Challenge Math Challenge - February 2019

[QuantumQuest]

What is the solution of this recursive series?

Although I think that you can find it easily on the net, in any case, the explicit solution to the Fibonacci recurrence is $F_n = \frac{1}{\sqrt{5}}(\frac{1 + \sqrt{5}}{2})^n - \frac{1}{\sqrt{5}}(\frac{1 - \sqrt{5}}{2})^n$ known as Binet's formula.

 

[QuantumP7]

5. a) Find the last 3 digits of $3^{2405}$

Answer:
$3^{2405} = \left( 3^{5} \right) ^{481} = \left( 3^{5} \right) ^{400+80+1}$
The last 3 digits of $3^{2405}$ will then be $\left(3^{5} \right)^{1} = 243$.

 

[fresh_42]

5. a) Find the last 3 digits of $3^{2405}$

Answer:
$3^{2405} = \left( 3^{5} \right) ^{481} = \left( 3^{5} \right) ^{400+80+1}$
The last 3 digits of $3^{2405}$ will then be $\left(3^{5} \right)^{1} = 243$.

Why?

 

[QuantumP7]

Why?

$3^{2405} = 3^{5+2400} = (3^{5})(3^{2400})$
$3^{400} \equiv 1(mod 1000)$ so $3^{2400} \equiv 1(mod 1000)$
Then, $3^{2405} = (3^{5})(3^{2400}) \equiv (243*1)(mod 1000)$, so that the last 3 digits of $3^{2405}$ are 243.

 

[QuantumP7]

5.(b) Show that there is an integer $a \in \mathbb{Z}$ such that 64959| $(a^{2} - 7)$

Answer:

We can see that 3 divides 64959 because its digits add up to a multiple of 3 (6+4+9+5+9 = 33 = 3+3=6). (If necessary, I can explain why. Just let me know.) So, we need to find an integer a such that $a^{2}-7$ is a number that is 3 or 3 x $10^{n}$. Honestly, I brute forced this, and came up with a = 2. $2^2-7 = 4-7=-3$ which definitely divides 64959 (64959 = -3 * -21653).

 

[fresh_42]

5.(b) Show that there is an integer $a \in \mathbb{Z}$ such that 64959| $(a^{2} - 7)$

Answer:

We can see that 3 divides 64959 because its digits add up to a multiple of 3 (6+4+9+5+9 = 33 = 3+3=6). (If necessary, I can explain why. Just let me know.) So, we need to find an integer a such that $a^{2}-7$ is a number that is 3 or 3 x $10^{n}$. Honestly, I brute forced this, and came up with a = 2. $2^2-7 = 4-7=-3$ which definitely divides 64959 (64959 = -3 * -21653).

$3\,|\,64959$ but not the other way around. The key words here are the Legendre symbol and the Chinese remainder theorem.

As of part 5.a.), your solution

$3^{2405} = 3^{5+2400} = (3^{5})(3^{2400})$
$3^{400} \equiv 1(mod 1000)$ so $3^{2400} \equiv 1(mod 1000)$
Then, $3^{2405} = (3^{5})(3^{2400}) \equiv (243*1)(mod 1000)$, so that the last 3 digits of $3^{2405}$ are 243.

is correct, but a hint to Euler's theorem would have been fine:
$$
3^{\varphi(n)} \equiv 1 \,\operatorname{mod} n \\[12pt]
\varphi(1000) = \# \text{ numbers from 1 to 1000 which are relatively prime to 1000 } = \varphi(8) \varphi(125)=\varphi(2^3) \varphi(5^3)=2^3 \left(1- \dfrac{1}{2} \right) 5^3 \left(1- \dfrac{1}{5} \right) = 400
$$
and thus $3^{400} \equiv 1 \,\operatorname{mod} 1000$ is the information which really counted.

 

[archaic]

Spoiler: Fragments of 6

Well, since there's an integer $n$ this looks like induction. (I know this isn't a nice way to spot induction, do you please know a website/paper/pdf which talks about the matter?)
What jumps to me is this :
For $n=1$ we have $f(x)=f(x+\frac{1}{n})=f(x+1)$ set $x=0$ and we get$f(0)=f(1)$which is true.
For $n=2$ we have $f(x)=f(x+\frac{1}{n})=f(x+\frac{2}{n})$ set $x=0$ and we get$f(0)=f(1)$which is true.
For $n=3$ we have $f(x)=f(x+\frac{1}{n})=f(x+\frac{2}{n})=f(x+\frac{3}{n})$ set $x=0$ and we get$f(0)=f(1)$which is true.
...
For $n=n$ we have $f(x)=f(x+\frac{1}{n})=f(x+\frac{2}{n})=...=f(x+\frac{n}{n})$ set $x=0$ and we get$f(0)=f(1)$which is true.
For $n=n+1$ we have $f(x)=f(x+\frac{1}{n})=f(x+\frac{2}{n})=...=f(x+\frac{n}{n})=f(x+\frac{n+1}{n})$ set $x=0$ and we get$f(0)=f(1)$which is true.
I am unsure about the last two line and I do not know how to proceed from the first lines.

 

[QuantumQuest]

Well, since there's an integer $n$ this looks like induction. (I know this isn't a nice way to spot induction, do you please know a website/paper/pdf which talks about the matter?)

Do you mean a resource talking about induction in general or something more specific?

For $n=2$ we have $f(x)=f(x+\frac{1}{n})=f(x+\frac{2}{n})$ set $x=0$ and we get$f(0)=f(1)$which is true

This is not correct. Check again this (and the lines below it).

Also, is it necessary to use induction?

 

[QuantumP7]

$3\,|\,64959$ but not the other way around. The key words here are the Legendre symbol and the Chinese remainder theorem.

As of part 5.a.), your solution

is correct, but a hint to Euler's theorem would have been fine:
$$
3^{\varphi(n)} \equiv 1 \,\operatorname{mod} n \\[12pt]
\varphi(1000) = \# \text{ numbers from 1 to 1000 which are relatively prime to 1000 } = \varphi(8) \varphi(125)=\varphi(2^3) \varphi(5^3)=2^3 \left(1- \dfrac{1}{2} \right) 5^3 \left(1- \dfrac{1}{5} \right) = 400
$$
and thus $3^{400} \equiv 1 \,\operatorname{mod} 1000$ is the information which really counted.

Thank you, sir! Even though I misread part b, I knew that there was some deeper theory and a more elegant solution that I was unaware of. If no one else attempts part b within a few days, then I will redo it, myself.

I look like a complete idiot in this thread, but I really am learning a lot.

 

[archaic]

or something more specific?

Yes, on how to recognize it and why it works. (a farmer which has $n+1$ sheeps of which $n$are black doesn't imply all are, I probably have a very shallow understanding of it)

Also, is it necessary to use induction?

To put it sarcastically, "I'm a simple man, I see $n$ I use induction".

 

[QuantumQuest]

Yes, on how to recognize it and why it works. (a farmer which has $n+1$ sheeps of which $n$are black doesn't imply all are, I probably have a very shallow understanding of it)

As you already know, mathematical induction is a mathematical proof technique that is used to prove that some property $P(n)$ holds true for every natural number. I think that the key here is to understand what is induction exactly and do as many examples / exercises as you can. This will vastly help you gain the required momentum in order to recognize where is induction needed and apply the concept correctly. There is a multitude of examples you can find using a simple google search. Also, you can take a look at this cmu.edu https://www.cs.cmu.edu/~adamchik/21-127/lectures/induction_1_print.pdf for some problems beyond the usual (very basic) ones - it is Computer Science notes so it has some relevant to CS examples.

To put it sarcastically, "I'm a simple man, I see $n$ I use induction".

Mathematical induction is a very important method regarding proofs but (obviously) it is not a litmus test. For this particular question, I would hint to think of the problem in terms of functions i.e. using mostly Calculus.

 

[Buzz Bloom]

Hi @Infrared:

I gather you are the reviewer for problem #4. I have just completed my second try at a solution in post #30. Please take a look at it and let me know if I made any errors, or if it still needs some additional presentation.

Regards,
Buzz

 

[Infrared]

Could you please clarify the following step for me? Your answer is close, but it's not quite right (also, you should give a closed form instead of leaving it as an infinite sum).

(1/k!)[(1-1/n)+(1-2/n)+...+(1-(n-k+1)]
= (1/k!) [ 1 - (1/n)Σ{j=1,k+1}j + O(1/n²)]

​

Also, 1+2+\ldots+(k+1)=(k+1)(k+2)/2, not k(k+1)/2, but just fixing this doesn't give the right answer either.

 

[Buzz Bloom]

1+2+…+(k+1)=(k+1)(k+2)/2

but just fixing this doesn't give the right answer either.

it's not quite right (also, you should give a closed form instead of leaving it as an infinite sum).

Hi Infrared:

Thank you very much for your response to my solution attempt. The first error is just carelessness on my part. As an octogenarian I have found that happens much more often than it did a decade or two ago. I will now make an effort to find the second error to make the answer "quite right". It is not immediately obvious to me, but I should be able to find it. I expect that putting the answer into closed form will be the most difficult part of the necessary fix, but I am hopeful that when I find the second error, that step will become easier.

Could you explain a bit what needs clarification in the step

(1/k!)[(1-1/n)+(1-2/n)+...+(1-(n-k+1)]
= (1/k!) [ 1 - (1/n)Σ_{j=1,k+1}j + O(1/n²) ]​

I am guessing it is related to O(1/n²), but I am not sure about that.

ADDED
I now see that the sum
(1-1/n)+(1-2/n)+...+(1-(n-k+1)
should be a product. I will also fix that careless error in Post #30.

Regards,
Buzz

 

[fresh_42]

High School 5:

It suffices to show that any prime $p \in \{1, 7, 11, 13, 17, 19, 23, 29\}_{\mod(30)}$. Notice that this is true for otherwise we would always have another prime dividing $p$. Done.

I cannot follow your argumentation. Could you be a bit more detailed? And what about 60?

 

[Buzz Bloom]

Hi @Infrared:

I believe I have edited post #30 to correct all my errors for solving problem #4, and I think it is now OK. Please take another look at it.

Regards,
Buzz

 

[fresh_42]

High school 4:

Let $a, b$ be the two numbers. If anyone of these is divisible by $3$ we are done as $3$ also divides their product. Hence assume that $3 \nmid a, b$. This implies $a, b \equiv 1, 2 \mod(3)$. It is easy to see that any choice of $a, b$ such that $3\nmid a, b$ gives either their sum or difference as a multiple of $3$.

If it is so easy to see, why don't you write it?

 

[Ujjwal Basumatary]

I cannot follow your argumentation. Could you be a bit more detailed? And what about 60?

I have edited my posts. Sorry about the brief post I thought it would be unnecessary to post trivial facts.

 

[fresh_42]

I have edited my posts. Sorry about the brief post I thought it would be unnecessary to post trivial facts.

Sloppiness has to be earned. Easy to see, trivial or obvious are not terms I want to see in the high school section, since they all are.

 

[fresh_42]

High School 5:

It suffices to show that any prime $p$ is congruent modulo $1, 3, 7, 11, 13, 17, 19, 23, 29$ modulo $30$. Notice that this is true for otherwise we would always have another prime dividing $p$. For example if $p\equiv 6 \mod(30)$ we would have $6$ divide $p$.

Why does $30 \,|\, (p-6)$ imply $6\,|\,p\,?$ Sorry, but this needs an explanation. As it stands, it is wrong.

As for $60$, any prime is congruent to $1, 7, 11, 13, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59$ modulo $60$ and hence the statement holds good. In general a prime $p$ can be congruent to only $1$ or another prime $q$ modulo a composite number $a$ such that $q \nmid a$

Same here. Why are only those primes possible remainders? I do not need a solution, I have a better one, but you should demonstrate your reasoning on a high school level. And you did not answer the question: what about $60$? What does "holds good" mean?

 

[Buzz Bloom]

Although I think that you can find it easily on the net, in any case, the explicit solution to the Fibonacci recurrence is Fn=1√5(1+√52)n−1√5(1−√52)nF_n = \frac{1}{\sqrt{5}}(\frac{1 + \sqrt{5}}{2})^n - \frac{1}{\sqrt{5}}(\frac{1 - \sqrt{5}}{2})^n known as Binet's formula.

Hi QQ:

I get that the "solution" of the Fibonacci number recursive definition is the closed form formula that calculated the value of the nth number in the series. What I do not get is the relationship of this "solution" to algorithms and the time it takes to perform a calculation using an algorithm. hat I am guessing is there are more than one definition for what it means to solve a recursion expression. One meaning is the closed form that represents the nth term of the recursive sequence, and another meaning has something to do with algorithms and time to perform calculations. If I get the time, I will try to read the 30 page article you cited, and perhaps then I will understand this second meaning.

Regards,
Buzz

 

[Buzz Bloom]

For which values of φ\varphi are f⊥gf \perp g in L2([1,∞))?

Show that there is an integer a∈Za \in \mathbb{Z} such that 64959|(a2−7).

For how many (a1,…,a6)∈{0,1}6(a_1,\ldots,a_6)\in \{0,1\}^6 is it true that Q(a1ζ+…+a6ζ6)=Q(ζ)?

I am hoping that someone can educate me the meanings of some notation that was not included in any math courses I took as a math major over 50 years ago.
In the first quote, what does the symbol ⊥ mean, and what does L²([1,∞)) mean?
In the second quote, what does the vertical bar in 64959|(a²−7) mean?
In the third quote, what does special font Q mean in Q(a₁ζ+…+a₆ζ⁶)=Q(ζ)?

Regards,
Buzz

 

[Infrared]

@Buzz Bloom I am still concerned by (1-1/n)\ldots (1-(k+1)/n)=1-\frac{1}{n}(1+2+\ldots+ (k+1))+O(1/n^2). I agree that you can expand the lefthand side in powers of 1/n and that the first two coefficients are what you have written. But the coefficients for 1/n^k are going to be polynomials in k and k is not bounded independent of n since k can be as large as n. For example if your next term was k^3/n^2, then for the terms when k is close to n, this is O(n).

 

[Buzz Bloom]

But the coefficients for 1/n^k are going to be polynomials in k and k is not bounded independent of n since k can be as large as n.

Hi Infrared:

Thank you for your reply. You definitely got me. I will have to explore this issue in detail, and it will take quite a while.

Regards,
Buzz

 

[Infrared]

I am hoping that someone can educate me the meanings of some notation that was not included in any math courses I took as a math major over 50 years ago.
In the first quote, what does the symbol ⊥ mean, and what does L²([1,∞)) mean?
In the second quote, what does the vertical bar in 64959|(a²−7) mean?
In the third quote, what does special font Q mean in Q(a₁ζ+…+a₆ζ⁶)=Q(ζ)?

Regards,
Buzz

The first two problems aren't mine, but I'm pretty sure the following is intended (and @fresh_42 can clarify what he means if I get it wrong):
L^2([1,∞)) means the set of real-valued functions f defined on [1,\infty) such that \int_1^\infty f(x)^2 dx exists and is finite (technically, it is defined as equivalence classes of such functions up to equality almost everywhere, and the integral means the Lebesgue integral instead of the (improper) Riemann integral, but neither of these facts are important to the problem). There is an inner product on this space defined by
(f,g)=\int_1^\infty f(x)g(x)dx. Then f\perp g means (f,g)=0. Oftentimes, you work with complex-valued functions instead and then the inner product would be (f,g)=\int_1^\infty f(x)\overline{g(x)}dx.

The vertical line means "divides". For example, 3\vert 9 is true but 3\vert 10 is not.

\mathbb{Q} is the field of rational numbers. If \alpha is an element of some field K that contains \mathbb{Q}, then \mathbb{Q}(\alpha) is the smallest subfield of K that contains all elements of \mathbb{Q} as well as \alpha. For example, you can check \mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}: a,b\in\mathbb{Q}\}.

 

[QuantumQuest]

I get that the "solution" of the Fibonacci number recursive definition is the closed form formula that calculated the value of the nth number in the series. What I do not get is the relationship of this "solution" to algorithms and the time it takes to perform a calculation using an algorithm. hat I am guessing is there are more than one definition for what it means to solve a recursion expression. One meaning is the closed form that represents the nth term of the recursive sequence, and another meaning has something to do with algorithms and time to perform calculations. If I get the time, I will try to read the 30 page article you cited, and perhaps then I will understand this second meaning.

I gave you the (explicit) solution because you asked for the solution of the equation i.e. for the $F(n)$.

Perhaps you can help me gain an insight by advising me regarding a simple recursive function. For this example I have chosen the
Fibonacci numbers as a recursive function.

F(0) = F(1) = 1
For n>1, F(n) = F(n-1) + F(n-2)

What is the solution of this recursive series?

In any case, what you ask in post #50 is something outside the questions $1$ and $2$ of the challenge themselves but I'll try to explain it as it is something inside the context of these questions.

The explicit solution of the Fibonacci series gives you the $n$-th term of the series. If you want to calculate this by developing / using a program on a computer, you have to implement an algorithm which in this case can be iterative or recursive. The iterative version is based on having a repetition structure (mostly known as a loop) which computes the $n$-th term of the series in an imperative repetition fashion. Now, the formula $F_n = F_{n-1} + F_{n-2}$ for Fibonacci series, gives us a natural example of recursion which can be written as a recursive function like this

function fib(n) {
    if n = 0 return 0
    if n = 1 return 1
    return fib(n-1) + fib(n-2)
    }

In order to measure the time it takes for the above algorithm to execute when implemented as a program, approximately, we are measuring in terms of lines of code. Let $T(n)$ denote the number of steps needed to compute fib(n). If $n \lt 2$ the function runs for just a couple of steps i.e. for $n \leq 1$ we have $T(n) \leq 2$. Now, if $n$ gets larger there are two recursive invocations of the function fib. The time they take is $T(n -1)$ and $T(n - 2)$ respectively. We must also add the time taken for the checks on the value of $n$ and for the final addition. So, we have $T(n) = T(n - 1) + T(n - 2) + 3$ for $n \gt 1$. Now, if we compare this to the recurrence relation for $F_n$ we easily see that $T(n) \geq F_n$. $T(n)$ is exponential in $n$.

In general, any recursive function such as the above gives us a recurrence relation for time: the time for any routine is the time within the routine itself, plus the time for the recursive calls.

 

[Buzz Bloom]

Hi @Infrared and @QuantumQuest:

Thank you both very much for your posts #55 and #56. I feel your posts have taught me some very useful notations and concepts.

Regards,
Buzz

 

[Buzz Bloom]

Hi @Infrared:

I have fixed another error in post #30, but I still have more work to do on it.

Regarding your issue about O(1/n), I have thought about a method to explain why this concept is correct, but I had not yet added that explanation because I have not yet decided on how to present this concept mathematically in the solution. The concept involves the terms (all multiplied by n) of the form

G_j,k(x,n) = a_j,k (1/n^j) x^k/k! for 1<j<k.​

The limit

lim_n→∞ nG_j,k (x,n) = 0​

because the values of a_j,k are all finite and are independent of n. Do you think a detailed mathematical explanation of this would be sufficient to fix the solution?

Regards,
Buzz

 

[lpetrich]

I will solve Problem 4.

Spoiler

It is to find $L = \lim_{n\to\infty} n ((1 + x/n)^n - e^x)$.

The first term in parentheses, $(1 + x/n)^n$, is well-known for having $e^x$ as a limit as $n\to\infty$. So to find L, we must evaluate $\lim_{n\to\infty} n (\text{that term's difference from } e^x)$. So in order to have a more easily-expanded expression, we do
$$ \left( 1 + \frac{x}{n} \right)^n = \exp \log \left( \left( 1 + \frac{x}{n} \right)^n \right) = \exp \left( n \log \left( 1 + \frac{x}{n} \right) \right) $$
We next expand the logarithm in an infinite series, and then do that to the exponential:
$$ \left( 1 + \frac{x}{n} \right)^n = \exp \left( \sum_{k=1}^\infty \frac{(-1)^{k+1} x^k}{k n^{k-1}} \right) = \exp \left( x - \frac{x^2}{2n} + O\left( \frac{1}{n^2} \right) \right) = e^x \left( 1 - \frac{x^2}{2n} + O\left( \frac{1}{n^2} \right) \right)$$
Plugging this into the expression for L, we find
$$ L = \lim_{n\to\infty} n e^x \left( 1 - \frac{x^2}{2n} + O\left( \frac{1}{n^2} \right) - 1 \right) = \lim_{n\to\infty} e^x \left( - \frac{x^2}{2} + O\left( \frac{1}{n} \right) \right) = - \frac{x^2}{2} e^x$$
Thus, the answer is
$$ L = - \frac{x^2}{2} e^x $$

 

[Infrared]

@lpetrich Nice work! It's all right.