convert inf value to integer value
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Walter Roberson on 4 Apr 2014
b = 11239388;
e = 375;
m = 751;
feval(symengine, 'powermod', b, e, m)
See also John's contribution as mentioned by Nitin, and see http://www.mathworks.com/matlabcentral/fileexchange/38516-powermod
Roger Stafford on 5 Apr 2014
Edited: Roger Stafford on 5 Apr 2014
John's function 'powermod' is a very efficient method of solving your problem, but it is possible to solve it in such a way that the solution is eminently clear to you without taking 'powermod' on trust.
To begin with, if you do p = mod(11239388,751), you will know that
11239388 = p + some integral multiple of 751
where p is a certain integer less than 751. Therefore
11239388^375 = (p + that integral multiple of 751)^375
= p^375 + some other larger multiple of 751
by the binomial theorem. Hence the problem is reduced to finding just the quantity mod(p^375,751), which is a lot easier.
You can solve that by starting with 1 and repeatedly multiplying it by p, then taking its mod(~,751), in a for-loop 375 times. Each time the 'mod' operation will keep the number within the necessary upper limit of accuracy for 'double' values before proceeding with the next multiplication. At the end you will have your desired answer.
(Actually in this particular problem you can no doubt arrive at an answer faster than than you could with John's 'powermod' if you only do this last operation 15 instead of 375 times and take a careful look at the answer, remembering that 375 is a multiple of 15.)