matlab loops break and continue commands

In number theory, a semi-perfect number is a positive integer number that is equal to the sum of its largest three positive divisors, excluding the number itself. For example; 18 is a semiperfect number, the positive divisors of this number are 1,2,3,6,9,18, and the sum of the three largest integer divisors (excluding the number itself) is 3 + 6 + 9 = 18. Write a MATLAB program that checks whether a given number is semi-perfect number.

2 Kommentare

Ameer Hamza
Ameer Hamza am 9 Apr. 2020
This seems like a homework question. What have you already tried?
mehmet kiraccakali
mehmet kiraccakali am 9 Apr. 2020
Bearbeitet: mehmet kiraccakali am 9 Apr. 2020
n=input('Enter a number ');
sum=0;
for i=n-1:-1:1
if mod(n,i)==0
sum=sum+i;
end
end
if sum==n
fprintf('%d is a semi-perfect number \n',n);
else
fprintf('%d is not a semi-perfect number \n',n);
end
I couldn't solve this question either, I tried to make the solution above. but i don't know how to find the three biggest divisors.

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Antworten (1)

Peng Li
Peng Li am 9 Apr. 2020

0 Stimmen

You are almost there. Just need a seperate counter that force the loop to break if it accumulates to 3.
n = input('Enter a number ');
sum = 0;
ind = 0;
for i = n-1:-1:1
if mod(n, i) == 0
sum = sum + i;
ind = ind + 1;
if ind == 3
break;
end
end
end
if sum == n
fprintf('%d is a semi-perfect number \n',n);
else
fprintf('%d is not a semi-perfect number \n',n);
end

7 Kommentare

thank you so much
Ayberk Dönmez
Ayberk Dönmez am 9 Apr. 2020
write a matlab program that calculates the sum of the largest positive divisor (excluding itself) and the smallest positive divisor (excluding 1) of a given number? do you have any idea about this question ?
You can keep doing your way of using a loop; store all divisors found; and do a min/max after that.
A better work around:
n = input('Enter a number ');
k = 2:ceil(n/2); % you don't need to search through bigger numbers, as they won't be a divisor anyway
d = k(rem(n, k) == 0); % you get all divisors
if isempty(d)
disp('no divisors found except the number itself and 1');
else
yourNeedResult = d(1) + d(end);
fprintf("the largest positve divisor (excluding itself) of %d is %d\r", n, d(end));
fprintf("the smallest positive divisor (excluding 1) of %d is %d\r", n, d(1));
fprintf("there sum is %d\r", d(1)+d(end));
end
Enter a number 18
the largest positve divisor (excluding itself) of 18 is 9
the smallest positive divisor (excluding 1) of 18 is 2
there sum is 11
Walter Roberson
Walter Roberson am 9 Apr. 2020
Hint: mod(number, potential_factor)
Gamze Elmastas
Gamze Elmastas am 9 Apr. 2020
What is the mean of if ind==3? I dont know the mean of ind.
Ayberk Dönmez
Ayberk Dönmez am 9 Apr. 2020
thank you so much
Walter Roberson
Walter Roberson am 9 Apr. 2020
sum of its largest three positive divisors
That is why 3 was chosen to compare against.
ind is being used to count how many factors have been found so far.

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