Let m and n be one digit integers (0 to 9) with m>n
1a. Evaluate the mean value of (m!-n!)mod(10)
1b. Evaluate the mean value of (m!-n!)mod(3)
Let m and n be two digit integers (10 to 99) with m>n
2a. Evaluate the mean value of (m!-n!)mod(10)
2b. Evaluate the mean value of (m!-n!)mod(3)
3. Find recursive formulas for n digit integers.
n
m 0 1 2 3 4 5 6 7 8
1 0 mod 10
2 1 1
3 5 5 4 mean = 4.71111111111111
4 3 3 2 8 ( 212/45 )
5 9 9 8 4 6
6 9 9 8 4 6 0
7 9 9 8 4 6 0 0
8 9 9 8 4 6 0 0 0
8 9 9 8 4 6 0 0 0 0
1 0 mod 3
2 1 1
3 2 2 1
4 2 2 1 0 mean = 0.822222222222222
5 2 2 1 0 0 ( 37/45 )
6 2 2 1 0 0 0
7 2 2 1 0 0 0 0
8 2 2 1 0 0 0 0 0
9 2 2 1 0 0 0 0 0 0
2-digit m and n are all zero
Mod 10 becomes zero when n! starts to include 5.
Mod 3 becomes zero when n! starts to include 3.
clearvars,clc
for k=[10 3];
for m=1:9
for n=0:m-1
f=F(m,n,k);
fprintf('%3d',f)
end
fprintf('\n')
end
fprintf('\n')
end
for k=[10 3];
for m=sym(11):30
for n=sym(10):m-1
f=F(m,n,k);
fprintf('%3d',f)
end
fprintf('\n')
end
fprintf('\n')
end
function v=F(m,n,k)
v=mod(factorial(m)-factorial(n),k);
end
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Posted by Charlie
on 2024-03-16 13:20:43 |
thoughts and calculations
