Two n-digit integers (leading zeros allowed) are considered equivalent if one is a rearrangement of the other (for example 12033, 20331 and 01332 are considered equivalent five digit integers.)  a) We need to find the number of five digit integers that are not equivalent. b)We need to find the number of non equivalent five digit integers if the digits 1, 3 and 7 can appear at most once. c) When all three digits  (1,3,7) appear once.

y f(x)=x+4 10 Use the figures to calculate the left and right Riemann sums for f on the given interval and the given value of n. f(x)=x+4 on [2,7]; n = 5 The left Riemann sum is The right Riemann sum is 864 C 2 4 6 (Simplify your answer.) (Simplify your answer.) LV f(x)=x+4 N- 2 4 6 Ax

(a) Given an n−bit integer number a = an−1an−2 . . . a0 define a polynomial A(x) satisfying A(2) = a. (b) Given two n−bit integers a and b, give an algorithm to multiply them in O(n log(n)) time. Use the FFT algorithm from class as a black-box (i.e. don’t rewrite the code, just say run FFT on ...). Explain your algorithm in words and analyze its running time. (Think/investigate why this is not enough to claim a fast multiplication algorithm).

Problem 7 [Algorithms integers m and k such that n = m × 24, where m is the smallest integer. For any even integer n, it is always possible to find a pair of 1. Write an algorithm that finds a factorization of any even integer n as stated above. For instance, we have the following two factorizations: 48 = 3 × 24 instead of 48 = 12 × 22 52 = 13 x 22 instead of 52 = 26 × 2. 2. Analyze the time of your algorithm by computing the number of its multiplications. Show your work step by step. Otherwise, your solution is incorrect.

Suppose that you want to calculate 1434661 · 3785648743 − 100020304 · 54300201. You are told thatthe answer is a positive integer less than 90. Compute the answer mod 10 and mod 9, then combineto get the answer. Show all the necessary work to obtain your answers. Note: This technique hasoften been employed to do calculations with large integers because most of the computations use smallintegers, with only the Chinese Remainder Theorem step requiring large integers. For example, if theanswer is known to be less than 1036, the calculations can be done mod p for each prime less than100. Since the product of the primes less than 100 is approximately 2.3 × 1036, the Chinese RemainderTheorem gives the answer exactly

For c), is it possible? Using ISBN: 9780756649791 from a)