Let
N=36p-5q
a. What values of positive integers p & q trigger the lowest positive N?
b. What if the 2 words"positive " of a. were erased?
As p and q get larger (they get larger together to get N as close as possible to zero) the resulting smallest N gets larger and larger:
clc
digits(2000)
for p=1:200
term1=vpa(36)^p;
q=floor((log(term1)/log(vpa(5))));
disp([p q term1-vpa(5)^q])
end
p q N
[ 1, 2, 11.0]
[ 2, 4, 671.0]
[ 3, 6, 31031.0]
[ 4, 8, 1288991.0]
[ 5, 11, 11638051.0]
[ 6, 13, 956079211.0]
[ 7, 15, 47846585971.0]
[ 8, 17, 2058170454331.0]
[ 9, 20, 6192525027791.0]
[ 10, 22, 1271972649047351.0]
[ 11, 24, 72017059066876511.0]
[ 12, 26, 3248265218936851271.0]
[ 13, 28, 133328825194959067631.0]
[ 14, 31, 1484329341387422919091.0]
[ 15, 33, 104658597893798543446651.0]
[ 16, 35, 5048278064273030523063811.0]
The numbers continue through (p,q) = (200,445) and presumably thereafter.
So the answer to part a. is p=1, q=2, yield the lowest positive N, which is 11.
In part b:
If we don't need N to be positive, we can get N as low as we want in the negative direction just by increasing q indefinitely without increasing p.
N can also be made indefinitely smaller in absolute value by making both p and q get lower and lower indefinitely (more and more negative), always keeping q approximately equal to p*log(36)/log(5).
Edited on February 11, 2021, 8:30 am
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Posted by Charlie
on 2021-02-11 08:29:18 |
solutions
