Determine all
possible integers x that satisfy this equation:
⌊x/2⌋·⌊x/3⌋·⌊x/4⌋·⌊x/5⌋=x3
Reminder: ⌊z⌋ denotes the floor function (the greatest integer less than or equal to z.)
Initially it would seem 5!=120 would satisfy the bill, as indeed it does as the x^4 / x = x^3, as everything is an integer including the fractions, as 5! is divisible by any of them.
Also trivial is x=0.
To guard against somehow, the floors producing some other answer,
clc
for x0=0:30000
for x=[-x0 x0]
lhs=floor(x/2)*floor(x/3)*floor(x/4)*floor(x/5);
rhs=x^3;
if lhs==rhs
disp([x lhs rhs])
end
end
end
lhs
rhs
checks until, as seen by the final lhs and rhs, the fact the 4th degree of the lhs vs the 3rd degree of the rhs overwhelms the flooring, so zero and 120 are the only integral solutons.
The output of the above is
x lhs rhs
0 0 0
0 0 0
120 1728000 1728000
lhs =
6.75e+15
rhs =
27000000000000
Zero appears twice as each number and its negation is tried.
|
|
Posted by Charlie
on 2022-10-24 11:32:53 |
answers
