$n,p>0$
such that a binary repunit with $n$
digits is equal (in value) to a decimal repunit with $p$
digits. Prove your answer.

 What is the value of
$2222_3$
in base$10?$
 Let
$a$
be a number in base$b$
with digits$a_na_{n1}\ldots a_0.$
Show that$a \equiv a_0 \mod b.$
 What is the value of

Which values of
$n$
and$p$
work and which don’t? 
Are there any constraints relating
$n$
and$p?$

More specifically, pay attention to the number of digits used to express a number in base
$2$
compared to in base$10.$

You should have found that
$n=p=1$
is a solution. If there are other solutions, could you try to establish a lower bound on$p?$

Are any other solutions? Could you prove your hypothesis?

How can you mathematically express that “a binary repunit with
$n$
digits is equal in value to a decimal repunit with$p$
digits”? 
… as an equality of two geometric series?

Have you tried simplifying and rearranging the expression into a form which you can apply your constraints to get a contradiction?

We can easily see that
$n=p=1$
works. What about other values? We can also see that$n>p,$
since any number greater than$1$
needs more binary digits than base$10$
digits to be represented.By inspection it’s also clear that
$p=2,3$
would not work since those are the decimal repunits$11$
and$111,$
neither of which are equal to any binary repunits. The closest are$111_2=8$
and$1111111_2=127.$
Therefore$p>3.$
The original condition can be written as
$\sum_{i=0}^{n1}2^i=\sum_{i=0}^{p1}10^i,$
which gives$2^n1=$
${(10^p1) \over 9}.$
We can rearrange this to get$9\cdot2^{np}5^p=2^{3p}.$
Together with the above conditions,
$n>p$
and$p>3,$
we get a contradiction, since$2^{3p}$
is a rational less than 1 and$9\cdot2^{np}5^p$
is an integer. Therefore$n=p=1$
is the only solution.