Pro_GRMR wrote: ↑Mon Mar 29, 2021 11:51 pm
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?
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.
- We are given that:
$\overline{aa}$ $\vert$ $\overline{abba}$
$\Longleftrightarrow$ $a$ $\vert$ $\overline{abba}$ and $11$ $\vert$ $\overline{abba}$
$\Longleftrightarrow$ $a$ $\vert$ $\overline{abba}$
$\Longleftrightarrow$ $a$ $\vert$ $\overline{bb0}$
$\Longleftrightarrow$ $a$ $\vert$ $b*2*5*11$
Since $a$ is a digit $a\in\{0,1,2,3,4,5,6,7,8,9\}$
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;
}