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How much zeros has the number $1000!$ at the end? 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!
Solution Verification: How many positive integers less than $1000$ have . . . A positive integer less than 1000 1000 has a unique representation as a 3 3 -digit number padded with leading zeros, if needed To avoid a digit of 9 9, you have 9 9 choices for each of the 3 3 digits, but you don't want all zeros, so the excluded set has count 93 − 1 = 728 9 3 1 = 728 Hence the count you want is 999 − 728 = 271 999 728 = 271
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
For each integer - Mathematics Stack Exchange For each integer 2 ≤ a ≤ 10 2 ≤ a ≤ 10, find the last four digits of a1000 a 1000 [[Hint: We need to calculate a1000 a 1000 mod 10000 10000 Use Euler’s theorem and Chinese remainder theorem For example, 10000 =24 ⋅54 10000 = 2 4 ⋅ 5 4; 21000 ≡ 0 2 1000 ≡ 0 mod 24 2 4, and 2500 ≡ 1 2 500 ≡ 1 mod 54 5 4 ]]