Challenge Math Challenge - February 2021

[martinbn]

Sure. I think it was even mentioned in the early days.

Sorry, that I don't remember this, so I can't say what happened. I only need to know who will moderate it: send me a PM if solved by whom and in which post, or give me the solution (for dummies) so that I can moderate it.

Yes, the problem should come with a detailed solution, so that you an deside if it fits the chalange.

 

[Office_Shredder]

There are a bunch of unsolved questions from previous threads, maybe have a challenge month where you just repost them and we work together to try to crack them.

 

[wrobel]

big gun hint for 3b

Spoiler

 

[julian]

There are a bunch of unsolved questions from previous threads, maybe have a challenge month where you just repost them and we work together to try to crack them.

I went back and solved a couple of the previous problems I just didn't post the answers. I'd like another crack at some of the other previous problems myself. I think some of the questions require me learning a bit of new maths, and for that reason I'd like to have a go at them myself.

I thought problems originally were being posted by various members. Has @fresh_42 been setting all the problems of late? Impressive.

Now and then I've been putting together a few problems when I can, which I was going to give @fresh_42 at some point.

I posted one problem last month because the problems got done early. You can have a go at that if you want . It doesn't require that you learn new maths, you solve the problem by taking some quantity to be an integer and expressing it another way that allows you to make estimations (hint: integrals) and making postulates.

 

[Keith_McClary]

unsolved questions from previous threads

I spent some time on November 2018 #8 .

 

[julian]

I just posted my solution to September 2020 #6:

https://www.physicsforums.com/threads/math-challenge-september-2020.993056/page-3

Got fed up with tweaking it so have posted it as it is. Sorry it so long.

 

[julian]

To elaborate on problem 4.

It was obvious that the AM-GM inequality would be involved, but like I said, a simple application leads to the Harmonic series:

\begin{align*}
\sum_{n=1}^\infty \prod_{k=1}^n (a_k)^{1/n} & \leq \sum_{n=1}^\infty \sum_{k=1}^n \frac{a_k}{n} \\
& = a_1 \sum_{n=1}^\infty \frac{1}{n} + a_2 \sum_{n=2}^\infty \frac{1}{n} + \cdots
\end{align*}

but the Harmonic series diverges, so I decided to consider a "weighted" version of the AM-GM inequality. It is not uncommon to see "weighted" versions of inequalities.

Replacing $\sum_{n=1}^\infty \sum_{k=1}^n \rightarrow \sum_{k=1}^\infty \sum_{n=k}^\infty$ is playing around with double sums, and has a simple interpretation in terms of dots on a 2d grid. You need the sum over $k$ to go from 1 to $\infty$.

Putting $b_k = \frac{c_k}{c_{k-1}}$ and getting $c_1 c_2 \cdots c_n = b_n$ is simply telescoping products and obviously simplifies things.

And $\left( 1 + \frac{1}{k} \right)^k \leq e$ is a well known inequality.

This

\begin{align*}
\sum_{n=k}^\infty \frac{1}{n(n+1)} = \sum_{n=k}^\infty \left( \frac{1}{n} - \frac{1}{n+1} \right) = \frac{1}{k}
\end{align*}

is a very well known example of telescoping for series.

 

[fresh_42]

To elaborate on problem 4.

It was obvious that the AM-GM inequality would be involved, but like I said, a simple application leads to the Harmonic series:

\begin{align*}
\sum_{n=1}^\infty \prod_{k=1}^n a_k & \leq \sum_{n=1}^\infty \sum_{k=1}^n \frac{a_k}{n} \\
& = a_1 \sum_{n=1}^\infty \frac{1}{n} + a_2 \sum_{n=2}^\infty \frac{1}{n} + \cdots
\end{align*}

but the Harmonic series diverges, so I decided to consider a "weighted" version of the AM-GM inequality. It is not uncommon to see "weighted" versions of inequalities.

Replacing $\sum_{n=1}^\infty \sum_{k=1}^n \rightarrow \sum_{k=1}^\infty \sum_{n=k}^\infty$ is playing around with double sums, and has a simple interpretation in terms of dots on a 2d grid. You need the sum over $k$ to go from 1 to $\infty$.

Putting $b_k = \frac{c_k}{c_{k-1}}$ and getting $c_1 c_2 \cdots c_n = b_n$ is simply telescoping products and obviously simplifies things.

And $\left( 1 + \frac{1}{k} \right)^k \leq e$ is a well known inequality.

This

\begin{align*}
\sum_{n=k}^\infty \frac{1}{n(n+1)} = \sum_{n=k}^\infty \left( \frac{1}{n} - \frac{1}{n+1} \right) = \frac{1}{k}
\end{align*}

is a very well known example of telescoping for series.

I'm having trouble sweeping this up to a proof, starting with a forgotten root in the very first term. It looks somehow right (even though I used Stirling), but can you line up your inequalities`?

 

[julian]

Sorry, I already gave the proof of everything on page 1. Here I'm just commenting on my methodology. (I've put the root in now).

 

[fresh_42]

Sorry, I already gave the proof on page 1. Here I'm just commenting on my methodology.

Oops! Your proof on page 1 was fine, no need to explain it. It is indeed Carleman's inequality. It is even a strict one.

The problems which have names can certainly be found somewhere, but that does not mean that they are famous enough to be automatically known, plus it means they are worth tackling them! I think it is still better to have a "I already have proven this" moment when encountering those statements in a book, than having solved a complicated but useless integral. I do not assume cheating in such cases. Why should I? Anyone who writes down an answer that they found elsewhere only cheats themselves, but at least has made the effort to learn a new theorem. I always try to put some value to the problems even if it means that they can be found somewhere.

 

[julian]

I thought it was strict inequality. I used that $1 + x \leq e^x$ to derive $\left( 1 + \frac{1}{k} \right)^k \leq e$ with equality holding only when $k \rightarrow \infty$.

 

[fresh_42]

I thought it was strict inequality. I used that $1 + x \leq e^x$ to derive $\left( 1 + \frac{1}{k} \right)^k \leq e$ with equality holding only when $k \rightarrow \infty$.

You can simply consider GM=AM to get a contradiction. That's easier than to deal with the exponential function.

 

[julian]

In May 2018 I found that I could prove the inequality: $(\theta^2 +x_1^2) \dots (\theta^2 +x_n^2) \geq [\theta^2 + (x_1^2 \dots x_n^2)^{1/k}]^k$ with Cauchy's forward-backward induction proof, which I was pleased about:

https://www.physicsforums.com/threa...challenge-may-2018.946386/page-3#post-6006214

@StoneTemplePython pointed out a simpler proof: that the inequality is basically the superadditivity of the Geometric Mean. Something I wasn't aware of but I reckon might possibly be the mainstream proof. I have just now done a search for "superadditivity of the Geometric Mean" and the Cauchy's forward-backward induction proof and couldn't find anything (well, aside from the physicsforums "Math Challenge - May 2018" webpage). However, there is most be somebody else who has also had this idea.

Also, I have realized that the inequality can be derived by solving a constrained extremal problem. Which by the way is one of the challenge problems that I have come up with.

I have also come up with a combinatorial proof which was fairly easy to understand but takes ages to explain. It is also a good moment when you come up with proofs that might not be the mainstream proof.

p.s. Sorry if got defensive. It is just that there is a bunch of idiots following me around, but you don't want to know about that.

 

[julian]

Anyway.

 

[fresh_42]

I have just now done a search for "superadditivity of the Geometric Mean" and the Cauchy's forward-backward induction proof and couldn't find anything (well, aside from the physicsforums "Math Challenge - May 2018" webpage). However, there is most be somebody else who has also had this idea.

Also, I have realized that the inequality can be derived by solving a constrained extremal problem. Which by the way is one of the challenge problems that I have come up with.

If you look at the previous solutions (currently 4 pdf files), you will find a lot of useful inequalities and equations around symmetries. Whenever possible I added their name.

 

[mathwonk]

In reference e.g. to problems generally, I suggest that the goal of this exercise might be to illuminate some key point, rather than to quote a powerful result that implies the problem. Problem #9 e.g., which is indeed essentially the maximum modulus principle, has several proofs, with different morals, each with something to teach us. The mean value property, which follows from integral representations, and I think shows that the value of an appropriate function at the center of a circle equals the average of its values on the boundary, is one that will appeal to analytically minded persons. The approach I hinted at is more geometric, and shows that every analytic function which is non constant near 0, (and with value zero there), looks locally, in some coordinates, like the function z^n, for some n ≥ 1. This makes the local geometric behavior of analytic complex functions very transparent, and implies also the useful "open mapping principle". To each his/her own. The mean value property also has the virtue of remaining true also for harmonic functions, and even in characterizing them, as I recall.

Remrk on #10: most group theory problems I solve follow from the Sylow theorems. But there are some more powerful results out there due to Burnside and others which some of us are not aware of. This problem #10 seems to yield to his "transfer theorem" or "N/C theorem", one that I only learned today. Is there a more clever, less advanced solution, does anyone know, using only Sylow? E.g. as an example of this sort, just try to show that every group of order 30 = 2.3.5 is solvable.

 

[suremarc]

3. Prove or find a counterexample to:
(a) $L^2$ convergence implies pointwise convergence.
(b) $\displaystyle{\lim_{n \to \infty}\int_{0}^{\infty}\dfrac{\sin x^n}{x^n}\,dx}=1$
(c) Let $(f_n)$ be a sequence of measurable functions which converge uniformly to zero on $[0,\infty).$ Then
$$
\lim_{n \to \infty}\int_{[0,\infty)} f_n(x)\,dx=\int_{[0,\infty)} \lim_{n \to \infty} f_n(x)\,dx\,.
$$

Spoiler: Solution

(a) False; consider a trivial sequence of null sets $X_n=Y$ for some null set $Y$, whence $\int |\mathbf{1}_{X_n}|^2\,d\mu=0$. Thus $\mathbf{1}_{X_n}$ converges to zero in $L^2(\mathbb{R})$ (in fact $X_n=0$ wrt the $L^2(\mathbb{R})$ seminorm), but the pointwise limit is $\mathbf{1}_Y$.

(b) First consider the integral $\int_0^1\frac{\sin x^n}{x^n}dx$. From the alternating Maclaurin series of $\sin x$ one has the bound $0\leq 1-\frac{x^{2n}}{6}\leq\frac{\sin x^n}{x^n}\leq 1$ for $x\in(0,1)$. This gives us bounds $$1-\frac{1}{12n}\leq\int_0^1\frac{\sin x^n}{x^n}dx\leq 1$$ so by the squeeze theorem, the limit as $n\rightarrow\infty$ is $1$.

Now consider $\int_1^\infty\frac{\sin x^n}{x^n}dx$. We have the simple chain of inequalities $$\Bigg|\int_1^\infty\frac{\sin x^n}{x^n}dx\Bigg|\leq \int_1^\infty\Bigg|\frac{\sin x^n}{x^n}\Bigg|dx\leq\int_1^\infty\frac{1}{x^n}dx=\frac{1}{n+1}$$ so our integral is bounded between $0$ and $\frac{1}{n+1}$, and shrinks to $0$ as $n\rightarrow\infty$.

Since $$\int_0^\infty\frac{\sin x^n}{x^n}dx=\int_0^1\frac{\sin x^n}{x^n}dx+\int_1^\infty\frac{\sin x^n}{x^n}dx,$$ the integral is equal to $0+1$ or $1$.

(c) False: consider the fn. $$f_n(x)=\frac{(-1)^n\mathbf{1}_{[0,n)}}{n}.$$ $f_n$ converges uniformly to $0$, but $\int_0^\infty f_n(x)\,dx=(-1)^n$.

 

[PeroK]

View attachment 277251

High Schoolers only

11. Let $z$ be a natural number with $1995$ decimal digits and $1\leq n\leq 1994.$ Then we note the number, which we get by cutting off the first $n$ digits and append them in the same order at the end of $z$ by $z^{[n]}.$ Show that if $z$ is divisible by $27,$ then all $z^{[n]}$ are divisible by $27,$ too.

12. Let $a,b,c,d$ be positive real numbers. Prove (in the logically correct order)
$$
\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{d}}\leq \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}
$$

13. Let $m\geq 2$ be a given natural number. We define a sequence $(x_0,x_1,x_2,\ldots)$ of numbers by $x_0=0,x_1=1,$ and for $n\geq 0$ we set $x_{n+2}$ to be the remainder of $x_{n+1}+x_n$ by division by $m,$ chosen such that $0\leq x_{n+2} < m.$ Decide whether for every $m\geq 2$ there exists a natural number $k\geq 1,$ such that $x_{k+2}=1\, , \,x_{k+1}=1\, , \,x_k=0.$

14. We define real functions
$$
f_n(x):=x^3+(n+3)\cdot x^2+2n\cdot x -\dfrac{n}{n+1}
$$
for every non-negative integer $n\geq 0.$ Determine all values of $n,$ such that all zeros of $f_n(x)$ are contained in an interval of length $3.$

15. (a) Determine the number of all pairs of integers $(x,y) \in \mathbb{N}_0^2$ with $\sqrt{x}+\sqrt{y}=1993.$
15. (b) Determine for every $n\in \mathbb{N}$ the greatest power of $2$ which divides $[( 4+\sqrt{18} )^n ].$

I guess it's clear that the high-school questions have become too hard. Given that none of them has been solved!

These look like a sufficient challenge for anyone without the pure maths knowledge to tackle the main questions.

 

[fresh_42]

I guess it's clear that the high-school questions have become too hard. Given that none of them has been solved!

These look like a sufficient challenge for anyone without the pure maths knowledge to tackle the main questions.

I won't mind if anyone else tries to solve them in this case, i.e. at the end of month. But I am not sure whether your hypothesis (too hard) holds. There could be other reasons. I admit that some are tricky, but #11 and #12 are quite easy. Especially #12 is straight forward: start with the inequality, end up with something like $f(a,b,c,d)^2\geq 0$ and write it down backwards. It's more a question about proof logic than about arithmetic.

The questions in March will be easier, or at least significantly shorter to answer.

 

[fishturtle1]

Spoiler: Problem 10

Lemma: If $a, a+k$ are integers with $a \ge 2$ and $k \ge 1$, we have $a + (a+k) < a(a+k)$

Proof: We have $a(a+k) \ge 2(a+k) = (a+k) + (a+k) \ge (a+1) + (a+k) > a + (a+k)$. []Suppose $G$ is a group of order $pqr$ where $p < q < r$ are distinct primes. Then $G$ is solvable.

Proof: We note that a group of order $ab$ where $a$ and $b$ are distinct primes is solvable. So, it's enough to show $G$ has a normal subgroup of prime order. Let $n_p$ denote the number of Sylow p subgroups of $G$. Define $n_q, n_r$ similarly. Sylow's theorems tells us
\begin{align*}
n_p &= 1 (\text{ mod p })\\
n_q &= 1 (\text{ mod q })\\
n_r &= 1 (\text{ mod r }) \\
\end{align*}
Sylow's theorems also tells us $n_p \vert qr$, so $n_p = 1, q, r,$ or $qr$. Similarly, $n_q = 1, p, r$ or $pr$ and $n_r = 1, p, q$ or $pq$. If $n_r = 1$, we're done. If $n_r = p$, then $p = 1 (\text{ mod r })$. But that means $p - 1 = rk$ for some positive integer $k$. But $p-1 < p < r \le rk$. We may conclude $n_r \neq p$ and $n_r \neq q$. So we must have $n_r = pq$. This shows $G$ has $pq(r-1) = pqr - pq$ elements of order $r$. Subtracting that from $pqr$, we have $pqr - (pqr - pq) = pq$. Subtracting $1$ for the identity element, we have $pq - 1$ elements unaccounted for. Assume by contradiction $n_p, n_q \neq 1$. We'll show that no matter how small $n_p$ and $n_q$ are, there are still too many elements. If $n_q = p$, then $p - 1$ is a positive multiple of $q$ which is a contradiction. So, the smallest $n_q$ can be is $n_q = r$.


The smallest $n_p$ can be is $n_p = q$. So, if $n_q = r$ and $n_p = q$, we have $r(q-1)$ elements of order $q$ and $q(p-1)$ elements of order $p$. But we only have $pq - 1$ elements left. We see
$$(pq - 1) - q(p-1) - r(q-1) = pq - 1 - pq + q - qr + r = q + r - qr - 1 < q + r - qr < 0$$
We've reached a contradiction and can conclude $n_p = 1$ or $n_p = 1$. This gives a normal subgroup $H$ of prime order, say $p$. Then $G/H$ is a group of order $qr$. So, $G/H$ and $H$ are solvable. We may conclude $G$ is solvable.

 

[Not anonymous]

Spoiler: Question 12

Define $x = (a + b)$ and $y = (c + d)$. Then the LHS of the inequality in the question, i.e. $$
\dfrac {1} {\dfrac {1} {a} + \dfrac {1} {b}} + \dfrac {1} {\dfrac {1} {c} + \dfrac {1} {d}}
$$
is equivalent to $$
\dfrac {ab} {a+b} + \dfrac {cd} {c+d} = \dfrac {ab} {x} + \dfrac {cd} {y} = \dfrac {aby + cdx} {xy}
$$

Similarly, the RHS can be expressed in terms of $x$ and $y$ as
$\dfrac {1} {\dfrac {1} {a+c} + \dfrac {1} {b+d}} = \dfrac {(a+c)(b+d)} {a+b+c+d} = \dfrac {(a+c)(b+d)} {x+y}$

$\Rightarrow LHS - RHS = \dfrac{aby+cdx} {xy} - \dfrac{(a+c)(b+d)} {x+y} =$

$\dfrac{(abxy + aby^{2} +cdx^{2} + cdxy) - (abxy + adxy + bcxy + cdxy)} {xy(x+y)} =$

$\dfrac {aby^{2} + cdx^{2} - adxy - bcxy} {xy(x+y)} = \dfrac {(ay - cx)(by - dx)} {xy(x+y)} =$
$\dfrac {(ac + ad - ca - cb)(bc + bd - da - db)} {xy(x+y)} =$
$\dfrac {(ad - bc)(bc - ad)} {xy(x+y)} = \dfrac {-(ad - bc)^{2}} {xy(x+y)}$

The denominator of the final expression $xy(x+y)$ must be positive since the $a, b, c, d$ are all positive and therefore $x, y$ too are positive. And $-(ad - bc)^{2} \leq 0$ for any real-valued $a, b, c, d$. Hence $LHS - RHS \leq {0} \Rightarrow \text{LHS} \leq \text{RHS}$, hence proving the inequality.

 

[fresh_42]

Spoiler: Question 12

Define $x = (a + b)$ and $y = (c + d)$. Then the LHS of the inequality in the question, i.e. $$
\dfrac {1} {\dfrac {1} {a} + \dfrac {1} {b}} + \dfrac {1} {\dfrac {1} {c} + \dfrac {1} {d}}
$$
is equivalent to $$
\dfrac {ab} {a+b} + \dfrac {cd} {c+d} = \dfrac {ab} {x} + \dfrac {cd} {y} = \dfrac {aby + cdx} {xy}
$$

Similarly, the RHS can be expressed in terms of $x$ and $y$ as
$\dfrac {1} {\dfrac {1} {a+c} + \dfrac {1} {b+d}} = \dfrac {(a+c)(b+d)} {a+b+c+d} = \dfrac {(a+c)(b+d)} {x+y}$

$\Rightarrow LHS - RHS = \dfrac{aby+cdx} {xy} - \dfrac{(a+c)(b+d)} {x+y} =$

$\dfrac{(abxy + aby^{2} +cdx^{2} + cdxy) - (abxy + adxy + bcxy + cdxy)} {xy(x+y)} =$

$\dfrac {aby^{2} + cdx^{2} - adxy - bcxy} {xy(x+y)} = \dfrac {(ay - cx)(by - dx)} {xy(x+y)} =$
$\dfrac {(ac + ad - ca - cb)(bc + bd - da - db)} {xy(x+y)} =$
$\dfrac {(ad - bc)(bc - ad)} {xy(x+y)} = -(ad - bc)^{2}$

The denominator of the final expression $xy(x+y)$ must be positive since the $a, b, c, d$ are all positive and therefore $x, y$ too are positive. And $-(ad - bc)^{2} \leq 0$ for any real-valued $a, b, c, d$. Hence $LHS - RHS \leq {0} \Rightarrow \text{LHS} \leq \text{RHS}$, hence proving the inequality.

I'm not sure whether $x,y$ are really making things easier, but o.k. The usual way to solve such problems is to start with the final statement, change that until something true comes up, and finally write them down backward! Like $0 \geq -(ad - bc)^{2} \Longrightarrow 0\geq \dfrac {(ad - bc)(bc - ad)} {xy(x+y)} \Longrightarrow \ldots$

 

[Not anonymous]

Spoiler: Question 11

Let $x$ be the number formed by the first (leftmost) $n$ digits of $z$ and let $y$ be the number formed by removing those first $n$ digits from $z$. For example, if $z=54321$ (this is not a 1995-digit number but using this just for illustration) and $n=2$, then $x$ and $y$ would be 54 and 321 respectively. A 1995-digit $z$ and the corresponding $z^{[n]}$ can be expressed in terms of $x, y$ as $z = 10^{1995-n}x + y$, and $z^{[n]} = 10^{n}y + x$. The difference between the 2 numbers is #### $d = z^{[n]} - z = (10^{n} - 1)y - (10^{1995-n} - 1)x$.

Now consider positive integers that are powers of 10 and find their modulos w.r.t. 27. $10^0 \mod {27} = 1, 10^1 \mod {27} = 10, 10^2 \mod {27} = 19, 10^3 \mod {27} = 1, ...$ and we note that the modulo sequence ${1, 10, 19}$ repeats. Thus, we can write $10^{n} \equiv (1 + 9(n \mod{3})) \mod 27$ for any non-negative integer $n$. (Eq. 1)

By the same logic, we get $10^{1995-n} \equiv (1 + 9((1995 - n) \mod{3}))\mod 27 \equiv (1 + 9((3 - n) \mod{3}))\mod 27$ for any non-negative integer $n \leq 1995$. (Eq. 2)

If $z$ is divisible by 27, then we get $z \mod 27 = 0 \Rightarrow (10^{1995-n}x + y) \mod 27 = 0$
$\Rightarrow y \mod 27 = -(10^{1995-n}x \mod 27) = -(1 + 9((3 - n) \mod{3}))x \mod 27 =$
$-(1 + 27 - 9(n \mod 3))x \mod 27 \equiv -(1 - 9(n \mod 3))x \mod 27$ (Eq. 3)

Similarly, $d \mod 27$ can be expressed in terms of modules of $x, y, n$ w.r.t. 27.
$d \mod 27 = ((10^{n} - 1)y + (10^{1995-n} - 1)x) \mod 27 =$
$(((10^{n} - 1) \mod 27)(y \mod 27) + ((10^{1995-n} - 1) \mod 27)x) \mod 27$ (Eq. 4)

Applying equations (1), (2) and (3) on (Eq. 4) gives:
$d \mod 27 \equiv ((9(n \mod{3}) \times -(1 - 9(n \mod{3}))x) - (9((3 - n) \mod{3}))x) \mod 27$ (Eq. 5)

Defining $a \equiv n \mod 3$ and substituting accordingly in equation (5) gives:
$d \mod 27 \equiv ((9a \times -(1 - 9a)x) - (9(3 - a)x)) \mod 27 \equiv$
$(-9ax + 81a^{2}x - 27x + 9ax) \mod 27 \equiv (81a^{2} - 27x) \mod 27 \equiv 0 \mod 27$.

But $d \mod 27 \equiv 0 \Rightarrow z^{[n]} \mod 27 = z \mod 27$.
In other words, $z^{[n]} \equiv z \mod 27$ if $z \equiv 0 \mod 27$, and hence $z^{[n]}$ too is divisible by 27 if $z$ is.
####

 

[fresh_42]

Spoiler: Question 11

Let $x$ be the number formed by the first (leftmost) $n$ digits of $z$ and let $y$ be the number formed by removing those first $n$ digits from $z$. For example, if $z=54321$ (this is not a 1995-digit number but using this just for illustration) and $n=2$, then $x$ and $y$ would be 54 and 321 respectively. A 1995-digit $z$ and the corresponding $z^{[n]}$ can be expressed in terms of $x, y$ as $z = 10^{1995-n}x + y$, and $z^{[n]} = 10^{n}y + x$. The difference between the 2 numbers is #### $d = z^{[n]} - z = (10^{n} - 1)y - (10^{1995-n} - 1)x$.

Now consider positive integers that are powers of 10 and find their modulos w.r.t. 27. $10^0 \mod {27} = 1, 10^1 \mod {27} = 10, 10^2 \mod {27} = 19, 10^3 \mod {27} = 1, ...$ and we note that the modulo sequence ${1, 10, 19}$ repeats. Thus, we can write $10^{n} \equiv (1 + 9(n \mod{3})) \mod 27$ for any non-negative integer $n$. (Eq. 1)

By the same logic, we get $10^{1995-n} \equiv (1 + 9((1995 - n) \mod{3}))\mod 27 \equiv (1 + 9((3 - n) \mod{3}))\mod 27$ for any non-negative integer $n \leq 1995$. (Eq. 2)

If $z$ is divisible by 27, then we get $z \mod 27 = 0 \Rightarrow (10^{1995-n}x + y) \mod 27 = 0$
$\Rightarrow y \mod 27 = -(10^{1995-n}x \mod 27) = -(1 + 9((3 - n) \mod{3}))x \mod 27 =$
$-(1 + 27 - 9(n \mod 3))x \mod 27 \equiv -(1 - 9(n \mod 3))x \mod 27$ (Eq. 3)

Similarly, $d \mod 27$ can be expressed in terms of modules of $x, y, n$ w.r.t. 27.
$d \mod 27 = ((10^{n} - 1)y + (10^{1995-n} - 1)x) \mod 27 =$
$(((10^{n} - 1) \mod 27)(y \mod 27) + ((10^{1995-n} - 1) \mod 27)x) \mod 27$ (Eq. 4)

Applying equations (1), (2) and (3) on (Eq. 4) gives:
$d \mod 27 \equiv ((9(n \mod{3}) \times -(1 - 9(n \mod{3}))x) - (9((3 - n) \mod{3}))x) \mod 27$ (Eq. 5)

Defining $a \equiv n \mod 3$ and substituting accordingly in equation (5) gives:
$d \mod 27 \equiv ((9a \times -(1 - 9a)x) - (9(3 - a)x)) \mod 27 \equiv$
$(-9ax + 81a^{2}x - 27x + 9ax) \mod 27 \equiv (81a^{2} - 27x) \mod 27 \equiv 0 \mod 27$.

But $d \mod 27 \equiv 0 \Rightarrow z^{[n]} \mod 27 = z \mod 27$.
In other words, $z^{[n]} \equiv z \mod 27$ if $z \equiv 0 \mod 27$, and hence $z^{[n]}$ too is divisible by 27 if $z$ is.
####

Looks o.k.

You could have saved a lot of text by assuming $n=1$. Why is that sufficient?

 

[Not anonymous]

Spoiler: Question 13

$(x_{0}, x_{1}, x_{2}, ...)$ is an infinite sequence but since $x_{i}$'s can take only a finite number of values ($x_{i} \in \{0, 1, ..., m-1\} \, \forall i$) there can only be a finite number of distinct values for pairs of adjacent elements in the sequence. Also, the complete sequence, or any subsequence of contiguous elements is completely determined by the first 2 elements. Thus there must be some $p \geq 0, q > 0$ such that the finite subsequence $(x_{p}, x_{p+1}, ..., x_{p+q})$ starts repeating to form the rest of the sequence, i.e. $x_{p+q+1} = x_{p}, x_{p+q+2} = x_{p+1}$ and so on.

Let $t$ be the smallest such value of $p$, i.e. the smallest among start indices of repeating subsequences in the original sequence, and let $j$ be the length of corresponding repeating subsequence.

Suppose $t \geq 1$, i.e. the repeating subsequence starts no earlier than at the second element. Then the original sequence will be of the form $(0, 1, ..., x_{t-1}, x_{t}, x_{t+1}, ..., x_{t+j}, x_{t+j+1}=x_{t}, x_{t+j+2}=x_{t+1}, ..., x_{t+2j} = x_{t+j}, ...)$.

By definition of how the sequence is formed and the pattern of repeating subsequences, we must have
$x_{t+1}= (x_{t-1} + x_{t}) \mod m$ (Eq. 1)
$x_{t+j+1} = (x_{t+j-1} + x_{t+j}) \mod m \Rightarrow x_{t} = (x_{t+j-1} + x_{t+j}) \mod m$ (Eq. 2)
$x_{t+j+2} = (x_{t+j} + x_{t+j+1}) \mod m \Rightarrow x_{t+1} = (x_{t+j} + x_{t}) \mod m$ (Eq. 3)
(with $\mod m$ denoting the remainder of division by $m$)

Combining equations (1) and (3) we see that $x_{t-1} \mod m = x_{t+j} \mod m$. Since $x_{t-1}, x_{t+j} \in \{0, 1, .., m-1\}$, it follows that $x_{t-1} = x_{t+j}$. This implies that the subsequence $(x_{t-1}, x_{t}, x_{t+1}, ..., x_{t+j-1})$ is a repeating subsequence in the original sequence. But this contradicts the assumption that $t > 1$. Hence we are left with 0 as the only valid value for $t$. In other words, the earliest repeating subsequence start at $i=0$, i.e. the repeating subsequence starts with the pair $(x_{0} = 0, x_{1} = 0)$. Hence the ordered triple $(0, 1, 1)$ too must repeat in the original sequence, thereby proving that there exists $k \geq 1$ such that $x_{k} = 0, x_{k+1} = 1, x_{k+2} = 1$

Note: More precisely, $k \geq 2$ since we need $x_{k} = 0$ whereas $x_{1} = 1 \neq 0$

 

[fresh_42]

Spoiler: Question 13

$(x_{0}, x_{1}, x_{2}, ...)$ is an infinite sequence but since $x_{i}$'s can take only a finite number of values ($x_{i} \in \{0, 1, ..., m-1\} \, \forall i$) there can only be a finite number of distinct values for pairs of adjacent elements in the sequence. Also, the complete sequence, or any subsequence of contiguous elements is completely determined by the first 2 elements. Thus there must be some $p \geq 0, q > 0$ such that the finite subsequence $(x_{p}, x_{p+1}, ..., x_{p+q})$ starts repeating to form the rest of the sequence, i.e. $x_{p+q+1} = x_{p}, x_{p+q+2} = x_{p+1}$ and so on.

Let $t$ be the smallest such value of $p$, i.e. the smallest among start indices of repeating subsequences in the original sequence, and let $j$ be the length of corresponding repeating subsequence.

Suppose $t \geq 1$, i.e. the repeating subsequence starts no earlier than at the second element. Then the original sequence will be of the form $(0, 1, ..., x_{t-1}, x_{t}, x_{t+1}, ..., x_{t+j}, x_{t+j+1}=x_{t}, x_{t+j+2}=x_{t+1}, ..., x_{t+2j} = x_{t+j}, ...)$.

By definition of how the sequence is formed and the pattern of repeating subsequences, we must have
$x_{t+1}= (x_{t-1} + x_{t}) \mod m$ (Eq. 1)
$x_{t+j+1} = (x_{t+j-1} + x_{t+j}) \mod m \Rightarrow x_{t} = (x_{t+j-1} + x_{t+j}) \mod m$ (Eq. 2)
$x_{t+j+2} = (x_{t+j} + x_{t+j+1}) \mod m \Rightarrow x_{t+1} = (x_{t+j} + x_{t}) \mod m$ (Eq. 3)
(with $\mod m$ denoting the remainder of division by $m$)

Combining equations (1) and (3) we see that $x_{t-1} \mod m = x_{t+j} \mod m$. Since $x_{t-1}, x_{t+j} \in \{0, 1, .., m-1\}$, it follows that $x_{t-1} = x_{t+j}$. This implies that the subsequence $(x_{t-1}, x_{t}, x_{t+1}, ..., x_{t+j-1})$ is a repeating subsequence in the original sequence. But this contradicts the assumption that $t > 1$. Hence we are left with 0 as the only valid value for $t$. In other words, the earliest repeating subsequence start at $i=0$, i.e. the repeating subsequence starts with the pair $(x_{0} = 0, x_{1} = 0)$. Hence the ordered triple $(0, 1, 1)$ too must repeat in the original sequence, thereby proving that there exists $k \geq 1$ such that $x_{k} = 0, x_{k+1} = 1, x_{k+2} = 1$

Note: More precisely, $k \geq 2$ since we need $x_{k} = 0$ whereas $x_{1} = 1 \neq 0$

I don't get it. $t$ is irrelevant to the question. We already have a determined subsequence $(0,1,1)$ and it doesn't matter where it is. The $j$ is the crucial point since the problem statement requires $j=1,$ i.e. that we actually get $(0,1,1)$ as subsequence and not $(0,\ldots,1,\ldots,1)$.

You must lead $j>1$ to a contradiction, not $t>1$.

 

[Not anonymous]

I don't get it. $t$ is irrelevant to the question. We already have a determined subsequence $(0,1,1)$ and it doesn't matter where it is. The $j$ is the crucial point since the problem statement requires $j=1,$ i.e. that we actually get $(0,1,1)$ as subsequence and not $(0,\ldots,1,\ldots,1)$.

You must lead $j>1$ to a contradiction, not $t>1$.

Sorry, I don't understand why $t$ is not relevant to the proof. The question essentially asks whether the contiguous subsequence $(0, 1, 1)$ occurs again in the full sequence. And I attempted to prove that by showing that the sequence can be viewed as consisting purely of a repeating subsequence (I consider only subsequences of length at least 2, hence $q >0$, and $j$ is the value of $q$ for the specific case of $p = t$) after some offset (at this point, the proof leaves open the possibility of the original sequence consisting of a prefix subsequence that does not repeat itself) and and then showing that $t=0$ is the smallest offset of such a repeating subsequence. And $t=0$ implies that the ordered contiguous pair $(0, 1)$ occurs again.

We already have a determined subsequence $(0,1,1)$ and it doesn't matter where it is

There is specifically a mention of $k \geq 1$ in the question, meaning we a looking for a repetition of $(0, 1, 1)$ after the subsequence starting $(x_{0}, x_{1}, x_{2})$. So shouldn't the proof show that $(x_{i}=0, x_{i+1}=1)$ occurs (again) for some $i > 0$?

The $j$ is the crucial point since the problem statement requires $j=1,$ i.e. that we actually get $(0,1,1)$ as subsequence and not $(0,\ldots,1,\ldots,1)$.

You must lead $j>1$ to a contradiction, not $t>1$.

There seems to be some confusion regarding how the $j$ in my proof was interpreted. I don't get why you have mentioned $(0,\ldots,1,\ldots,1)$. The proof only talks about subsequences of contiguous elements, so it only attempts to show repetition of $(0, 1, 1)$, not $(0,\ldots,1,\ldots,1)$ with gaps between the indices of those 3 elements.

I have already stated in the proof (though indirectly, via $q$) that only consider subsequences where $j \geq 1$. But I don't require $j$ to be strictly equal to 1, nor do I attempt to prove it since I think it does not matter. Suppose $m = 2$, then the full sequence would be of the form $(0, 1, 1, 0, 1, 1, 0, 1, 1, \ldots)$. We can view this as a repeating subsequence in many different ways, for e.g. as a repetition of $(0, 1, 1)$ (3 elements) or of $(0, 1, 1, 0, 1, 1)$ (6 elements), or of $(0, 1, 1, 0, 1, 1, 0, 1, 1)$ (9 elements), and so on. Or it can be vied as $(0,1)$ followed by the repeating srquence $(1,0,1)$. Or as $(0,1,1,0)$ followed by the repeating sequence $(1,1,0)$. And so on other possibilities. In the proof, I don't impose a constraint to consider only the shortest repeating subsequence. But whatever be the length of the repeating subsequence, I show that the earliest repeating subsequence is a subsequence that starts with the pair $(x_{0}=0, x_{1}=1)$, thus proving that $(0, 1, 1)$ will occur again for some $k \geq 1$. This being the case, it is not clear why it is necessary to show that $j > 1$ leads to a contradiction. Please let me know if you need more clarification on some part of the proof or if I have got something wrong in the proof.

 

[Not anonymous]

A slightly different approach compared to my earlier answer for question 13. I hope the proof presented here will be clearer as it considers only repetition of pairs (instead of arbitrary length subsequences as in the earlier answer) in order to show the repetition of $(0, 1, 1)$.

Spoiler: Question 13

Consider all non-overlapping pairs of adjacent elements in the given sequence starting with $(x_{0}, x_{1})$, i.e. $\{(x_{i}, x_{i+1}) \, \forall i \in \{0, 2, 4, ...\}\}$. The number of pairs in the set if infinite, yet the number of distinct values a pair can take is finite - there are at most $m^{2}$ distinct pair values possible in any sequence consisting only of integers that are remainders of division by $m$. Thus there must be at least one pair, say $(x_{p}, x_{p+1}), p \in \{0, 2, 4, ...\}$, in the infinite pair set that repeats, i.e. there exists some $q \in \{2, 4, 6, ...\}$ such that $x_{p+q} = x_{p}$ and $x_{p+q+1} = x_{p+1}$.

Let $t$ be the smallest such value of $p$ for which $(x_{p}, x_{p+1})$ repeats as described above, and let $j$ be the corresponding value of $q$. Suppose $t \geq 2$. We note the following:

1. $x_{t+j+1} = x_{t+1} \Rightarrow (x_{t+j-1} + x_{t+j}) \mod m = (x_{t+j-1} + x_{t} \mod m = x_{t+1})$
(where $\mod m$ denotes remainder of division by $m$)

But since $x_{t+1} = (x_{t-1} + x_{t}) \mod {m}$, it follows that
$(x_{t-1} + x_{t}) \mod {m} = (x_{t+j-1} + x_{t}) \mod {m} \Rightarrow x_{t-1} \mod {m} = x_{t+j-1} \mod m$
$\Rightarrow x_{t+j-1} = x_{t-1}$ (Eq. 1)
where we use the fact that $x_{i} \mod m = x_{i}$ for all $x_{i}$'s in the sequence as they are all values from $\{0, 1, ..., m-1\}$.

2. $x_{t+j} = (x_{t+j-2} + x_{t+j-1}) \mod m \Rightarrow x_{t} = (x_{t+j-2} + x_{t+j-1}) \mod m$. Using (Eq. 1) and the fact that $x_{t+j} = x_{t}$ by definitions of $t, j$, we get $x_{t} = (x_{t+j-2} + x_{t-1}) \mod m$.

Since $x_{t}= (x_{t-2} + x_{t-1}) \mod m$ by definition of the original sequence, it follows that
$(x_{t+j-2} + x_{t-1}) \mod m = (x_{t-2} + x_{t-1}) \mod m \Rightarrow x_{t-2} \mod {m} = x_{t+j-2} \mod m$
$\Rightarrow x_{t+j-2} = x_{t-2}$ (Eq. 2)

From (Eq. 1) and (Eq. 2) it follows that $(x_{t-2}, x_{t-1})$ is also a repeating pair, i.e. a repeating pair is found at index $t-2$ that is smaller than $t$, contradicting the assumption that $t \geq 2$ is the smallest start index of a repeating pair. Hence we must have $t=0$ as the smallest start index of repeating pairs in the sequence, i.e. the pair value $(x_{0}=0, x_{1}=1)$ must occur again in the sequence. Thus there must be some $k \geq 2$ such that $x_{k}=0, x_{k+1}=1$. By definition of the sequence, $x_{k+2} = (x_{k}+x_{k+1}) \mod m = 1$ (as $m > 1$), and hence it follows that there exists a $k \geq 2$ such that $x_{k}=0, x_{k+1}=1, x_{k+2}=1$

 

[fresh_42]

Sorry, I don't understand why $t$ is not relevant to the proof.

Let's make an example and set $m=7$. Then we get as $(x_0,x_1,x_2,\ldots)$
$$
0,1,1,2,3,5,1,6,0,6,6,5,4,2,6,1,0,1,1,\ldots
$$
and

Decide whether for every $m\geq 2$ there exists a natural number $k\geq 1,$ such that $x_{k+2}=1\, , \,x_{k+1}=1\, , \,x_k=0.$

asks for $k=16$, the subsequence $x_{16},x_{17},x_{18}.$However, you considered

By definition of how the sequence is formed and the pattern of repeating subsequences, we must have
$x_{t+1}= (x_{t-1} + x_{t}) \mod m$ (Eq. 1)
$x_{t+j+1} = (x_{t+j-1} + x_{t+j}) \mod m \Rightarrow x_{t} = (x_{t+j-1} + x_{t+j}) \mod m$ (Eq. 2)
$x_{t+j+2} = (x_{t+j} + x_{t+j+1}) \mod m \Rightarrow x_{t+1} = (x_{t+j} + x_{t}) \mod m$ (Eq. 3)

Now, if $t=15$, then you are looking for $x_{16},x_{16+j},x_{17+j}$ which is irrelevant for all cases $j>0$. However, you didn't consider $j$ at all, only $t$.

Next, if $t=7$ and $j=9$ we get the subsequence $(x_{t+1},x_{t+j+1},x_{t+j+2})=(x_{8},x_{17},x_{18})=(0,1,1)$ which is in the range of what you investigated. But it is not asked for, because we want to have three consecutive sequence members, i.e. $j=0$.

I will read your post #58 in a while, but I am a bit skeptical since you only consider pairs ...
... and used far too many indices $i,j,m,p,q,t.$ I will need a machete to get through.

Are you interested in a hint because you were on the right track with your original solution?

 

[Not anonymous]

asks for $k=16$, the subsequence $x_{16}, x_{17}, x_{18}$.

Yes, that is my understanding too.

Now, if $t=15$, then you are looking for $x_{16}, x_{16+j}, x_{17+j}$ which is irrelevant for all cases $j > 0$. However, you didn't consider $j$ at all, only $t$.

Not exactly. While it is true that I do consider only $j > 0$ (since I look for repeating subsequences of length at least 2), your interpretation of how I use $j$ is not the same as what I intended to convey. So if $t=15$, what the proof says is that the $(j+1)$-tuple $(x_{15}, x_{15+1}, ..., x_{15+j})$ is a subsequence that first occurs at starts at index 15 and repeats itself in the original sequence right after the end of the first occurrence, i.e. starting again at index (15+j+1), i.e. $x_{15+j+1} = x_{15}, x_{16+j+1} = x_{16}, ..., x_{15+2j+1} = x_{15+j+1}$. Or more generally $x_{15+u(j+1)} = x_{15}, x_{16+u(j+1)} = x_{16}, ..., x_{15+j+u(j+1)} = x_{15} \forall u \in \{1, 2, ...\}$, though the generalization for $u > 1$ is not really required for the proof.