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So, to calculate,
(499! * 500!)(P - 2) % P,
if I store the value of, ((499! % P) * (500! % P)) in a variable 'temp', then my expression reduces to,
(temp)(P - 2) % P,
Am I right, Sir...??
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Sherlock and Permutations
You are viewing a single comment's thread. Return to all comments →
So, to calculate,
(499! * 500!)(P - 2) % P,
if I store the value of, ((499! % P) * (500! % P)) in a variable 'temp', then my expression reduces to,
(temp)(P - 2) % P,
Am I right, Sir...??