Consider this expression:
1/2/3/4/5/6
Using one or more pairs of open and closed parentheses, that is: (), determine the total number of resulting
distinct positive integers.
clearvars,clc
global src stackN stackS ct
ct=0;
src=1:6;
stackN=[];
stackS=string.empty;
addon();
ct
function addon()
global src stackN stackS ct
if length(stackN)>1
ssN=stackN; ssrc=src; ssS=stackS;
stackN(1)=stackN(2)/stackN(1);
stackN(2:end-1)=stackN(3:end);
stackN(end)=[];
if length(char(stackS(1)))>1
if length(char(stackS(2))) > 1
stackS(1)=['(' char(stackS(2)) ')/(' char(stackS(1)) ')'];
else
stackS(1)=[ char(stackS(2)) '/(' char(stackS(1)) ')'];
end
else
if length(char(stackS(2))) > 1
stackS(1)=['(' char(stackS(2)) ')/' char(stackS(1)) ];
else
stackS(1)=[ char(stackS(2)) '/' char(stackS(1)) ];
end
end
stackS(2:end-1)=stackS(3:end);
stackS(end)=[];
if isempty(src) && length(stackS)==1
fprintf('%-30s %17.14f\n',stackS,stackN)
ct=ct+1;
else
addon
end
stackN=ssN; src=ssrc; stackS=ssS;
end
if ~isempty(src)
ssN=stackN; ssrc=src; ssS=stackS;
stackN(2:end+1)=stackN(1:end);
stackS(2:end+1)=stackS(1:end);
stackN(1)=src(1); src(1)=[];
stackS(1)=string(stackN(1));
addon
stackN=ssN; src=ssrc; stackS=ssS;
end
end
((((1/2)/3)/4)/5)/6 0.00138888888889
(((1/2)/3)/4)/(5/6) 0.05000000000000
(((1/2)/3)/(4/5))/6 0.03472222222222
((1/2)/3)/((4/5)/6) 1.25000000000000
((1/2)/3)/(4/(5/6)) 0.03472222222222
(((1/2)/(3/4))/5)/6 0.02222222222222
((1/2)/(3/4))/(5/6) 0.80000000000000
((1/2)/((3/4)/5))/6 0.55555555555556
(1/2)/(((3/4)/5)/6) 20.00000000000000
(1/2)/((3/4)/(5/6)) 0.55555555555556
((1/2)/(3/(4/5)))/6 0.02222222222222
(1/2)/((3/(4/5))/6) 0.80000000000000
(1/2)/(3/((4/5)/6)) 0.02222222222222
(1/2)/(3/(4/(5/6))) 0.80000000000000
(((1/(2/3))/4)/5)/6 0.01250000000000
((1/(2/3))/4)/(5/6) 0.45000000000000
((1/(2/3))/(4/5))/6 0.31250000000000
(1/(2/3))/((4/5)/6) 11.25000000000000
(1/(2/3))/(4/(5/6)) 0.31250000000000
((1/((2/3)/4))/5)/6 0.20000000000000
(1/((2/3)/4))/(5/6) 7.20000000000000
(1/(((2/3)/4)/5))/6 5.00000000000000
1/((((2/3)/4)/5)/6) 180.00000000000000
1/(((2/3)/4)/(5/6)) 5.00000000000000
(1/((2/3)/(4/5)))/6 0.20000000000000
1/(((2/3)/(4/5))/6) 7.20000000000000
1/((2/3)/((4/5)/6)) 0.20000000000000
1/((2/3)/(4/(5/6))) 7.20000000000000
((1/(2/(3/4)))/5)/6 0.01250000000000
(1/(2/(3/4)))/(5/6) 0.45000000000000
(1/((2/(3/4))/5))/6 0.31250000000000
1/(((2/(3/4))/5)/6) 11.25000000000000
1/((2/(3/4))/(5/6)) 0.31250000000000
(1/(2/((3/4)/5)))/6 0.01250000000000
1/((2/((3/4)/5))/6) 0.45000000000000
1/(2/(((3/4)/5)/6)) 0.01250000000000
1/(2/((3/4)/(5/6))) 0.45000000000000
(1/(2/(3/(4/5))))/6 0.31250000000000
1/((2/(3/(4/5)))/6) 11.25000000000000
1/(2/((3/(4/5))/6)) 0.31250000000000
1/(2/(3/((4/5)/6))) 11.25000000000000
1/(2/(3/(4/(5/6)))) 0.31250000000000
ct =
42
Of the 42 possibilities, 4 are integers: 20, 5, 180 and 5 again. Since we want distinct integers, there are 3 of these: 5, 20 and 180.
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Posted by Charlie
on 2023-08-16 14:03:26 |
computer solution
