A. Basic Vocabulary

We assume WLOG that $$$b_1>b_2$$$. The number of digits in the base-$$$b_1$$$ representation of $$$n$$$ is therefore not greater than that in base-$$$b_2$$$. Hence, if a solution exists, $$$(n)_{b_1}$$$ must be a prefix of $$$(n)_{b_2}$$$. Here $$$(x)_b$$$ represents the base-$$$b$$$ representation of the integer $$$x$$$.

Let's start off by observing that any number $$$n$$$ when represented in base $$$b$$$, where $$$2b>n$$$ is of the form $$$[1,n-b]_b$$$. Considering $$$2b_1>n$$$, what must be the constraint on $$$b_2$$$ for $$$(n)_{b_1}$$$ to be a prefix?

Let $$$n-b_1=x$$$, then $$$n=[1,x]_{b_1}$$$. Note that $$$x>0$$$ since $$$b_1<n$$$.

Thus we have,

Thus, $$$(b_2+x)$$$ is the quotient obtained on dividing $$$n$$$ by $$$b_2^k$$$.

Now, we note that $$$1 \leq x \leq b_2-1$$$, since $$$x$$$ is a non-zero digit in base-$$$b_2$$$. Therefore,

If there exists any $$$b_2$$$, $$$k$$$ satisfying the above inequality, then we can derive $$$b_1$$$ as follows:

• $$$q \leftarrow \Big\lfloor\frac{n}{b_2^k}\Big\rfloor$$$
• $$$x \leftarrow q-b_2$$$
• $$$b_1 \leftarrow n-x$$$

We observe that the upper bound of $$$(1)$$$ strictly increases with $$$b_2$$$ for a given $$$k$$$. Therefore, it is optimal to choose the largest value of $$$b_2$$$ such that $$$(b_2+x)b_2^k \leq n$$$, which can be found in $$$\mathcal{O}(\log(\sqrt{n}))$$$ by Binary Search. Since $$$b_2 \geq 2$$$, we must test at most $$$\log_2{(n)}$$$ values of $$$k$$$.

Note that all this while, we are operating under the assumption that $$$2b_1<n$$$. Let's observe the case when $$$k=1$$$. We have,

Now, consider how we find $$$b_2$$$. We choose the largest value that satisfies the lower bound and check whether it satisfies the upper bound. What if the upper bound for a certain $$$b_2$$$ went beyond the lower bound for $$$b_2+1$$$? For visualisation, one could try imagining that for every $$$b_2$$$ there is a certain interval in which $$$n$$$ must lie for the inequality to hold. What happens when these intervals start to overlap? Then we'd be guaranteed a solution for any value of $$$n$$$.

which is true $$$\forall b_2 \geq 4$$$.

Therefore, we conclude that $$$\forall n \geq (4+1)(4)=20$$$, $$$\exists b_1,b_2$$$ such that $$$2b_1>n$$$ and $$$(1)$$$ holds for $$$k=1$$$.

Thus, we finally have our complete solution. For $$$n<20$$$ the trivial brute force algorithm is fast enough. For $$$n\geq 20$$$, we test solely for $$$k=1$$$ as described before and report the answer obtained.

/* 
 * author: twoplusthree 
 * created: 11.04.2024 20:24:53 
 */
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define all(x) (x).begin(),(x).end()
using vi=vector<int>;
void solve(){
  int n;
  cin>>n;
  if(n>=20){
    int b;
    int low,hig,mid;
    low=2;
    hig=1e9+5;
    while(low<=hig){
      mid=(low+hig)/2;
      if((mid+1)*mid<=n){
        b=mid;
        low=mid+1;
      }else{
        hig=mid-1;
      }
    }
    assert(2*b*b>n);
    int q=n/b;
    int x=q-b; assert(1<=x&&x<=b-1);
    cout<<(n-x)<<" "<<b<<"\n";
  }else{
    auto tobase=[](int x,int b)->vi{
      vi res;
      while(x>0){
        res.push_back(x%b);
        x/=b;
      }
      reverse(all(res));
      return res;
    };
    vi base[n];
    for(int i=2;i<n;i++){
      base[i]=tobase(n,i);
    }
    auto isprefix=[](vi& v1,vi& v2)->bool{
      int n=(int)v1.size();
      if((int)v2.size()<n){
        return false;
      }
      for(int i=0;i<n;i++){
        if(v1[i]!=v2[i]){
          return false;
        }
      }
      return true;
    };
    for(int i=2;i<n;i++){
      for(int j=2;j<i;j++){
        if(isprefix(base[i],base[j])){
          cout<<i<<" "<<j<<"\n";
          return;
        }
      }
    }
    cout<<-1<<"\n";
  }
  return;
}
signed main(){
  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  int tt=1;
  cin>>tt;
  for(int i=1;i<=tt;i++){
    //cout<<"Case #"<<i<<": ";
    solve();
  }
  return 0;
}

E. Parker Rectangle

Given $$$r+c=n \implies c=n-r$$$. We assume WLOG that $$$r\leq c \implies 0<r\leq \frac{n}{2}$$$.

Let $$$M_{r \times c}$$$ be the constructed matrix. Let $$$S$$$ be the sum of all elements in $$$M$$$. Let $$$R_i$$$ denote the sum of elements in the $$$i^{th}$$$ row and $$$C_j$$$ denote the sum of elements in the $$$j^{th}$$$ column of $$$M$$$.

For convenience, we consider the matrices to be $$$0$$$-indexed, that is, the first element of the first row of $$$M$$$ is denoted as $$$M_{00}$$$.

Claim: There exists no Parker Rectangle of odd order greater than $$$5$$$.

By inspection, it is trivial to construct Parker Rectangles of order $$$3$$$ and $$$5$$$, such as the ones below:

$$$M_3=\begin{bmatrix}0&1\end{bmatrix}$$$

$$$M_5=\begin{bmatrix}0&3&4 \cr 5&2&1\end{bmatrix}$$$

For $$$n=2m$$$, we use the following method of construction:

Let $$$r=c=m$$$. We define two $$$m \times m$$$ matrices $$$A$$$ and $$$B$$$ as follows:

For example, when $$$m=3$$$,

And, when $$$m=4$$$,

Claim: For any given $$$i,j$$$ the ordered pair $$$(A_{ij},B_{ij})$$$ is unique.

We now define an injection $$$g:\mathbb{Z}_m\times \mathbb{Z}_m \rightarrow \mathbb{Z}$$$ as $$$g(x,y)=\alpha x+y$$$ where $$$\alpha \geq m.$$$ For the purposes of this construction let us consider $$$\alpha = m$$$.

We then construct the Parker Rectangle $$$M_n$$$ as $$$M_{ij}=g(A_{ij},B_{ij})$$$.

For example, when $$$m=3$$$,

And, when $$$m=4$$$,

Proof of correctness

Note that by construction, each row in matrices $$$A$$$ and $$$B$$$ contains the numbers $$$0 \ ... \ m-1$$$ exactly once (can be proved using the Pigeonhole principle).

Hence $$$\forall i$$$,

When $$$m$$$ is odd, each column in $$$A$$$ and $$$B$$$ also contains contains the numbers $$$0 \ ... \ m-1$$$ exactly once. Thus $$$C_i=\frac{(m-1)m}{2}(\alpha +1) \ \forall i$$$, and so the maximum absolute difference $$$=0 \leq \lceil \frac{n}{2}\rceil$$$.

When m is even, the even-indexed columns in $$$B$$$ contain the even numbers $$$0,2,\ ... \ m-2$$$ repeated twice, and the odd-indexed columns contain the odd numbers $$$1,3, \ ... \ m-1$$$ repeated twice.

Hence,

And therefore,

Hence, the maximum absolute difference $$$=\frac{m^2}{2}-\frac{m(m-2)}{2}=m \leq \lceil \frac{n}{2}\rceil$$$.