Mathematical

1. Mathematical

2. Approach • Many of these problems read as brain teasers at first, but can be worked through in a logical way. • Just remember to rely on the rules of mathematics to develop an approach, and then to carefully translate that idea into code.

3. Example • Given two numbers m and n, write a method to return the first number r that is divisible by both (e.g., the least common multiple).

4. hints • What does it mean for r to be divisible by m and n? • It means that all the primes in m must go into r, and all primes in n must be in r. • What if m and n have primes in common? • For example, if m is divisible by 3^5 and n is divisible by 3^7, what does this mean about r? • It means r must be divisible by 3^7. • The Rule • For each prime p such that p^a \ m (e.g., m is divisible by p^a) and p^b \ n, r must be divisible by p^max(a, b).

6. Find the LCM of these sets of numbers. • 3, 9, 21Solution: List the prime factors of each. 3: 3 9: 3 × 3 21: 3 × 763 can be divided evenly by 3, 9, and 21. • 12, 80 Solution: List the prime factors of each.12: 2 × 2 × 3 80: 2 × 2 × 2 × 2 × 5 = 80 240 can be divided by both 12 and 80.

8. Algorithm

9. Prime • A number is prime if it is only divisible by 1 and itself. So for example 2, 3, 5, 79, 311 and 1931 are all prime, while 21 is not prime because it is divisible by 3 and 7. • To find if a number n is prime we could simply check if it divides any numbers below it. • We can use the modulus (%) operator to check for divisibility:

10. Solution • for (inti=2; i<n; i++) • if (n%i==0) • return false; • return true; • We can make this code run faster by noticing that we only need to check divisibility for values of i that are less or equal to the square root of n 

11. Implementation • public booleanisPrime (int n) { • if (n<=1) • return false; • if (n==2) • return true; • if (n%2==0) • return false; • int m=Math.sqrt(n); • for (inti=3; i<=m; i+=2) • if (n%i==0) • return false; • return true; • }

12. Problem • Design an algorithm to find the kth number such that the only prime factors are 3, 5, and 7.

13. Hints

14. Hints • 3 * (previous number in list) • 5 * (previous number in list) • 7 * (previous number in list) • How would we find the next number in the list? • Well, we could multiply 3, 5 and 7 times each number in the list and find the smallest element that has not yet been added to our list. • This solution is O(n^2). • Not bad, but I think we can do better

15. Hints Red: duplications

16. Hints • In our current algorithm, we’re doing 3*1, 3*3, 3*5, 3*7, 3*9, 3*15, 3*21, 3*25 …, and the same for 5 and 7.We’ve already done almost all this work before—why are we doing it again? • We can fix this by multiplying each number we add to our list by 3, 5, 7 and putting the results in one of the three first-in-first-out queues. • To look for the next “magic” number, we pick the smallest element in the three queues.

17. Solution