# Basic Math Challenge - June 2018

• Challenge
• Featured
No criticism, I only thought, as you didn't explicitly mention the theorem, that Fermat's little theorem is faster to find for our readers than to search for Euler's theorem. Euler has left so many theorems ...
I sincerely mean this, I always appreciate your comments and had not thought of it from that perspective, that writing Euler's theorem is ambiguous and that writing FLT could save time for the reader. I definitely think the proof I wrote could be improved by being more specific.

• fresh_42
Gold Member
I definitely think the proof I wrote could be improved by being more specific.

The steps you have in the proof are correct and lead to the correct result. But @fresh_42's comment is really worthwhile because the point in a proof is besides being correct to be also easily followed. Fermat's little theorem is a special case of Euler's theorem ##a^{\phi (n)} \equiv 1(\mod n)## where ##\phi (n)## is the Euler totient function, so you're essentially right in saying that you use these two things but as you apply FLT in your proof it is necessary for your readers in order to easily follow along, to state that explicitly.

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• fishturtle1
opus
Gold Member
What level of math is typically sufficient for these basic challenges? Im always interested but they still look pretty heavy!

fresh_42
Mentor
What level of math is typically sufficient for these basic challenges? Im always interested but they still look pretty heavy!
It is the same as it is so often: once you have the correct entry point, the proofs themselves aren't hard. I only solved mine, but I think it's similar with all of them. Do you have one particular in mind? Maybe we can give some hints.

• opus
opus
Gold Member
It is the same as it is so often: once you have the correct entry point, the proofs themselves aren't hard. I only solved mine, but I think it's similar with all of them. Do you have one particular in mind? Maybe we can give some hints.
I don't even think a hint would be beneficial at this point for me. I'm taking Calculus I in the fall, so I don't have any of those tools in my belt yet. What's a common requisite for being able to solve these? Calculus? Discrete math?

fresh_42
Mentor
I don't even think a hint would be beneficial at this point for me. I'm taking Calculus I in the fall, so I don't have any of those tools in my belt yet. What's a common requisite for being able to solve these? Calculus? Discrete math?
The examples are quite a variety of very different problems, i.e. not subject to a single area.
1. calculus
2. combinatorics
3. geometry (area formulas) with linear algebra (equations of straight lines)
4. geometry (of a circle)
5. combinatorics
6. discrete mathematics, resp. abstract algebra
7. differentiations and solving a system of linear equations
8. calculus
9. simply a riddle
10. integration, i.e. calculus
Farmer Joe's field wasn't very difficult, the key was basically to find the height of the yellow area, and the key was to use a coordinate system, find the equations of the lines and solve it. The coding was a bit like a crossword puzzle in the weekend edition of a newspaper, and for #7 one basically only needed to know what the differentials of the functions ##e^x\; , \;\sin(x)\; , \;\cos(x)## are.

I choose the first problem, because I found that those inequalities are useful to know. And, yes, it's again about differentiation, integration and the power series expansion of ##\log \,##. The first part can be solved by differentiation. The second part is a bit tricky, and the third is actually the easiest of them, although it might not appear as such, but one doesn't even have to know a lot about integration. They all can be solved within a few lines, but I admit one has to find an idea.

The basic difficulty for us is, that there are so many different people around. We already introduced this two weeks ban for members, we expected them to find the answers, in order to give the younger ones a chance. So if we added easier questions, then there are definitely some people who first had to promise us not to spoil those.

Why don't you try number 5? This one looks fun. I would start by an example with smaller numbers to get an idea about the difficulty, but I would expect, that it has something to do with symmetries. It reminds me a bit of my very first exercise at university: "Can there be a walk through Königsberg which crosses all seven bridges exactly once, and if so, can there also be a round trip?"

#### Attachments

• QuantumQuest and StoneTemplePython
4.4. Using only geometric reasoning, calculate the average value of f(x)=√2x−x2f(x) = \sqrt{2x - x^2} on [0,2] [0,2]
(by @QuantumQuest)

I am not sure really what the question is requiring ^^".

4) I assume the question just doesn't want me to use calculus/integration. If so then:
the equation is just half a circle:
$$f(x)=\sqrt{2x-x^2}=\sqrt{2x-x^2+1-1}=\sqrt{1-(x^2-2x+1)^2}=\sqrt{1-(x-1)^2} \rightarrow y^2+(x-1)^2=1$$

Thus the answer is half the half area of the circle. $$Average=\frac{\pi}{4}$$

Note: y is restricted to be positive.

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• QuantumQuest
StoneTemplePython
Gold Member
I don't even think a hint would be beneficial at this point for me. I'm taking Calculus I in the fall, so I don't have any of those tools in my belt yet. What's a common requisite for being able to solve these? Calculus? Discrete math?

I think @fresh_42's response is on point. A couple things I would add.

1.) To the extent you are in the US, I am sympathetic. I don't think I would have been able to solve much (or any?) of these in highschool even after I took calc 1. But so what -- if there are problems you can at least visualize, then try to struggle to the understand the associated solutions and you'll learn something interesting in the process. For example, I'm not sure I would have solved the cypher problem but it certainly is easy to visualize what is being discussed and follow the solution.

2.) I think we have a slight bias toward geometric problems in here as geometry in various forms is quite important in math and physics. It is also has a certain visual flavor. (Also reference part 1.)

• QuantumQuest
Gold Member
I am not sure really what the question is requiring ^^".

4) I assume the question just doesn't want me to use calculus/integration

The point in this problem is to translate "using only geometric reasoning" correctly. You interpreted it correctly and your solution is right. For completeness, one thing that I want to add is that what we must find is ##\frac{1}{2}\int_{0}^{2}\sqrt{2x - x^2} dx##. Then, using the way you used in your solution, we finally find that the graph of ##y = f(x)## between ##x = 0## and ##x = 2## is the upper half of the circle you found. So, the aforementioned integral is the area ##\frac{\pi}{2}## of that semicircle, so the average value is ##\frac{\pi}{4}##.

opus
Gold Member
QUOTE REMOVED TO SAVE SPACE

QUOTE REMOVED TO SAVE SPACE

Thanks guys. I'll try my hand at that problem fresh_42.
And yes I'm a U.S student- a horrendous public education, but that's no excuse for me. My lack of mathematical skills is my own doing as I didn't care in the least for it when I was in high school. Trying to make up for that now!
I think these problems are really great and helpful and I'm going to spend some time on trying to understand them so I can have meaningful participation down the road.

• QuantumQuest
Gold Member
And yes I'm a U.S student- a horrendous public education, but that's no excuse for me. My lack of mathematical skills is my own doing as I didn't care in the least for it when I was in high school. Trying to make up for that now!
I think these problems are really great and helpful and I'm going to spend some time on trying to understand them so I can have meaningful participation down the road.

Well, I don't live in US but as far as I know, public education is not horrendous compared to other countries in the world. What I can tell you for sure is that at College and University level is great and personally I really owe a lot of my knowledge in US resources - online classes, books, University resources, just to name a few of them. In any case, it's really good that you recognize that it is essentially your efforts that can make the difference - I was in this same position in my country as I was finishing High School, when I prepared on my own for what was missing in my knowledge in order to have a good head start when getting to University.

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• opus and Phylosopher
fresh_42
Mentor
Thread is now open to all. We have still four unsolved problems: 1,2,5,8.

So Homework Helpers, Science Advisors and Mentor can you help?

5.5. Can one pack 53 bricks, each of dimension 1 x 1 x 4\text{1 x 1 x 4} in to a 6 x 6 x 6\text{6 x 6 x 6} box?
(by @StoneTemplePython)

Thread is now open to all. We have still four unsolved problems: 1,2,5,8.

So Homework Helpers, Science Advisors and Mentor can you help?

I always wanted to solve such problem. Mr. fresh_42 suggested that this is a combinatorics problem. I didn't studied such problems before. From my own search on the web, this is called an optimization problem.

At first I tried to solve it directly by thinking. I remained with one brick and a volume of a brick if I am not mistaken (If I remember correctly, I had a volume of a brick but not its shape). I was going to make a program that solves the problem, but I felt Like cheating ^^".

since it is Jun 16. Can anyone give us a hint?

PeroK
Homework Helper
Gold Member
2020 Award
##2.## One card out of a deck has been lost. The other 51 cards are repeatedly shuffled and then thirteen cards are dealt, face up. The cards are 2 spades, 3 clubs, 4 hearts, and 4 diamonds. What is the chance that the missing card is (a) spade, (b) club, (c) heart, (d)diamond ##\space## ##\space## (by @StoneTemplePython)

We can use the idea of relative frequecies. If we repeat the experiment many times, then the missing card will initially be a S, C, H or D with equal frequency. So, we need to calculate how often in each of these cases we get 2S, 3C, 4H and 4D.

First, we can simplify things by noting that the frequency of getting those cards exactly in that order is related to the frequency of getting those cards in any order by a common factor (regardless of which card is missing). The common factor is ##\binom{13}{2} \binom{11}{3} \binom{8}{4}##

Second, the frequency is related by another common factor in all cases of ##\frac{1}{51} \frac{1}{50} \dots \frac{1}{39}##.

Omitting these common factors, we have the frequency of being dealt SSCCCHHHHDDDD:

If a Spade is missing: ##12 \times 11 \times 13 \times 12 \times 11 \times 13 \times 12 \times 11 \times 10 \times 13 \times 12 \times 11 \times 10 = 13^3 \times 12^4 \times 11^4 \times 10^2##

If a Club is missing: ##13 \times 12 \times 12 \times 11 \times 10 \times 13 \times 12 \times 11 \times 10 \times 13 \times 12 \times 11 \times 10 = 13^3 \times 12^4 \times 11^3 \times 10^3##

If a Heart is missing: ##13 \times 12 \times 13 \times 12 \times 11 \times 12 \times 11 \times 10 \times 9 \times 13 \times 12 \times 11 \times 10 = 13^3 \times 12^4 \times 11^3 \times 10^2 \times 9##

And the same for Diamonds.

Finally, taking out the common factors again we get the relative frequencies:

If a Spade is missing: ##11##

If a Club is missing: ##10##

If a Heart is missing: ##9##

If a Diamond is missing: ##9##

This means that out of every ##39## times we get dealt the cards 2S, 3C, 4H and 4D: ##11## times a Spade was the missing card, ##10## times the Club was the missing card etc.

Given we have been dealt 2S, 3C, 4H and 4D, therefore:

The probability a Spade is the missing card is ##11/39##; a Club the missing card ##10/39##; and a Heart or Diamond each ##9/39##.

• StoneTemplePython, QuantumQuest and mfb
StoneTemplePython
Gold Member
I always wanted to solve such problem. Mr. fresh_42 suggested that this is a combinatorics problem. I didn't studied such problems before. From my own search on the web, this is called an optimization problem.

At first I tried to solve it directly by thinking. I remained with one brick and a volume of a brick if I am not mistaken (If I remember correctly, I had a volume of a brick but not its shape). I was going to make a program that solves the problem, but I felt Like cheating ^^".

since it is Jun 16. Can anyone give us a hint?

I don't think writing a program would get credit, but to be honest if you make one, and you don't mind posting code at the end of the month, I'd be interested in seeing it if you're ok with sharing.

There's a few different ways to view the problem. Let me think on possibility of a hint.

StoneTemplePython
Gold Member
I always wanted to solve such problem. Mr. fresh_42 suggested that this is a combinatorics problem. I didn't studied such problems before. From my own search on the web, this is called an optimization problem.

At first I tried to solve it directly by thinking. I remained with one brick and a volume of a brick if I am not mistaken (If I remember correctly, I had a volume of a brick but not its shape). I was going to make a program that solves the problem, but I felt Like cheating ^^".

since it is Jun 16. Can anyone give us a hint?

This is a 3-D problem. Maybe it is tough... so consider a simpler, 2-D problem first. For the hint, I'll give a suggestive 2-D problem:

consider packing ##\text{1 x 2}## blocks on an 8 x 8 chessboard that has two squares at opposite corners removed. Can you pack 31 blocks into this slightly modified board? Why or why not?

It then takes a bit of visualization to make the leap from 2-D to 3-D.

• Phylosopher
pbuk
Gold Member
As its nearly the end of June, I'll bite...
consider packing ##\text{1 x 2}## blocks on an 8 x 8 chessboard that has two squares at opposite corners removed. Can you pack 31 blocks into this slightly modified board? Why or why not?
No, because each block covers 1 black square and 1 white square but the modified board has 32 black squares and 30 white squares (or vice versa).
It then takes a bit of visualization to make the leap from 2-D to 3-D.
A bigger leap than I have managed in the last fortnight...

StoneTemplePython
Gold Member
As its nearly the end of June, I'll bite...

No, because each block covers 1 black square and 1 white square but the modified board has 32 black squares and 30 white squares (or vice versa).

A bigger leap than I have managed in the last fortnight...

I think we're posting solutions to unsolved problems mid-month, so stay tuned.

##5.## Can one pack 53 bricks, each of dimension ##\text{1 x 1 x 4}## in to a ##\text{6 x 6 x 6}## box?
(by @StoneTemplePython)

Assuming that each brick is ##1m * 1m * 4m##, and the box is ##6m * 6m * 6m##;
The volume taken by 1 brick is equal to ##1m * 1m * 4m = 4m^3## and the volume of said box would be ##6m * 6m * 6m = 216m^3##, then;
##216m^3 / 4m^3 = 54##
You can fit 54 bricks into said box, therefore the answer is yes.

EDIT: I may be a bit late but the topic wasn't locked and the answer wasn't posted so I figured I would do it.

fresh_42
Mentor
Assuming that each brick is ##1m * 1m * 4m##, and the box is ##6m * 6m * 6m##;
The volume taken by 1 brick is equal to ##1m * 1m * 4m = 4m^3## and the volume of said box would be ##6m * 6m * 6m = 216m^3##, then;
##216m^3 / 4m^3 = 54##
You can fit 54 bricks into said box, therefore the answer is yes.

EDIT: I may be a bit late but the topic wasn't locked and the answer wasn't posted so I figured I would do it.
Problem is: you have to keep the bricks in one piece!

• PeroK
mfb
Mentor
@FelixLudi: With that argument you could also fit two 4x1x1 bricks in a 2x2x2 box. But you cannot.

I overlooked that somehow, but I'm out of solutions now since I can't get the right distribution of bricks to fit in (not even with an online 3d modeler).

Assuming that each brick is ##1m * 1m * 4m##, and the box is ##6m * 6m * 6m##;
The volume taken by 1 brick is equal to ##1m * 1m * 4m = 4m^3## and the volume of said box would be ##6m * 6m * 6m = 216m^3##, then;
##216m^3 / 4m^3 = 54##
You can fit 54 bricks into said box, therefore the answer is yes.

EDIT: I may be a bit late but the topic wasn't locked and the answer wasn't posted so I figured I would do it.

I did the same thing first to verify whether it is impossible to fit them or not. If the total volume of the bricks is larger than ##6*6*6## then it is impossible to fit the bricks inside such space. If total volume of the bricks is equal or smaller than the volume of the space, then it might be possible to do so.

It can be impossible to fit them even if they have equal or smaller volume.

Counterexample: Fitting ##2*2*2## brick in ##1*1*5## space (Im not sure, but I think this is a valid counterexample somehow!)

mfb
Mentor
I overlooked that somehow, but I'm out of solutions now since I can't get the right distribution of bricks to fit in (not even with an online 3d modeler).
"No" is a valid answer if you can prove that they cannot fit.
I'm quite sure they cannot fit but the approaches I tried to prove it don't work (or work only in two dimensions).

"No" is a valid answer if you can prove that they cannot fit.
I'm quite sure they cannot fit but the approaches I tried to prove it don't work (or work only in two dimensions).

Can you share your trials with us? It might give us a new perspective.

I tried to approach the problem mathematically but I failed. I tried to think of it outside the mathematical formulas, but to be honest I am not satisfied with the such method (Its not giving me an answer anyway ^^").