Sum the digits of a number?

Hi, I'd like to know how can someone add the digits of a number to the final point in matlab.

For example 525 --> 5 + 2 + 5 = 12 --> 1 + 2 = 3

I'm thinking about dividing by ten and adding the digits after the decimal point while at the same time round the number.

Any ideas on how to add the digits would be really helpfull.

Also how to identify a digit by knowing its position in a number

i.e. the 5th digit of 9483672 is 6. Thanks in advance

[José-Luis]

Is this homework?

[tzenh karetzh]

Not exactly.My homework is to find the prime numbers in a certain space. This is just a question that came to me because we know that numbers like 711 that their digits add up to 3,6 or 9 can be divided by 3.For the prime numbers it's much easier to go for mod(number,3). And maybe it can be useful to someone else for other use

Answer 1

[Les Beckham]

I know it sounds too easy to be true but this manipulation is actually the same as modulo 9. No loops or string conversions needed.

>> mod(525,9)

ans =

     3

>> mod(9483672,9)

ans =

     3

For the second part of your question, try this:

function [out] = extract_digit(num,digit)
    %EXTRACT_DIGIT Return the specified digit from a number
    out = num2str(num);
    out = out(digit);
end

[Stephen23]

Neat idea, needs special-case handling for 9 itself.

See also: https://www.sjsu.edu/faculty/watkins/Digitsum00.htm

[Les Beckham]

Good catch. I forgot about that special case.

[Dimitri]

This won't work...mod(45,9) give zero..but what he wants is 9.

[John D'Errico]

@Dimitri Assuming you want to keep on re-summing the digits until the sum is a single digit number... then special case handling still will work. You just need to think about what happens.

The ONLY case where that digit sum is zero is when the number is itself zero. Therefore, if the number is NOT zero, but the modulus was zero, then the digit sum would have been 9.

As such we can see a simple solution:

N = 12345678;
if N == 0
  digsum = 0;
else
  digsum = mod(N,9);
  if digsum == 0
    digsum = 9;
  end
end
digsum

digsum = 9

Is that algorithm correct for N larger than 0? Clearly it works when N == 9. We can write a really simple code to compute the sum of the digits directly, for just one iteration, and then just iterate until it is done.

dsum = @(n) sum(dec2base(n,10) - '0');
digsum = N;
while digsum > 9
  digsum = dsum(digsum);
end
digsum

digsum = 9

Answer 2

[Azzi Abdelmalek]

a=525;
b=num2str(a);
while numel(b)>1
 a=sum(str2double(regexp(b,'\d','match')));
 b=num2str(a);
end
out=str2num(b)

% -------------------------------

a=9483672;
b=num2str(a);
b(5)

[rohan dhane]

you are fanta.......

[Stephen23]

Simpler without regexp and str2double:

a = 525;
b = num2str(a);
while numel(b)>1
    a = sum(b-'0');
    b = num2str(a);
end
out = a;

Answer 3

[Sean de Wolski]

http://www.mathworks.com/matlabcentral/cody/problems/2040-additive-persistence

Answer 4

[Ans sadiq]

function out=digit_sum(in)

q=in;

a=q/10;

b=floor(a);

c=q-b*10;

w=c;

if q>0

w=w+digit_sum(b);

end

out=w;

end

Answer 5

[Pablo López]

You can try this:

n = 525;
sum(str2num(num2str(sum(str2num(num2str(n)')))'))

[Stephen23]

@Pablo López: Did you try your code with longer numbers, e.g. n=9483672 ?

[Pablo López]

Hi @Stephen Cobeldick!

Yes, I've tested it and it works correctly since 9483672 = 9+4+8+3+6+7+2 = 39 = 3+9 = 12

And if you enter the following code:

n = 9483672;
sum(str2num(num2str(sum(str2num(num2str(n)')))'))

The result is ans = 12

Hope that helps!

[Stephen23]

The question gave this example:

"For example 525 --> 5 + 2 + 5 = 12 --> 1 + 2 = 3"

which indicates that the 12 digits should also be summed to 3. Usually when this task is given, it is required to continue summing until only one digit is reached (and this is what the other answers do). In contrast, your code sums twice only, regardless of how many digits remain. This may or may not be the intended behavior, it depends entirely on the specifications requested by the assignment. I just wanted to note the distinction.

[Pablo López]

Well seen! Thank you for your observation.

Answer 6

[Image Analyst]

Here is how I did it:

fprintf('Beginning to run %s.m ...\n', mfilename);
% Get a random integer.
originalNumber = int64(randi(2^53-1, 1, 1))
% Do the first iteration.
strNumbers = num2str(originalNumber);
intNumbers = strNumbers - '0'
loopCounter = 1;
maxIterations = 100; % The Failsafe (so we never get an infinite loop due to a logic error).  Every while loop should always have a failsafe.
while length(intNumbers) >= 2 && loopCounter < maxIterations
	theSum = sum(intNumbers);
	fprintf('After %d iterations, the number is %s and the sum of its digits is %d\n',...
		loopCounter, strNumbers, theSum);
        % Prepare the next iteration:
	strNumbers = num2str(theSum);
	intNumbers = num2str(theSum) - '0';
	loopCounter = loopCounter + 1;
end
fprintf('Done running %s.m ...\n', mfilename);

I get:

int64

7208285642958972

intNumbers =

7 2 0 8 2 8 5 6 4 2 9 5 8 9 7 2

After 1 iterations, the number is 7208285642958972 and the sum of its digits is 84

After 2 iterations, the number is 84 and the sum of its digits is 12

After 3 iterations, the number is 12 and the sum of its digits is 3

Answer 7

[Javier Echanobe]

n=round(1000*rand(1))
SUM=sum(num2str(n))-48*length(num2str(n)) % 48 is the ASCII code for 0

________________

When I execute this code I obtain:

n =

633

SUM =

12

[Dyuman Joshi]

This method has not been mentioned in this thread anywhere.

Why not use the methods that have been suggested?

[Stephen23]

@Dyuman Joshi: it does something different. The question and the other answers repeatedly sum until there is only one digit remaining. This answer only sums once and... returns something (potentially) with multiple digits.

[Dyuman Joshi]

@Stephen23, I also figured that it does something different.

But from @Javier Echanobe's line "When I execute this code I obtain", I assumed that they were trying to do the same as the original question, but could not achieve it.

Hence my comment.

[Javier Echanobe]

n=round(1000*rand(1))

n = 985

SUM=sum(num2str(n))-48*length(num2str(n)); % 48 is the ASCII code for 0
while SUM>9
    SUM=sum(num2str(SUM))-48*length(num2str(SUM));
end
SUM

SUM = 4

You are right.

This code does make it.