3
$\begingroup$

Prove that for every positive integer $n$, there exist a $n$ digit number, divisible by $5^n$, whose all digits are odd. for example,

for $n=1, 5$

$n=2,75$

$n=3, 375$.......

I have no idea how to solve it. please help.

$\endgroup$
  • $\begingroup$ I don't have an answer yet, but I notice that the solutions for increasing $n$ are all the same, except that each one has an additional digit on the front. For example, you have 5, then 75, then 375, and the next one is 9​​375. So there's probably some way to say that if $s_n$ is an $n$-digit solution, then you can choose $k$ from $1,3,5,7,9$ so that $k\cdot10^{n-1} + s_n$ is an $n+1$-digit solution. You already know that it has all odd digits, so the only matter is showing that it is divisible by $5^n$, and you automatically have that it is divisible by $5^{n-1}$. $\endgroup$ – MJD Jun 28 '14 at 13:48
3
$\begingroup$

We prove this by induction. You have already shown that it is true for $n=1,2,3$.

Suppose that for some specific $n\ge3$ we have an integer $a=5^nb$ such that $a$ has $n$ digits and every digit is odd. Consider the number $a'=a+10^nk$, where $1\le k<10$. Clearly this is a number of $n+1$ digits; all of them, except perhaps the first, are odd; we shall show that it is possible to choose $k$, also odd, such that $a'$ is a multiple of $5^{n+1}$.

To make $a'$ a multiple of $5^{n+1}$ we require $5^{n+1}\mid 5^nb+10^nk$, that is, $$5\mid b+2^nk\ .$$ Since $5$ and $2^n$ are coprime, this has a unique solution with $1\le k\le5$. If $k=2$ or $k=4$ we can replace it by $k=7$ or $k=9$ respectively; therefore we can always satisfy the divisibility condition with an odd value of $k$, and the proof is complete.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.