2026/043) Show that there exists an n digit number each digit being odd and the number is divisible by $5^n$

We shall prove the same  by construction. But before that set us try to understand the pattern

One digit number divisible by 5there is only one number 5 and the digit is odd 

2 digit number divisible by $5^2=25$ the numbers are  25,50,75 and 75 has both digits odd 

3 digit number divisible by $5^3=125$ the numbers are  125,250,375 and so on 375 has all there digits odd

4 digit number divisible by $5^4= 625$ I am not enumerating  and a number 9375

We shall use this as a basis for construction of number by induction we shall expand the number from n digits to n+1 digits by adding a a digit to the left.

Let there be an n digit number with all n digits odd and divisible by  $5^n$ and let it be $k*5^n$. Note that k has to be odd else digit in unit place shall be zero which i even.

Now we know   that $10^n$ is divisible by $5^n$

So adding $p *10^n$ we can convert the n digit number to n+1 digit number and this is divisible by $5^n$.

We have n+1 digit number $p * 10^n  + k * 5^n= (p *2 ^n + k) 5^n$

Now we require and do we have $p *2^n + k$ divisible by 5

That is $p * 2^n \equiv -k \pmod 5$

As 3 is multiplicative inverse of 2 we get

$p  \equiv -k * 3^n \pmod 5$

p cannot be zero as $gcd(3,5) = 1$

So p is 1 or 2 or 3 or 5

If p is 1 or 3 then we are done

If p is 2 or 4 add 5 to p to get p single digit and odd