Sum the digits of a number?
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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
2 Kommentare
Antworten (7)
Les Beckham
am 2 Mai 2020
Bearbeitet: Les Beckham
am 2 Mai 2020
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
4 Kommentare
John D'Errico
am 6 Apr. 2021
@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
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
Azzi Abdelmalek
am 17 Jan. 2014
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)
Ans sadiq
am 19 Aug. 2021
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
0 Kommentare
Pablo López
am 5 Feb. 2019
You can try this:
n = 525;
sum(str2num(num2str(sum(str2num(num2str(n)')))'))
4 Kommentare
Stephen23
am 6 Feb. 2019
Bearbeitet: Stephen23
am 6 Feb. 2019
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.
Image Analyst
am 7 Apr. 2021
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
0 Kommentare
Javier Echanobe
am 15 Sep. 2023
Bearbeitet: Javier Echanobe
am 15 Sep. 2023
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
4 Kommentare
Dyuman Joshi
am 15 Sep. 2023
@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
am 18 Sep. 2023
Verschoben: Dyuman Joshi
am 18 Sep. 2023
n=round(1000*rand(1))
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
You are right.
This code does make it.
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