Let $d=\gcd(n,10)$. Take $n=d\cdot\frac{10^{\varphi\left(\frac{n}{d}\right)}-1}{9}$. It's easy to see that this satisfies the problem's condition.

why does it have to be distinct primes

MarkBcc168 wrote: Let $d=\gcd(n,10)$. Take $n=d\cdot\frac{10^{\varphi\left(\frac{n}{d}\right)}-1}{9}$. It's easy to see that this satisfies the problem's condition. FTFY

Math-Ninja wrote: MarkBcc168 wrote: Let $d=\gcd(n,10)$. Take $n=d\cdot\frac{10^{\varphi\left(\frac{n}{d}\right)}-1}{9}$. It's easy to see that this satisfies the problem's condition. FTFY What does that stand for?

achen29 wrote: Math-Ninja wrote: MarkBcc168 wrote: Let $d=\gcd(n,10)$. Take $n=d\cdot\frac{10^{\varphi\left(\frac{n}{d}\right)}-1}{9}$. It's easy to see that this satisfies the problem's condition. FTFY What does that stand for? $\textrm{FTFY=Fixed that for you}$.

ythomashu wrote: why does it have to be distinct primes because n=25 etc doesn't work

ythomashu wrote: ythomashu wrote: why does it have to be distinct primes because n=25 etc doesn't work You just answered your self...

cjquines0 wrote: A palindrome is a number that remains the same when its digits are reversed. Let $n$ be a product of distinct primes not divisible by $10$. Prove that infinitely many multiples of $n$ are palindromes. Let $n=\prod_{i=1}^{k}{p_i}$ where $p_1,p_2,...,p_k$ are distinct primes. Then $$10^{\ell \prod_{i=1}^{k}{(p_i-1)}}-1$$is a multiple of $n$ that is also a palindrome for all $\ell \in \mathbb{Z}^+$.