Challenge Math Challenge - February 2021

[fresh_42]

That's why I said "I don't get it" and not that it is wrong. And I still can't follow you. You have all the necessary ingredients but your conclusion is strange. You assume a minimal starting point $t$ for a repetition and found a contradiction unless $t=0,$ i.e. $(x_1,x_2,x_3)=(0,1,1)$ which is also wrong.

How can you conclude that

Hence the ordered triple $(0, 1, 1)$ too must repeat in the original sequence, thereby proving that there exists $k \geq 1$ such that $x_{k} = 0, x_{k+1} = 1, x_{k+2} = 1$

I can only see contradictions all over the place.

E.g.

... i.e. the repeating subsequence starts with the pair $(x_{0} = 0, x_{1} = 0)$.

is one contradiction. If we had two successive zeros, then we would be stuck on zeroes.

If you use your method without the minimality requirement you can show that
$$
x_{t+j+1} = x_{t}, x_{t+j+2} = x_{t+1}
$$
leads you to
$$
x_{t+j}=x_t
$$
which you can repeat until
$$
0=x_0=x_t\; , \;1=x_1=x_{t+1}\; , \;1=x_2=x_{t+2}
$$
I think you tried essentially this, just with a minimality argument instead of a repetition argument. But I think you got lost in your zoo of indexes. E.g. simply consider $x_t=x_{t+j}$ instead of $x_t=x_{t+j+1}$. This little shift makes the entire consideration hard to read.

I'll give you the credit because you have used all necessary steps: pigeon hole principle, modular arithmetic, iteration, and add my solution which I think is easier to read:

There are at most $m^3$ possible triplets $(x_t,x_{t+1},x_{t+2})_{t\in \mathbb{N}}\in \{0,1,\ldots,m-1\}^3$ so there have to be repetitions. Hence there are $t\geq 0,k\geq 1$ with
$$
x_t=x_{k+t}\, , \,x_{t+1}=x_{k+t+1}\, , \,x_{t+2}=x_{k+t+2}
$$
Assume $t>0.$ Per construction we know that $x_{t+1}\equiv x_t+x_{t-1} \mod m$ and $x_{k+t+1}\equiv x_{k+t}+x_{k+t-1} \mod m,$ hence subtraction yields
\begin{align*}
x_{t+1}-x_{k+t+1}&=0=x_t+x_{t-1}-x_{k+t}-x_{k+t-1} \equiv x_{t-1}-x_{k+t-1} \mod m \\
\Longrightarrow m\,&|\,(x_{t-1}-x_{k+t-1})
\end{align*}
which is only possible if $x_{t-1}=x_{k+t-1}$ for a number between $0$ and $m-1.$ We can repeat this argument until
$$
0=x_0=x_k\; , \;1=x_1=x_{k+1}\; , \;1=x_2=x_{k+2}
$$

 

[Not anonymous]

E.g.
... i.e. the repeating subsequence starts with the pair $(x_{0}=0, x_{1}=0)$.

is one contradiction. If we had two successive zeros, then we would be stuck on zeroes.

Sorry, there was a typo in that statement in the proof and that would have added to the confusion. I meant to say $(x_{0}=0, x_{1}=1)$ (and not $x_{1}=0$).

Here is a sequence or state diagram illustrating what the 1st proof means when it says "repeating subsequence". The backarrow from $x_{t+j}$ to $x_{t}$ indicates that the $(j+1)$-tuple $(x_{t}, .., x_{t+j})$, consisting entirely of consecutive elements, repeats itself to fill the original sequence. I had pictured the sequence in mind in this diagrammatic way while arriving at 1st proof.

Using pigeonhole principle, the proof said that such a repeating subsequence must exist. And by proving (or rather by trying to prove, if the proof is indeed wrong for some reason even after correcting the above-mentioned typo) that $t=0$, it establishes that the pair-value $(0, 1)$ (again considering only a pair of adjacent elements) must occur more than once in the full sequence. And by definition of how the sequence is formed, if the adjacent elements pair $(0, 1)$ occurs at index pair $(k, k+1)$, then the 3-tuple $(0, 1, 1)$ occurs at the index triple $(k, k+1, k+2)$.

and used far too many indices $i, j, m, p, q, t$. I will need a machete to get through.

Sorry again for complicating the proof, but to me showing the recurrence of a pair of consecutive elements or of an arbitrarily long subsequence of consecutive elements came to mind more naturally first and hence I attempted to prove using those. I agree that proving the repetition of a triple instead could have made the proof (and also verifying if there are errors in the proof) easier to follow for others.

 

[Not anonymous]

Looks o.k.

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

Yes, the proof of modulo equivalence could have been given for just $n=1$ and then by showing how numbers obtained for other values of $n$ could be obtained by repeating the 1-digit cut-from-left, append-to-right operation.

Spoiler: Question 11

Say $a_{1994}a_{1993}..a_{0}$ is the decimal representation of $z$. Applying the cut and append operation with $n=1$ (i.e. cutting from the leftmost position and appending it to the right just one digit) on $z^{[1]}$ gives $a_{1992}a_{1991}..a_{1}a_{0}a_{1994}a_{1993}$ and this is the same as $z^{[2]}$. Repeating the same cut-and-append operation on $z^{[2]}$ gives $a_{1991}a_{1990}...a_{0}a_{1994}a_{1993}a_{1992}$ and this happens to be the same as $z^{[3]}$. We can generalize this and say that repeating $k$ times the cut-and-append with $n=1$ on successive numbers from this transformation, starting with $z$, yields $z^{[k]}$.

Suppose $y$ is the rightmost digit of $z$. Then $z = 10x+y$ where $x$ is the number formed by removing the rightmost digit from $z$. It is easy to see that $z^{[1]} = 10^{1994}y + x$.

If $27 \mid z$, we have $(10x + y) \mod 27 = 0 \Rightarrow y \mod 27 = -10x \mod 27$ (Eq. 1)

$d = (z^{[1]} - z) \mod 27 = (10^{1994}y + x - 10x - y) \mod 27 = ((10^{1994}-1)y - 9x) \mod 27 =$
$(((10^{1994}-1) \mod 27)(y \mod 27) - (9x mod 27)) \mod 27 \Rightarrow$
$d = (9(1994 \mod 3)(y \mod 27) - (9x \mod 27)) \mod 27$ (Eq. 2)
Here, for $10^{1994} \mod 27$, I use an equivalence proved in my earlier proof for this question.

Applying (Eq. 1) on (Eq. 2) and simplifying gives:
$d = ((-180x \mod 27) - (9x \mod 27)) \mod 27 = -189x \mod 27 = 0$ (Eq. 3)
(since 27 divides 189)

(Eq. 3) implies that $27 \mid z^{[1]}$ if $27 \mid z$, and this applies to any 1995 digit positive integer. Hence since $z^{[2]} = (z^{[1]})^{[1]}$ and $z^{[2]}$ too is a 1995-digit integer, it follows that:
$27 \mid z \Rightarrow 27 \mid z^{[1]} \Rightarrow 27 \mid z^{[2]}$. This can be generalized to say that $27 \mid z \Rightarrow 27 \mid z^{[n]}$ for any integer $n$, $0 < n < 1995$.

Thanks for the hint. When I started on the problem, I did try to establish some sort of an equivalence in modulo between different numbers obtained by permuting digits in the given number, so I came somewhat close, but not close enough, to the more elegant approach which you have alluded to. Sometimes, we overlook simpler approaches thinking them to be complex and only a little help from an expert makes us see the easier path.

 

[fresh_42]

Yes, the proof of modulo equivalence could have been given for just $n=1$ and then by showing how numbers obtained for other values of $n$ could be obtained by repeating the 1-digit cut-from-left, append-to-right operation.

Spoiler: Question 11

Say $a_{1994}a_{1993}..a_{0}$ is the decimal representation of $z$. Applying the cut and append operation with $n=1$ (i.e. cutting from the leftmost position and appending it to the right just one digit) on $z^{[1]}$ gives $a_{1992}a_{1991}..a_{1}a_{0}a_{1994}a_{1993}$ and this is the same as $z^{[2]}$. Repeating the same cut-and-append operation on $z^{[2]}$ gives $a_{1991}a_{1990}...a_{0}a_{1994}a_{1993}a_{1992}$ and this happens to be the same as $z^{[3]}$. We can generalize this and say that repeating $k$ times the cut-and-append with $n=1$ on successive numbers from this transformation, starting with $z$, yields $z^{[k]}$.

Suppose $y$ is the rightmost digit of $z$. Then $z = 10x+y$ where $x$ is the number formed by removing the rightmost digit from $z$. It is easy to see that $z^{[1]} = 10^{1994}y + x$.

If $27 \mid z$, we have $(10x + y) \mod 27 = 0 \Rightarrow y \mod 27 = -10x \mod 27$ (Eq. 1)

$d = (z^{[1]} - z) \mod 27 = (10^{1994}y + x - 10x - y) \mod 27 = ((10^{1994}-1)y - 9x) \mod 27 =$
$(((10^{1994}-1) \mod 27)(y \mod 27) - (9x mod 27)) \mod 27 \Rightarrow$
$d = (9(1994 \mod 3)(y \mod 27) - (9x \mod 27)) \mod 27$ (Eq. 2)
Here, for $10^{1994} \mod 27$, I use an equivalence proved in my earlier proof for this question.

Applying (Eq. 1) on (Eq. 2) and simplifying gives:
$d = ((-180x \mod 27) - (9x \mod 27)) \mod 27 = -189x \mod 27 = 0$ (Eq. 3)
(since 27 divides 189)

(Eq. 3) implies that $27 \mid z^{[1]}$ if $27 \mid z$, and this applies to any 1995 digit positive integer. Hence since $z^{[2]} = (z^{[1]})^{[1]}$ and $z^{[2]}$ too is a 1995-digit integer, it follows that:
$27 \mid z \Rightarrow 27 \mid z^{[1]} \Rightarrow 27 \mid z^{[2]}$. This can be generalized to say that $27 \mid z \Rightarrow 27 \mid z^{[n]}$ for any integer $n$, $0 < n < 1995$.

Thanks for the hint. When I started on the problem, I did try to establish some sort of an equivalence in modulo between different numbers obtained by permuting digits in the given number, so I came somewhat close, but not close enough, to the more elegant approach which you have alluded to. Sometimes, we overlook simpler approaches thinking them to be complex and only a little help from an expert makes us see the easier path.

The "correct" way to write the modular arithmetic is
$$
10z-z^{[1]}=y\cdot(10^{1995}-1)=y\cdot (1000^{665}-1)
$$
From $1000=37 \cdot 27+1$ we get $1000^{665}\equiv 1^{665}\equiv 1 \mod 27.$ Hence $27\,|\,(10z-z^{[1]})$ and if $27\,|\,z$ then $27\,|\,z^{[1]}=10z-(10z-z^{[1]}).$

 

[Not anonymous]

Spoiler: 15a

We need $\sqrt{x} + \sqrt{y} =1993$ with $x, y$ being non-negative integers.

$\sqrt{x} = 1993 - \sqrt{y} \Rightarrow x = 1993^2 + y - 3986\sqrt{y} \Rightarrow \sqrt{y} = \dfrac {1993^2 + y - x} {3998}$

The RHS of the final equation above is a rational number whereas the LHS is the square root of a non-negative integer. It should be easy to show that the square root of a non-negative integer cannot be a rational number that is not also an integer ($\sqrt{y}$ will be an integer if $y$ is a perfect square, will be an irrational number otherwise). Hence it must be the case that $\sqrt{y}$ is also an integer. In the same way, we can show that $\sqrt{x}$ must also be an integer.

Therefore the possible values of $\sqrt{x}$ for the solution are $\{0, 1, ..., 1993\}$ and once the value of $\sqrt{x}$ is fixed, only one value is possible for each of $x$ and $y$. The possible solutions for the equation in question are therefore $(x, y)$ pairs from $\{(0^2, 1993^2), (1^2, 1992^2), (2^2, 1991^2), ..., (1993^2, 0^2)\}$ and so the number of pairs is 1994.

 

[fresh_42]

We need $\sqrt{x} + \sqrt{y} =1993$ with $x, y$ being non-negative integers.

$\sqrt{x} = 1993 - \sqrt{y} \Rightarrow x = 1993^2 + y - 3986\sqrt{y} \Rightarrow \sqrt{y} = \dfrac {1993^2 + y - x} {3998}$

The RHS of the final equation above is a rational number whereas the LHS is the square root of a non-negative integer. It should be easy to show that the square root of a non-negative integer cannot be a rational number that is not also an integer ($\sqrt{y}$ will be an integer if $y$ is a perfect square, will be an irrational number otherwise).

Yes, by the use of the fundamental theorem of arithmetic, i.e. by observing that squares have an even number of each prime.

Hence it must be the case that $\sqrt{y}$ is also an integer. In the same way, we can show that $\sqrt{x}$ must also be an integer.

Therefore the possible values of $\sqrt{x}$ for the solution are $\{0, 1, ..., 1993\}$ and once the value of $\sqrt{x}$ is fixed, only one value is possible for each of $x$ and $y$. The possible solutions for the equation in question are therefore $(x, y)$ pairs from $\{(0^2, 1993^2), (1^2, 1992^2), (2^2, 1991^2), ..., (1993^2, 0^2)\}$ and so the number of pairs is 1994.

 

[Not anonymous]

Does question 15(b) need an answer not expressed in terms of $(4+\sqrt{18})$? If yes, may I get some hint?

Spoiler: Question 15(b)

Let $m \in \mathbb{R}_{+}$ be the value for which $2^{m} = (4+\sqrt{18})^{n}$. Applying logarithm on both sides, we get $m \log {2} = n \log {(4+\sqrt{18})} \Rightarrow m = n \dfrac {\log {(4+\sqrt{18})}} {\log {2}}$. The largest integer power of 2 that divides $(4+\sqrt{18})^n$ must therefore be $\lfloor m \rfloor = \left\lfloor n \dfrac {\log {(4+\sqrt{18})}} {\log {2}} \right\rfloor \approx \lfloor 3.043106606 \times n \rfloor$

 

[fresh_42]

Does question 15(b) need an answer not expressed in terms of $(4+\sqrt{18})$? If yes, may I get some hint?

Spoiler: Question 15(b)

Let $m \in \mathbb{R}_{+}$ be the value for which $2^{m} = (4+\sqrt{18})^{n}$. Applying logarithm on both sides, we get $m \log {2} = n \log {(4+\sqrt{18})} \Rightarrow m = n \dfrac {\log {(4+\sqrt{18})}} {\log {2}}$. The largest integer power of 2 that divides $(4+\sqrt{18})^n$ must therefore be $\lfloor m \rfloor = \left\lfloor n \dfrac {\log {(4+\sqrt{18})}} {\log {2}} \right\rfloor \approx \lfloor 3.043106606 \times n \rfloor$

This one is rather tricky. E.g. for $n=3$ we get $\left[(4+\sqrt{18})^3\right]=560$ and $16\,|\,560$ so the solution in this case is $4.$

We are looking for a function $f:\mathbb{N}\longrightarrow \mathbb{N}$ such that $2^{f(n)}\,|\,\left[(4+\sqrt{18})^n\right]$ and where $f(n)$ can be calculated precisely and without calculator.

The standard approach if we see an expression like $a_n:=(4+\sqrt{18})^n$ is to consider $b_n:=(4-\sqrt{18})^n$ and get rid of what disturbes, namely $\sqrt{18}.$ In our case we define $c_n:=a_n+b_n\,,$ develop a recursion for $c_n$ and investigate which powers of $2$ divide $c_n.$ This statement can be proven by induction, and modular arithmetic yields the desired result.