How do you remove multiples of certain numbers from a row vector?

I am trying to create a for loop that finds all values in the range [1,100] that are also multiple of 3 or 5, but not 3 AND 5, and save these in a row vector labeled M1. It was suggested that I use mod()/rem() functions. I am struggling on how to remove the numbers that are multiples of 3 and 5.

M1=[];
for x = 1:100
    if mod(x,3)==0||mod(x,5)==0
      M1=[M1,x];
    end 
    if mod(x,3)==0 & mod(x,5)==0
      M1(x)=[]
    end
end

M1

Answer 1

[Max Alger-Meyer]

The easiest way to do this is with is just adding additional conditions to your first if statement, and deleting the second if statement alltogether. We do this by placing the "or" conditions in parantheses, and having a third condition that checks that the sum of the mod(x,3) and mod(x,5) isn't zero. If mod(x,3) and mod(x,5) sum to zero, then we know that the index must be wholly divisible by both.

M1=[];
for x = 1:100
    if (mod(x,3) == 0||mod(x,5) == 0) && sum(mod(x,3) + mod(x,5)) ~= 0
      M1=[M1,x];
    end 
end
M1

M1 = 1×41

3 5 6 9 10 12 18 20 21 24 25 27 33 35 36 39 40 42 48 50 51 54 55 57 63 65 66 69 70 72

[Garrett Viton]

How would one go about doing this if it was "3 and 5" instead of "3 or 5"?

[Max Alger-Meyer]

So if I understand you correctly, you want to remove only numbers that are divisible by both 3 and 5? If that's the case, you could just use the second condition in the if statement:

M1=[];
for x = 1:100
    if sum(mod(x,3) + mod(x,5)) ~= 0
      M1=[M1,x];
    end 
end
M1

M1 = 1×94

1 2 3 4 5 6 7 8 9 10 11 12 13 14 16 17 18 19 20 21 22 23 24 25 26 27 28 29 31 32

Alternatively, we can simplify this a little further by thinking about what is really happening. This didn't cross my mind the first time, but if you looks at the outputs from my original answer, you can see that the only numbers that are divisible by both 3 and 5 are just going to be multiple of 15, so we can use mod(x,15) instead of checking the sum of the mods of 3 and 5. That leaves us with:

M1=[];
for x = 1:100
    if mod(x,15) ~= 0
      M1=[M1,x];
    end 
end
M1

M1 = 1×94

1 2 3 4 5 6 7 8 9 10 11 12 13 14 16 17 18 19 20 21 22 23 24 25 26 27 28 29 31 32

Answer 2

[Jan]

You do not need a loop.

n = 100;
x = false(1, n);
x(3:3:n)   = true;
x(5:5:n)   = true;
x(15:15:n) = false;
Result     = find(x)

Result = 1×41

3 5 6 9 10 12 18 20 21 24 25 27 33 35 36 39 40 42 48 50 51 54 55 57 63 65 66 69 70 72

% Or:
x      = 1:100;
Result = x((mod(x, 3) == 0 | mod(x, 5) == 0) & mod(x, 15) ~= 0)

Result = 1×41

3 5 6 9 10 12 18 20 21 24 25 27 33 35 36 39 40 42 48 50 51 54 55 57 63 65 66 69 70 72

% Shorter:
Result = x(~(mod(x, 3) & mod(x, 5)) & mod(x, 15))

Result = 1×41

3 5 6 9 10 12 18 20 21 24 25 27 33 35 36 39 40 42 48 50 51 54 55 57 63 65 66 69 70 72

[Stephen23]

+1 very neat