$n,p>0$ such that a binary repunit with $n$ digits is equal (in value) to a decimal repunit with $p$ digits. Prove your answer.
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- What is the value of
$2222_3$in base$10?$ - Let
$a$be a number in base$b$with digits$a_na_{n-1}\ldots a_0.$Show that$a \equiv a_0 \mod b.$
- What is the value of
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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?
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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?
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Have you tried simplifying and rearranging the expression into a form which you can apply your constraints to get a contradiction?
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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
$\sum_{i=0}^{n-1}2^i=\sum_{i=0}^{p-1}10^i,$which gives$2^n-1=$${(10^p-1) \over 9}.$We can rearrange this to get$9\cdot2^{n-p}-5^p=2^{3-p}.$Together with the above conditions,
$n>p$and$p>3,$we get a contradiction, since$2^{3-p}$is a rational less than 1 and$9\cdot2^{n-p}-5^p$is an integer. Therefore$n=p=1$is the only solution.