Mathematical

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.

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).

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).

Find the LCM of these sets of numbers.  3, 9, 21 Solution: List the prime factors of each. 3: 3 9: 3 × 3 21: 3 × 7 63 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 = can be divided by both 12 and 80.

Algorithm

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:

Solution  for (int i=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

Implementation  public boolean isPrime (int n) {  if (n<=1)  return false;  if (n==2)  return true;  if (n%2==0)  return false;  int m=Math.sqrt(n);  for (int i=3; i<=m; i+=2)  if (n%i==0)  return false;  return true;  }

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

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

Hints *33*53*7 55*35*55*7 77*37*57*7 Red: duplications 3*33*53*75*57*7 33*3*33*3*53*3*73*5*53*7*7 55*3*35*5*35*3*75*5*55*7*7 77*3*37*3*57*3*77*5*57*7*7

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.

Solution