How many four-digit numbers $\overline{abba}$ are there such that $\overline{abba}$ is divisible by $\overline{aa}$?
Or,
Mursalin takes a four-digit palindromic number $n$ and deletes its middle two digits to obtain a two-digit number $m$. If $\frac{n}{m}$ is an integer, how many possible choices for $n$ are there?
BdMO Regional 2021 Secondary P7
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
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Re: BdMO Regional 2021 Secondary P7
If we assume $a$ can be equal to $b$ in the first variant then both are actually the same question. Now let's solve them.Pro_GRMR wrote: ↑Mon Mar 29, 2021 11:51 pmHow many four-digit numbers $\overline{abba}$ are there such that $\overline{abba}$ is divisible by $\overline{aa}$?
Or,
Mursalin takes a four-digit palindromic number $n$ and deletes its middle two digits to obtain a two-digit number $m$. If $\frac{n}{m}$ is an integer, how many possible choices for $n$ are there?
- We are given that:
$\overline{aa}$ $\vert$ $\overline{abba}$
$\Longleftrightarrow$ $a$ $\vert$ $\overline{abba}$ and $11$ $\vert$ $\overline{abba}$
$\Longleftrightarrow$ $a$ $\vert$ $\overline{abba}$
$\Longleftrightarrow$ $a$ $\vert$ $\overline{bb0}$
$\Longleftrightarrow$ $a$ $\vert$ $b*2*5*11$
We can do a case by case check:
$a \not= 0$
If $a = 1$ , $b\in\{0,1,2,3,4,5,6,7,8,9\}$
If $a = 2$ , $b\in\{0,1,2,3,4,5,6,7,8,9\}$
If $a = 3$ , $b\in\{0,3,6,9\}$
If $a = 4$ , $b\in\{0,2,4,6,8\}$
If $a = 5$ , $b\in\{0,1,2,3,4,5,6,7,8,9\}$
If $a = 6$ , $b\in\{0,3,6,9\}$
If $a = 7$ , $b\in\{0,7\}$
If $a = 8$ , $b\in\{0,4,8\}$
If $a = 9$ , $b\in\{0,9\}$
Adding them all up we get $\boxed{50}$ such palindromes
Here is a computational approach in C++:
Code: Select all
#include <bits/stdc++.h>
using namespace std;
bool ispalindrome(int i)
{
string str = to_string(i);
string rstr = str;
reverse(rstr.begin(), rstr.end());
if (rstr == str)
return true;
else
return false;
}
int main(void)
{
int count = 0, m;
for (int n = 1000; n <= 9999; n++)
{
if (ispalindrome(n)) // Checks if a number is palindrome
{
m = (n / 1000) * 11; // Makes a two-digit palindrome m out of the first digit of n
if (n % m == 0)
{
count++; // If m divides n add 1 to the tally
}
}
}
cout << count << endl;
}
"When you change the way you look at things, the things you look at change." - Max Planck