Answer:
public static boolean isDivisibleBy5(int number) {
boolean isTrue = true;
if (number % 5 == 0)
isTrue = true;
else
isTrue = false;
return isTrue;
}
Explanation:
- Create a method called <em>isDivisibleBy5</em> that takes one parameter, an integer number
- Inside the method, create a boolean variable, <em>isTrue</em>, that will be used to store the result
- Check if the number is divisible by 5, using the module operator. If it is, set the isTrue as true. Otherwise, set the isTrue as false
- Return the isTrue
I’d go with b. egotistical, i’m not entirely sure though. It seems like he is trying to appeal to the suspects EGO
That would be 13^4, or 13*13*13*13 = 28561
Answer:
Question is incomplete.
Assuming the below info to complete the question
You have a collection of n lockboxes and m gold keys. Each key unlocks at most one box. Without a matching key, the only way to open a box is to smash it with a hammer. Your baby brother has locked all your keys inside the boxes! Luckily, you know which keys (if any) are inside each box.
Detailed answer is written in explanation field.
Explanation:
We have to find the reachability using the directed graph G = (V, E)
In this V are boxes are considered to be non empty and it may contain key.
Edges E will have keys .
G will have directed edge b1b2 if in-case box b1 will have key to box b2 and box b1 contains one key in it.
Suppose if a key opens empty box or doesn’t contain useful key means can’t open anything , then it doesn’t belongs to any edge.
Now, If baby brother has chosen box B, then we have to estimate for other boxes reachability from B in Graph G.
If and only if all other boxes have directed path from box B then just by smashing box B we can get the key to box b1 till last box and we can unlock those.
After first search from B we can start marking all other vertex of graph G.
So algorithm will be O ( V +E ) = O (n+m) time.
