Here is the answer key 1.) See the bee! The bee is a busy bee. Oh, it's landing! 2.) Are you sleeping? Yes, I am. I am too. 3.) I envy you. 4.) Katie is examining the ants. 5.) The hen has five eggs. 6.) I am before you. 7.) I am a human being. You are an animal. 8.) The seal is in the sea. 9.) The elephant ate the hay. 10.) Why are you whining? I don't know. 11.) I don't know. 12.) Katie is dizzy. I see why. 13.) I hate you! I hate you too! 14.) If you are busy, I'll run away. 15.) Ham and eggs are healthy for you! 16.) Oh, for a tool. Here is a hammer. 17.) Old age.

@RZ923 That's a very funny incident! (The one about mixing up the Belgium and German flags.)

@The-Blade-Dancer Hi again! This is a good question! Even though guess-and-check sometimes works and can even be the quickest approach sometimes, there's something a bit unsatisfying about it. How do we really know that we'll get anywhere? What if we end up guessing forever? \( {\tiny \text{(The horror...) }} \) Well, in this case, we can use a bit of algebra along with the fact that \(a\) and \(b\) are integers in order to figure out that \(a = 7\) and \(b= 5.\) Let's first start with the problem statement. $$ \sqrt{a} + \sqrt{b} = \sqrt[3]{22 \sqrt{7} + 26\sqrt{5}} $$ Cube both sides of the equation. $$\begin{aligned} \left( \sqrt{a} + \sqrt{b} \right)^3 &= \left( \sqrt[3]{22 \sqrt{7} + 26\sqrt{5}} \right)^3 \\ a\sqrt{a} + 3a\sqrt{b} + 3b\sqrt{a} + b\sqrt{b} &= 22\sqrt{7} + 26\sqrt{5} \\ \left( 3b + a \right) \sqrt{a} + \left( 3a + b \right) \sqrt{b} &= 22\sqrt{7} + 26\sqrt{5} \\ \end{aligned} $$ That looks better..... right? (You don't agree it looks better?) Now let's use the fact that \(a\) and \(b\) are integers. We know the right hand side has some radicals, the \(\sqrt{7}\) and the \(\sqrt{5}\) terms. Where could they have come from, though? Could either radical have come from the \(\left( 3b + a\right) \) or the \(\left( 3a + b \right) \)? No, because \(a\) and \(b\) are both integers, so \(\left( 3b + a\right) \) and \(\left( 3a + b \right) \) are also both integers. Well, it looks like there's no doubt about where the \(\sqrt{5}\) and the \(\sqrt{7}\) came from.   $$ \sqrt{5} \text{ and } \sqrt{7} \text{ came from the } \sqrt{a} \text{ and the } \sqrt{b} $$ So there you have it! Now that you know that \(a = 5\) and \(b = 7\) (or vice versa... it doesn't actually matter), then you can figure out that their sum is \( 5 + 7 = \boxed{12}.\)

@The-Blade-Dancer Thank you for writing and for being willing to share your comments. Looking back on this explanation, I sort of want to smack my head ... You're right, it is a bit obtuse. (Okay not a bit, but very!) The question is asking whether \( a + b\sqrt{c} = x + y + 2\sqrt{xy} \) for integers \(a, b,\) and \(c\) implies that \(x\) and \(y\) are always rational. (Rational means that the number is of the form \(\frac{m}{n}\) where \(m\) and \(n\) are both integers.) Hmmm... it seems like this statement can't be true, just from using some common sense, because how can we be so lucky as to always get \(x\) and \(y\) rational... but how do we prove this? Luckily, with a question of the form "It's always true that.... ", all we have to do is find a single counterexample where it isn't true. I've updated the explanation to hopefully be a bit more simple and easy to understand. The counterexample I'm using is a case where \(x\) and \(y\) are both imaginary numbers but \(a,\) \(b,\) and \(c\) are integers.

@RZ923 This is cool!!

Are fictional characters good at maths? Not unless you Count Dracula. 🧛 🧮

@The-Blade-Dancer Thank you for asking and thanks for being curious. The reason choice i.) is incorrect is because in general, for any numbers \(a\) and \(b,\) the following isn't true: $$ \sqrt{ a + b } = \sqrt{a} + \sqrt{b} $$ You could try some values of \(a\) and \(b\) to see this. Like for example, \(a = 1\) and \(b = 4.\) Then let's check our statement. $$ \sqrt{ 1 + 4 } \stackrel{?}{=} \sqrt{1} + \sqrt{4} $$ $$ \sqrt{ 5} \stackrel{?}{=} 1 + 2 $$ $$ \sqrt{5} \neq 3 $$ The answer choice i.) is claiming that the following is true: $$ \sqrt{ 6 + 5\sqrt{2}} = \sqrt{6} + \sqrt{5\sqrt{2}} $$ However, this is just the same as the statement \(\sqrt{a + b} = \sqrt{a} + \sqrt{b} \) where \(a = 6\) and \(b = 5\sqrt{2}.\) And since we just showed that this statement is false, then choice i.) can't be true.

@Da-Parasite Math idiosyncrasies are the best!

@The-Blade-Dancer ...in a good way!

@The-Blade-Dancer Oh, did you see the "Review Results" button at the bottom of the screen that shows your score...? Clicking on this button will show you the page that has your answers and the solutions. If you missed that button and close the window by accident, don't worry, you can also review your results by clicking a link that is emailed to you. The email would look like this in your inbox: And inside that email would be a link that opens a browser window displaying your score and the "Review Results" button again. Could you please try checking your email to see if you got the link? If you really still can't get your results, then there is still a way I can get them to you.

@Kevin-Li said in AMC Mock Test 2020: for number 22, it says independently, 1/3 chance of getting COVID. It has not been specified that you have to first meet a person with covid, then have a 1/3 chance of getting it, in fact since it says INDEPENDENTLY, it means it cannot depend on another factor such as the 50 percent chance of getting COVID Hi Kevin, You are right on both of these comments! We're very sorry for the error in the diagram not showing up. I've fixed this and it should work now. (You can also the diagrams in this printable .pdf) Thanks so much for your patience. Thanks also for pointing out the unclear conditions in question #22. We will refine the question to indicate that we are concerned only with the probability of catching COVID from the three people in the problem. We really appreciate your patience and for letting us know. Please tell us if you see anything else!

@The-Blade-Dancer I forgot to formally announce this Thank you @The-Blade-Dancer for doing my job for me!!

@RZ923 I love this photo of your scratch work.. it is so tidy, unlike mine. Prof. Loh always says he likes to do the same problem in many ways, even for a small computational step. It helps you check your answer, and it also helps you to learn little tricks here and there! In case anyone is wondering how this way compares with the method that Prof. Loh used in the Day 2 Your Turn lesson, the difference is this: @RZ923 is adding the ways in these three sections to get the final total number of ways:   $$ 8100 + 900 + 8100 = 17100 $$   On the other hand, Prof. Loh calculates the ways using Inclusion-Exclusion. In fact, he can be quoted around 0:18 as saying, "Let me first not tell you how much is in this little piece (pointing to the left-side crescent moon shape), where the tens equals the hundreds and the hundreds does not equal the thousands... I will instead calculate the whole piece (the whole circle)." That's just what he does! He adds the two circles together, and subtracts the overcounted area in the middle.   $$ 9000 + 9000 - 900 = 17100 $$   Yay, we get the same answer! Math is consistent!

Also, debbie's 1000th post is two above this one!!!

@debbie Great advertisement

@RZ923 Visual math = awesome

Lol