Math Is Fun Forum

Hey, it's not the programmer's fault if he got bad instructions.

That worked out a lot nicer than I expected (assuming I haven't stumbled somewhere)

I like the value of the second answer, but I like the reasoning of the first.
Even though it's wrong, at least it explained itself a bit.

Maybe, but both are pretty quick.

for (int i = 0; i <= 9 ; i++)
{
     for (int j = i; j <= 9; j++)
     {
          if((i+j) % 9 == (i*j) % 9)
          {
               Console.WriteLine("(" + i + "," + j + ")")
          }
     }
}

That was around 2 minutes.

A function A -> B is surjective if every element of B is mapped to by some element of A.

For example, taking A = {1, 2, 3, 4, 5} and B = {7, 8, 9}...

The function with the following maps:

1 -> 7
2 -> 8
3 -> 9
4 -> 8
5 -> 7

is surjective, because 7, 8 and 9 all have something that maps to them.

This function:

1 -> 9
2 -> 8
3 -> 9
4 -> 8
5 -> 9

is not surjective, because 7 isn't mapped to by anything.

This may not be the best way, but I'd try to solve the problem by building up the function piece by piece.

f(1) doesn't really matter - whatever you assign it to, the value will be 'new'.
There are therefore 3 functions f: {1} -> {7, 8, 9} that map to one element.

f(2) might map to the same thing as f(1), or it might map to something different (there are two ways of doing this).

So there are 3 functions f: {1, 2} -> {7, 8, 9} that map to one element, and 6 functions that map to two.

Now we can consider f(3).
If you build on a function that maps to one element, there is one way of keeping it as a function that maps to one, and two ways to change it into a function that maps to two.

On the other hand, if you build on a function that maps to two elements, there are two ways to keep it that way, and one way of mapping it to all three.

Overall then, there are 3 functions f:{1,2,3} -> {7,8,9} that map to one element,
18 functions that map to two elements, and 6 functions that map to all three.

You can use the same inductive reasoning to find the numbers for after you add in f(4) and f(5), and it turns out that there are 150 surjective functions where |A| = 5 and |B| = 3.

---

Bit long-winded, but it works at least. Maybe someone else will come along with a much better way.

Nearly all of my University exams were like that. A typical question would look like:

(i) Define this/these term(s)

(ii) Prove this/these theorems

(iii) Answer this question that makes use of those definitions and theorems (you may remember doing something similar in lectures)

(iv) Answer this question that makes use of those definitions and theorems, but this time it's not like any exercise you've seen before. This time you have to think!

It's fair enough really. If you don't know your definitions then you won't get anywhere, so it's a good thing that it tests you on them.

All the steps are correct, but if the question was to simplify it then it's understandably not the right answer.

When I first read the question, I interpreted it as that you could roll the die twice and got paid the larger roll, without having to gamble.

If that's the case, the expected win is (1*1 + 3*2 + 5*3 + 7*4 + 9*5 + 11*6)/36 = 4.47222...

If having a second throw makes you lose the first though, then I agree with 4.25 being the best answer.

I got 54. I was shooting for 50 though, so I'm happy about that.

The easier way of adding the first 100 numbers is to group them like this:
1+2+3+...+98+99+100 = (1+100) + (2+99) + (3+98) + ... = 101+101+101+... = 50*101 = 5050.

The legend goes that Gauss's teacher set the class this problem to keep them quiet for a while, but Gauss discovered that method and had it solved in under a minute.

Since it tells you that the planets have been divided into inner and outer, I'd say that Pluto being in neither category means it's not counted as a planet (for the purposes of this question anyway).

It perhaps could have been clearer, but I wouldn't say it was ambiguous.