Problem 5
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
Solve:
-
clear
clc
for i = 100:20:1000000000
% if rem(i,11) == 0 && rem(i,12) == 0 && rem(i,13) == 0 && rem(i,14) == 0 && rem(i,15) == 0 &&rem(i,16) == 0 && rem(i,17) == 0 && rem(i,18) == 0 && rem(i,19) == 0 && rem(i,20) == 0
break
end
end
fprintf('The smallest evenly divisible number is %.0d\n',i) - clear
clc
for i = 100:20:1000000000
for j = 20:-1:1
if mod(i,j) ~= 0
break
end
end
if j == 1
answer = i;
break
end
end
fprintf('The smallest evenly divisible number is %.0d\n',answer)
本文通过两种不同的编程方法解决了寻找能被1到20所有整数整除的最小正整数的问题,该数也是这些数的最小公倍数。
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