University:
California State University, Northridge
Course:
MATH 102 | Pre-Calculus I
Academic year:

2024
Views:

66

Pages:

3
Author:

Joyce Guerra

Exercise 1: Prime Factorization a) 700: 2^2 * 5^2 * 7 b) 1560: 2^3 * 3 * 5 * 13 c) 3465: 3^2 * 5 * 7 * 11 d) 70840: 2^3 * 5 * 7 * 11 * 23 Exercise 2: Prime, Composite, or Neither a) 12: Composite b) 123: Neither c) 1234: Composite d) 12345: Neither e) 154: Composite f) 102302320: Composite g) 1: Neither h) 97: Prime Exercise 3: Prime Factorization a) 63: 3^2 * 7 b) 768: 2^8 * 3 c) 324: 2^2 * 3^4 d) 361: 19^2 e) 196: 2^2 * 7^2 f) 1024: 2^10 g) 480: 2^5 * 3 * 5 h) 10,000: 2^4 * 5^4 Alternative expression for 10,000: 10^4 = (2 * 5)^4 = 2^4 * 5^4 Exercise 5 (Continued): Part a: 10!: 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 Prime factorization of 10!: 2^8 * 3^4 * 5^2 * 7 Part b: Is 10! divisible by: 10? Yes, since 10 = 2 * 5 30? Yes, since 30 = 2 * 3 * 5 120? Yes, since 120 = 2^3 * 3 * 5 1000? No, because the power of 5 in 1000 (5^3) is larger than the power of 5 in 10! (5^2) Part c: Is 30! divisible by 2,400,000? Yes Explanation: 30! can be written as 24 * 2^5 * 5^5 * (remaining factors) 24 * 2^5 * 5^5 is equal to 2,400,000 Exercise 5 (Continued): Part d: Find the largest N such that 18! is divisible by 12^N. 18! = 1 * 2 * 3 * ... * 10 * 11 * 12 * 13 * ... * 18 From part a: 10! = 2^8 * 3^4 * 5^2 * 7 18! = (2^8 * 3^4 * 5^2 * 7) * 11 * (2 * 3) * (2 * 3) * (2 * 7) * (3 * 5) * (2^4) * 17 * (2 * 3^2) Simplified: 18! = 2^16 * 3^8 * 5^3 * 7^2 * 11 * 13 * 17 12^N = (2^2 * 3)^N = 2^(2N) * 3^N Based on the powers of 2 and 3 in 18!, the largest N for which 18! is divisible by 12^N is N = 8. Explanation: 12^8 = (2^2 * 3)^8 = 2^16 * 3^8, which is a factor of 18! 12^9 = (2^2 * 3)^9 = 2^18 * 3^9, which is not a factor of 18! because the powers of 2 and 3 in 12^9 are too large compared to those in 18!.