Question

Show that for every integer n there is a multiple of n that has only 0s and 1s in its decimal expansion. [duplicate]

Answer 1

To show there's a multiple of any integer n with only 0s and 1s in its decimal expansion, we apply the Pigeonhole Principle to numbers 1, 11, 111, etc., to find two with the same remainder when divided by n and subtract them to find the desired multiple.

Step-by-step explanation:

The student's question is asking to demonstrate that for any given integer n, there exists a multiple of n whose decimal representation is composed only of 0s and 1s. To showcase this, we will approach the problem using the Pigeonhole Principle. This principle states that if you have more pigeons than pigeonholes, then at least one pigeonhole must contain more than one pigeon. In this case, our pigeons are the numbers 1, 11, 111, etc., and our pigeonholes are the remainders when these numbers are divided by n.

Here is a step-by-step explanation:

1. Start by listing the numbers 1, 11, 111, etc., up to n+1 such numbers. These are numbers with each having one more 1 in their decimal expansion than the previous number.
2. Divide each of these numbers by n. According to the Pigeonhole Principle, at least two of these numbers must have the same remainder, as there are only n possible remainders (from 0 to n-1), but n+1 numbers.
3. Subtract the smaller number from the larger one where the remainders were equal. This difference is a multiple of n and will have only 0s and 1s in its decimal expansion.

By this logic, for any given n, you can always find a multiple that satisfies the condition of having only zeros and ones in its decimal representation.

Answer 2

To show that for every integer n there is a multiple of n that has only 0s and 1s in its decimal expansion, we can use the concept of integer powers.

Step-by-step explanation:

To show that for every integer n there is a multiple of n that has only 0s and 1s in its decimal expansion, we can use the concept of integer powers. For any positive integer n, let A = 111...1 (with n 1s in total). This number can be expressed as A = (10ⁿ - 1) / 9. Since n is an integer, 10ⁿ is a multiple of n. Therefore, (10ⁿ - 1) is also a multiple of n. Dividing by 9 doesn't change the fact that the number is a multiple of n, and the result has only 0s and 1s in its decimal expansion.