
In the case of multiplication by 8, I have found only one solution: are there any others?
The program does some base-100 arithmetic so that a conversion can be made one base-100 digit at a time via the flipper array.
clearvars
global flipper
for i=0:99
u=mod(i,10);
t=floor(i/10);
i2=10*u+t;
flipper(i+1)=char(i2);
flipper(i2+1)=char(i);
end
n=char(0);
while length(n)<4
psn=length(n);
n(psn)=n(psn)+1;
while n(psn)>99
n(psn)=char(0);
psn=psn-1;
if psn==0
n=[char(0) n];
psn=1;
end
n(psn)=n(psn)+1;
end
if isequal(flipflop(n),quadruple(n))
for i=1:length(n)
fprintf('%02d',double(n(i)));
end
fprintf('\n')
end
end
function f=flipflop(n)
global flipper
f=n;
for i=1:length(n)
f(i)=flipper(n(i)+1);
end
end
function q=quadruple(n)
q=n;
psn=length(n);
pr=q(psn)*4;
quot=floor(pr/100);
rem=mod(pr,100);
q(psn)=rem;
while quot>0 || psn>1
psn=psn-1;
if psn==0
q=[char(0) q];
psn=1;
end
pr=q(psn)*4+quot;
quot=floor(pr/100);
rem=mod(pr,100);
q(psn)=rem;
end
end
function o=octuple(n)
o=n;
psn=length(n);
pr=o(psn)*8;
quot=floor(pr/100);
rem=mod(pr,100);
o(psn)=rem;
while quot>0 || psn>1
psn=psn-1;
if psn==0
o=[char(0) o];
psn=1;
end
pr=o(psn)*8+quot;
quot=floor(pr/100);
rem=mod(pr,100);
o(psn)=rem;
end
end
is in the warmup mode and finds
1782
178200
179982
as each of these, when divided into its flipflop, gives a quotient of 4:
7128/1782 = 4
712800/178200 = 4
719928/179982 = 4
The octuple version required changing the length limit so that length of 4 base-100 digits would actually be processed, and of course the comparison had to be changed to call the octuple function instead of the quadruple function. With the changes, it found
02519748
20157984 / 02519748 = 8
It was run for a few minutes and when it was manually stopped, n was still only 4 base-100 digits, but the leftmost digit had reached 40 -- quite bit higher than the 02 in the solution.
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Posted by Charlie
on 2022-03-30 11:45:13 |
computer solution
