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How much zeros has the number $1000!$ at the end? 1 If a number ends with n n zeros than it is divisible by 10n 10 n, that is 2n5n 2 n 5 n A factorial clearly has more 2 2 s than 5 5 s in its factorization so you only need to count how many 5 5 s are there in the factorization of 1000! 1000!
algebra precalculus - Multiple-choice: sum of primes below $1000 . . . For example, the sum of all numbers less than 1000 1000 is about 500, 000 500, 000 So, 168 1000 × 500, 000 168 1000 × 500, 000 or 84, 000 84, 000 should be in the right ballpark 76127 76127 is the right answer, by this reasoning
Find the order of $\\sigma^{1000}$ where $\\sigma$ is the permutation . . . The order of a cycle is its length, and the order of a product of disjoint cycles is the l c m of their orders, i e the l c m of their lengths Hence σ σ has order 12 12, so that σ1000 =σ1000 mod 12 =σ4 σ 1000 = σ 1000 mod 12 = σ 4 So σ1000 σ 1000 has order 12 4 = 3 12 4 = 3
Repeating something with (1 n)th chance of success n times For an independent event (the results of past trials do not influence the probability of future ones), it's not guaranteed to happen exactly once per 1000 trials, but as the number of trials t grows large, the number of times an event with probability 1 n 1 n occurs will approach t n t n