MATLAB: Creating a formula for a question

codedaily contestinteger and positivenon-zerosymbolicvariable

Hello Dear Matlab users

I need some HELP.

I saw 'a "one-a-day" journal contest' question about so called 'perfect numbers'. (I don't know exact translation of the term). The question was as following:

The Number 18432 is a number that has a feature of non-zero digit in itself,non-repeating digit, also has the ability to split into their numbers sum and products. Also it's more one digit, positive number (it has a lot). Other examples are 12,432,3168,13248 and they go until the number reaches total 21 piece. Well the question asks:

what is the smallest one, greater than 18432??

I thought that I could write a m.file to answer this question.

I've tried to initiate the m.file as following

if true
A=sym('A%d%d',[1,5],'integer&&positive');
sum(A)
assumptions(A);
end

I know it needs a lot of work, but a general idea might spark some thoughts to reconfigure the necessary function.

Thank you already for your consideration.

Best Answer

So, you want to find a (nonrepeating) sequence of decimal digits, a vector D taken from the set [1:9], that have the two properties that

1. The product of the digits divides the number when viewed as an integer, thus converted to base 10.

2. The sum of the digits also divides the number when viewed as an integer.

N = (18434:99999)';D = dec2base(N,10) - '0';
ind = find((mod(N,sum(D,2)) == 0) & (mod(N,prod(D,2)) == 0) & all(D > 0,2) & all(diff(D,[],2) ~= 0,2));
N(ind)
ans =
   21216       
   21312       
   21672       
   24192       
   24912       
   26136       
   26712       
   27216       
   31212       
   32616       
   32832       
   34272       
   35175       
   41232       
   41616       
   42192       
   42624       
   42912       
   43632       
   51975       
   61824       
   71316       
   81216       
   83232

Seems a bit trivial. Unless by non-repeating digit, you mean that no single digit can appear more than once in the entire number. In that case...

ind = find((mod(N,sum(D,2)) == 0) & (mod(N,prod(D,2)) == 0) & all(D > 0,2) & all(diff(D,[],2) ~= 0,2) & all(diff(sort(D,2),[],2) > 0,2))
N(ind)
ans =
   61824

Still trivial.

Related Solutions

MATLAB: How to display the sum of the digits of a number entered by a user

n=length(num2str(x));%number of digits
s=sum(num2str(x)-'0');%sum of digits

MATLAB: Facing a problem in calculating the sum of digits of 100!.

It looks like this question has gotten enough interest that it deserves a complete answer on how to do it using MATLAB. And since you have done something reasonable...

You simply cannot compute factorial of a number as a double, if n is greater than 18.

factorial(1:20) < flintmax
ans =
  1×20 logical array
   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   0   0

Actually, it looks like you might get to 20, if you use uint64. But uint64 overflows beyond that point. So using a double will just fail for the sum of digits of factorial(100).

factorial(100)
ans =
  9.3326e+157

That is WAY past flintmax. Personally, I'd do it elegantly as:

sum(char(factorial(sym(100))) - '0')
ans =
   648

However nothing stops you from doing the multiplication the hard way, using carries.

factdigs = zeros(1,200);
factdigs(end) = 1;
N = 100;
for n = 2:N
    % multiply. note that you multiply ALL digits of
    % the number by n.
    factdigs = factdigs * n;
    % carry op
    carry = 1;
    % a while loop, since there may be multiple carries.
    % for example, a multiple carry would occur when you
    % multiply the number 6 times 100.
    while any(carry)
        % strip off the units digit
        R = rem(factdigs,10);
        % this divide will always be safe, resulting
        % in an exact integer.
        carry = (factdigs - R)/10;
        % The carry is just an add, with a shift.
        % if I knew the highest order digit will
        % always be zero, I could have used circshift.
        factdigs = R + [carry(2:end),0];
    end
end
sum(factdigs)
ans =
   648

This will work as long as I make the vector of digits long enough that factorial(100) does not overflow a 200 digit number. And since we know that factorial(100) is on the order of 9e157, 200 digits is fine.

reshape(factdigs,10,20)'
ans =
     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0
     0     0     9     3     3     2     6     2     1     5
     4     4     3     9     4     4     1     5     2     6
     8     1     6     9     9     2     3     8     8     5
     6     2     6     6     7     0     0     4     9     0
     7     1     5     9     6     8     2     6     4     3
     8     1     6     2     1     4     6     8     5     9
     2     9     6     3     8     9     5     2     1     7
     5     9     9     9     9     3     2     2     9     9
     1     5     6     0     8     9     4     1     4     6
     3     9     7     6     1     5     6     5     1     8
     2     8     6     2     5     3     6     9     7     9
     2     0     8     2     7     2     2     3     7     5
     8     2     5     1     1     8     5     2     1     0
     9     1     6     8     6     4     0     0     0     0
     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0

Of course, the code as I wote it would fail around n = 122.

factorial(122)
ans =
   9.875e+202

That would be fixable merely by preallocating factdigs to be a longer vector. Or, with a little more care in the code, I could have written it to dynamically expand the vector of digits, so factorials of any size would then theoretically work.

Best Answer

Related Solutions

MATLAB: How to display the sum of the digits of a number entered by a user

MATLAB: Facing a problem in calculating the sum of digits of 100!.

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