Take number like $\{112, 111122, 1111111111222,\ldots\}$ where the number comprises of $1$s at the beginning followed by a string of $2$s. The $n$th number would have $2^{n+1}-2n$ ones and $n$ twos. The sum of the digits is $2^{n+1}$ and the product of the digits is $2^n$.

iarnab_kundu wrote: Take number like $\{112, 111122, 1111111111222,\ldots\}$ where the number comprises of $1$s at the beginning followed by a string of $2$s. The $n$th number would have $2^{n+1}-2n$ ones and $n$ twos. The sum of the digits is $2^{n+1}$ and the product of the digits is $2^n$. But $S(a_n)=P(a_1)$...

The basic idea was right, just the implementation was wrong. Denote $N(a,b) = \underbrace{22\ldots 2}_{a \textrm{ times}}\underbrace{11\ldots 1}_{b \textrm{ times}}$, so $S(N(a,b)) = 2a+b$ and $P(N(a,b)) = 2^a$. For $n=1$ take $a_1=1$. For $n>1$ take some $m$ such that $2^m \geq m+n$; then take $a_1 = N(2^m,0)$, $a_k = N(m+k-1, 2^{m+k} - 2(m+k-1))$ for $2\leq k\leq n-1$, finally $a_n = N(m+n-1, 2^{2^m} - 2(m+n-1))$. Thus $S(a_1) = 2^{m+1}$ and $P(a_1) = 2^{2^m}$, $S(a_k) = 2^{m+k}$ and $P(a_k) = 2^{m+k-1}$ for $2\leq k\leq n-1$, finally $S(a_n) = 2^{2^m}$ and $P(a_n) = 2^{m+n-1}$.

I believe that the basic idea is to "artificially" make the product increase by a lot and then just add a lot of 1s to fix the sum. There are *many* ways to construct this. What do you think about this year's APMO compared to that of other years?

P4 and P5 was much more difficult.(may be 4th problem left less time for 5th problem)

How about the first 3 problems? I think that this year's Problem 1 is the hardest since 2011 (it is not straightforward to those who are experienced unlike 2011, 2012, 2013). Problem 2 is harder than all the other years since 2011, and Problem 3 is around same level with 2012 and harder than 2013 & 2011. Cutoff predictions for Bronze, Silver, Gold (not the one per country because of the award limit)? I'm expecting the mean to be around 12 with a significant STDEV, so I'm expecting something like 9, 15, 21 or 10, 16, 22 (same as 2011)

Yay! at least I got problem 1 correct but yeah I thought this year was harder than last year. I was able to do problem 1 and 2 without too much difficulty in 2013, but this year, the first two problems were not too easy. Hoping for an Honourable mention at least, although I know I have very little chance at the moment.

in Mavropenevma's solution we can replace 2 by any positive integer r greater than 2

Consider a sequence $(a_i)$, such that $a_i=\underbrace{22\ldots 2}_{i}\underbrace{11\ldots 1}_{2^{i+1}-2i}$, for every $i$. We get that $S(a_i)=2^{i+1}$ and $P(a_i)=2^i$. Therefore $S(a_1)<S(a_2)<\ldots$ and $P(a_i)=2^i=S(a_{i-1})$, hence the sequence satisfies conditions of given problem.

Solution: We can select sufficiently large positive integer $t$ such that $2^{t+1}>2(t+n)$ for fix $n.$ Then select $a_i=\underbrace{22...2}_{t+i-1}\underbrace{11...1}_{2^{t+i}-2(t+i-1)},$ where $i\not= 1,$ and for $i=1, a_1=\underbrace{22...2}_{t+n}\underbrace{11...1}_{2^{t+1}-2(t+n)}.$ Then we can see easily $S(a_1)<S(a_2)<...,S(a_n)$ and $S(a_i)=P(a_{i+1}).$

Good solution

We can do a algorithm: We are going to start with a induction: 1º) First case:$n=2\rightarrow S(a_1)=P(a_2)$, so, lets take $a_1=1$, so $P(a_2)=1$, then, a possible example would be $a_2=11$. 2º) Lets supoose that we can do what the problem says for $k$ numbers, $k \in \mathbb{N}$ 3º) Lets prove that it works for $k+1$ numbers So, lets take the same sequence from $a_1, a_2, \cdots, a_k$ and we are going to insert $a_{k+1}$ in the sequence with the following algorithm: 1- We have $S(a_k)=P(a_{k+1})$, so, lets take first $a_{k+1}=\bar{c_1}\bar{c_2}$ where $c_1c_2=S(a_k)$ 2- If $S(a_{k+1})>S(a_k)$, then we are done 3- Else, lets take now $a_{k+1}=\bar{c_1}\bar{c_2}\bar{c_3}\cdots \bar{c_t}\bar{1}$, then se sums go up and the product stay the same(See that $c_3= c_4 =\cdots =c_t = 1$) 4- Repeat the steps 2 and 3 until $S(a_{k+1})>S(a_k)$ Just one thing, lets take $\bar{a}\bar{b}=10a+b$ because i dont know hou to put a bar up of two numbers

Basically the same solutions as above except with powers of $2$ replaced with powers of $3$. Choose a sufficiently large $k_1$ such that $3(k_1+n-1)\leq 3^{k_1}$. Then consider the string of numbers $3^{k_1}, 3^{k_1+1}, \cdots, 3^{k_1+n-1}$. For brevity, let $s_k$ denote the $k$th element in this string, and let $s_0=s_n$. For $1\leq i\leq n,$ let $a_i$ be a number which has sum $s_i$ and product $s_{i-1}$, and which takes the form of $33\cdots 300\cdots 00$. Clearly, for $2\leq i\leq n,$ such a number $a_i$ exists because you can just make the first $k_1+i-1$ digits equal to $3$ and digits after that equal to $0$. It is also easy to see that $i=1$ exists because of the condition we imposed on $k_1$. It is easy to see that the $a_i$ generated satisfy the conditions. Remarks: Another one of those weird construction combo problems. Took me an embarrassingly long time to get the idea of adding a bunch of 1s, but at least it worked.

Exist $n$ numbers. $x_2 < x_3< … < x_n < x_1 < 9x_1 < 9^{x_2}$ $y_i = 9^(x_{i+1}) – 9x_i$ $a_i = 99…9111…1$ $9 \implies x_i$ and $1 \implies y_i$

Let $a$ be some sufficiently large constant such that $2^{a+2}\gg2n+2a+1002$ and $a_i=\underbrace{11\ldots1}_{2^{a+i+1}-2a-2i\text{ times}}\underbrace{22\ldots2}_{a+i\text{ times}}$ for $2\le i\le n$ while $a_1=\underbrace{11\ldots1}_{2^{a+2}-2n-2a-2\text{ times}}\underbrace{22\ldots2}_{n+a+1\text{ times}}$. We can verify that this satisfies the conditions of the problem for any $n\in\mathbb N$.

Observe that we can increase $S(x)$ by adding digits $1$ as $P(x)$ doesn't change at all, furthermore, if $S(x)=p^a$ for some prime $p$, it is comparebly easier to construct such digits. So, choose smallest prime $2$, we will construct numbers such that their digits are consists of $1,2$ only, and sum of digits equals to some power of $2$ i.e $S(x)=2^a$. The trickiest part is the $S(a_n)$ part. The condition is equivalent to $P(a_2)<P(a_3)<...<P(a_{n-1})<P(a_n)<P(a_1)$. Hence we should fix the greatest number to $P(a_1)$. Fixing all $P(a_i)$ fixes all $S(a_i)$. it is enough to show that we can construct $S$ by adding ones. Now, we fix $P(a_i)=2^{k-n-1+i}$ and let $P(a_1)=2^{k}$. But then $S(a_1)=2^{k-n+1},S(a_2)=2^{k-n+2}..., S(a_i)= 2^{k-n+i}$ and $S(a_n)=2^k$. We have a problem if number of $2$ exceeds our sum ( since smallest sum is number of $2$ $\cdot 2$, hence we can't add $1$.)i.e impossible to construct $S(x)$, which happens if $2(k-n-1+i)> 2^{k-n+i}$ which we can avoid by choosing arbitrarly bigger $k$. ( Another way to think is that the product increases faster than the sum , if we multiply "more", then we will have enough difference between multiplication and summation such that adding $1$s will make $S(a_i)$. Then, we can construct $S(a_i)$,hence we are done.

The construction is clear, just look at: $1111, 111122, 1111111111222$, upwards on powers of $2$.

Choose an integer \(a\) such that \(n - 1 \leq 2^{a - 1} - a\). Note that this is true for all sufficiently large \(a\). Then \begin{align*} a + n - 1 &\leq 2^{a - 1} \\ 2(a + n - 1) &\leq 2^a\text{.} \end{align*}Note that \(a \leq 2^a\) for all \(a\). Now, we choose \(a_1, a_2, \ldots, a_n\) such that \begin{align*} S(a_1) = \,\,&P(a_2) = 2^a \\ S(a_2) = \,\,&P(a_3) = 2^{a + 1} \\ S(a_3) = \,\,&P(a_4) = 2^{a + 2} \\ &\ \ \vdots \\ S(a_n) = \,\,&P(a_1) = 2^{a + n - 1}\text{.} \end{align*}We can achieve this by taking \(\log_2 P(a_i)\) twos and padding the number with \(S(a_i) - 2 \log_2 P(a_i)\) ones, for every \(a_i\). This is possible because \(2 \log_2 P(a_i) \leq S(a_i)\) for every \(a_i\): \begin{align*} 2 \log_2 2^{a + n - 1} &\leq 2^a \\ 2 \log_2 2^{a} &\leq 2^{a + 1} \\ 2 \log_2 2^{a + 1} &\leq 2^{a + 2} \\ &\vdots \\ 2 \log_2 2^{a + n - 2} &\leq 2^{a + n - 1} \end{align*}as desired.