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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!
probability of an event occuring with numerous attempts Often in calculating probabilities, it is sometimes easier to calculate the probability of the 'opposite', the technical term being the complement Because if something happens with probability p p, then it does not happen with probability 1 − p 1 p, e g if something happens with probability 0 40 0 40 (40% 40 %) then it does not happen with probability 1 − 0 40 = 0 60 1 0 40 = 0 60 (60%
Find the number of times - Mathematics Stack Exchange Question: Find the number of times 5 5 will be written while listing integers from 1 1 to 1000 1000 Now, it can be solved in this fashion The numbers will be of the form: 5xy, x5y, xy5 5 x y, x 5 y, x y 5 where x, y x, y denote the two other digits such that 0 ≤ x, y ≤ 9 0 ≤ x, y ≤ 9 So, x, y x, y can take 10 10 choice each
combinatorics - How many numbers between $1000$ and $100000$ that do . . . Think of all the numbers between 1000 and 100,000 as five digit numbers (i e 1000 is actually 01000) How many numbers five digit numbers are there, well the short answer is 100,000 and we can arrive at this by saying that we have 10 choices for the first digit, and 10 for the second, etc to find that we have 105 = 100, 000 10 5 = 100, 000 numbers
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
How many numbers between 1 and 1000 are divisible by 2, 3, 5 or 7? The of 210 210 - the number of values between 1 1 and 210 210 that are relatively prime to 210 210 - is (2 − 1)(3 − 1)(5 − 1)(7 − 1) = 48 (2 1) (3 1) (5 1) (7 1) = 48 Using this, we can say that there are 48 ⋅ 5 = 240 48 5 = 240 numbers not divisible by these four numbers up to 1050 1050 Some of these of course are out of range of the original question; we'll have to figure out