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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!
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
If the coefficient of $x^{50}$ in the expansion of $(1+x)^{1000}+2x(1+x . . . Problem : If the coefficient of x50 x 50 in the expansion of (1 + x)1000 + 2x(1 + x)999 + 3x2(1 + x)998 + ⋯ + 1001x1000 (1 + x) 1000 + 2 x (1 + x) 999 + 3 x 2 (1 + x) 998 + + 1001 x 1000 is λ λ then the value of 1952!50! 1001! λ 1952! 50! 1001! λ Please guide how to find the value of λ λ in this will be of great help as I am not getting any clue how to proceed further in this problem
Find $A^{1000}$ by using Cayley-Hamilton Theorem Find A1000 A 1000 by using the Cayley-Hamilton theorem I find the characteristic polynomial by P(A) = −A3 + 2A2 = 0 P (A) = A 3 + 2 A 2 = 0 (by Cayley-Hamilton) but I don't see how to find A1000 A 1000 by this characteristic polynomial