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# Practice Paper 4 Question 15

In base 10, or decimal, we use the digits from 0 to 9 to represent any positive integer. In base 2, or binary, we use the digits from 0 to 1 to do the same. An integer is called a repunit (repeated unit) if it can be written in some base using only the digit 1. Find all $$n,p>0$$ 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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## Warm-up Questions

1. What is the value of $$2222_3$$ in base $$10?$$
2. Let $$a$$ be a number in base $$b$$ with digits $$a_na_{n-1}\ldots a_0.$$ Show that $$a \equiv a_0 \mod b.$$

## Hints

• Hint 1
Which values of $$n$$ and $$p$$ work and which don't?
• Hint 2
Are there any constraints relating $$n$$ and $$p?$$
• Hint 3
More specifically, pay attention to the number of digits used to express a number in base $$2$$ compared to in base $$10.$$
• Hint 4
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 $$p?$$
• Hint 5
Are any other solutions? Could you prove your hypothesis?
• Hint 6
How can you mathematically express that "a binary repunit with $$n$$ digits is equal in value to a decimal repunit with $$p$$ digits"?
• Hint 7
... as an equality of two geometric series?
• Hint 8
Have you tried simplifying and rearranging the expression into a form which you can apply your constraints to get a contradiction?

## Solution

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.

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