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Monox D. I-Fly wrote:

(8 ^ 2) - [(1 * 3 + 4 + 5 + 7) + (9 + 6) * 0] = 45

I was hoping for 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45.

It does not work if you split 26 as 18 and 8.

8 × 10 + 18 = 98

98/2 = 49 (not prime)

The one with lots of terror bites.

Who is Nic Eone?

That's a physics pick-up line, not a math one.

A magnet goes into a bar and meets another magnet.

First magnet: I find you really attractive!

Second magnet: And I find you repulsive! Go away, we have nothing in common.

First magnet: Oh well, I guess we really are poles apart.

Super job!

Monox's solution is wrong; that solution is only for the case when the roots are real and *equal*.

Bamboozled wrote:

2. Find the set of values of k for which the roots of the equation x²+kx-k+3=0 are real and of the same sign.

For real roots we require

For the roots to be of the same sign, their product must be positive. So we want

Hence the set of values we want is

Sigh.

Okay, for people who still don't get it...let's just suppose the password is 133.

Person A comes along and tries to open the lock. They don't know the password but they assume the first digit is 3. So they go through all the combinations with fist digit 3: 311, 312, 313, 321, 322, 323, 331, 332, 333. The last one works. If they had tried it earlier they'd have opened the lock sooner. But these are the only combinations with first digit 3, so it is impossible to fail to open the lock after 9 tries (unless you're forgetful and repeat an already tried combination).

Person B comes along and thinks they can do better than A by assuming the first digit is 2. They go 211, 212, 213, 221, 222, 223, 231, 232, 233 – and find that they can't do better than the first person. The only way they can do it faster is by being a luckier guesser with the other two digits – but even without luck, the lock can still be opened after 9 tries at the very most.

Along comes person C and says, "I'll take the first digit to be 1." But this is just a fluke: they just happen to guess the first digit correctly. This way they can open the lock after at most four wrong tries (111, 112, 121, 122); the next one is bound to open the lock. So it is at most 5 steps for this lucky person.

This shows that by fixing the first digit and trying combinations for the other two, you can open the lock after no more than 9 steps. You can try any of 1, 2, 3 as the fixed first digit. If you're lucky and are correct with this digit, you should be able to open the lock after no more than 5 steps, but even without luck, and your guess for the first digit is wrong, you still shouldn't take more than 9 steps to open the lock.

Also it seems the LaTeX package in the forum software doesn't understand the \begin{align*} command. Instead of

```
\begin{align*}
x &= \cos(4t),\\
y &= \sin(6t).
\end{align*}
```

try \begin{array}:

```
\begin{array}{rcl}
x &=& \cos(4t),\\
y &=& \sin(6t).
\end{array}
```

The software doesn't understand \implies or \iff either, forcing the use of \Rightarrow and \Leftrightarrow instead.

phrontister wrote:

I may have misunderstood something here.

Yes.

phrontister wrote:

Are you saying that if the correct combination is 123 (with wild's position unknown), and you entered 111, 112, 131, 132, 211, 212, 213, 221, 222, 231, 232, 233, 311, 312, 313, 321, 322, 331, 332, and 333 (none of which will open the lock, as I understand it) before using one of the seven successful combinations you listed in post #5, that your method still holds?

No.