Consider a base n rational number α.β and it's reverse form β.α such that their product is an integer.
For example, in base ten: (3.5)*(5.3)= 18.55, which is NOT an integer.
Consider all positive integer bases n ≤ 36 and determine all valid triplets (α, β, n) of positive integers for which the product of (α.β)base n and (β.α)base n is an integer.
clearvars, clc
syms alpha beta base p
prev=0; digs='0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ';
for base=1:36
for alpha=sym(0):sym(base)-1
for beta=alpha:sym(base)-1
if beta==0
continue
end
p=(alpha+beta/base)*(beta+alpha/base);
ix=strfind(string(p),"/");
if isempty(ix)
if base>prev
disp(' ')
disp(base)
prev=base;
end
first=[digs(alpha+1) '.' digs(beta+1)];
second=[digs(beta+1) '.' digs(alpha+1)];
disp([first ' * ' second ' = ' char(string(p))])
end
end
end
end
shows that bases 4, 8, 9, 10, 12, 14, 16, 18, 20, 24, 25, 26, 27, 28, 30, 32, 33, and 36 have choices of alpha and beta that satisfy the criterion:
The product in each case is shown in decimal. Each set is preceded by a line showing the base.
4
0.2 * 2.0 = 1
8
0.4 * 4.0 = 2
9
0.3 * 3.0 = 1
0.6 * 6.0 = 4
10
2.5 * 5.2 = 13
12
0.6 * 6.0 = 3
14
7.A * A.7 = 81
16
0.4 * 4.0 = 1
0.8 * 8.0 = 4
0.C * C.0 = 9
18
0.6 * 6.0 = 2
0.C * C.0 = 8
20
0.A * A.0 = 5
24
0.C * C.0 = 6
C.G * G.C = 209
25
0.5 * 5.0 = 1
0.A * A.0 = 4
0.F * F.0 = 9
0.K * K.0 = 16
26
6.D * D.6 = 86
27
0.9 * 9.0 = 3
0.I * I.0 = 12
28
0.E * E.0 = 7
7.C * C.7 = 91
E.O * O.E = 364
30
A.F * F.A = 161
F.M * M.F = 354
32
0.8 * 8.0 = 2
0.G * G.0 = 8
0.O * O.0 = 18
33
3.M * M.3 = 81
36
0.6 * 6.0 = 1
0.C * C.0 = 4
0.I * I.0 = 9
0.O * O.0 = 16
0.U * U.0 = 25
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Posted by Charlie
on 2022-03-18 10:56:12 |
computer solutions
