This week, we learn about Modular Arithmetic, a system of arithmetic for the integers with many applications.

How would you use Modular Arithmetic to solve the following?

In the years 1600-1999, how many times was the first of January celebrated on a Sunday?

Share a question which can easily be approached by Modular Arithmetic.

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## Comments

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TopNewestIs there a positive integer \(n\) for which \(n^7 - 77\) is a Fibonacci number?

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\(n^7 \equiv 0, 1, 12, 17, 28 \pmod {29}\)

\(n^7 - 77 \equiv 9, 10, 11, 22, 27 \pmod {29}\)

The pisano period of \(29\) is \(1, 1, 2, 3, 5, 8, 13, 21, 5, 26, 2, 28, 1, 0\), none of the values are the same.

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Can you explain why you chose modulo 29? That seems like such a random number.

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Writing it this way takes away the mysticism of "why 29", and provides a path to approach the more general problem.

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What are the last three digits of \( 9^{9^9} \)?

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We have \(9^{50} \equiv 1\) (\(1000\)) and \(9^9 \equiv 39\) (\(50\)). Hence \(9^{9^9} \equiv 9^{39} \equiv 289\) (\(1000\)).

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Here are a few easy ones I know about modular arithmetic:

The fractional part of \(\frac {n}{7}\) is \(.142857\) repeating, or some variant thereof. What is the minimal positive integer N such that the fractional part of \(\frac{251}{7} \times n\) is \(.142857\) repeating?

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Sorry, I'm a new member, so I don't really know the formatting that well yet. We can start this problem by thinking about the number \(251\), which can be written as \(251 \equiv 6 \pmod{7}\). In order for \(\frac {251}{7} \times n\) to be \(0.142857\) repeating, \(251n \equiv 1 \pmod{7}\). Looking at the first congruence, we can start by realizing that \(251n\) will have a remainder of \(1\) iff \(6n \equiv 1 \pmod{7}\). We can see that the possible values for \(n\) will be of the form \(7k - 1\), where \(k\) is a positive integer. Therefore, the minimum possible value for \(n\) is \(6\).

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Your formatting is fine. Note, though, that you could simplify things a bit by making \(251 \equiv -1 \pmod 7\).

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Nice questions. I took the liberty of removing the answers and splitting out the questions, so that people can comment on them individually.

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Find all perfect squares such that they are divisible by 2 and not 4.

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All even numbers squared give an even number squared. An even number can be expressed by 2

n, where n is any integer. Thus, an even number squared can be expressed as 2^2n^2, which is 4n^2. Thus, no perfect squares are divisible by 2 and not 4.Log in to reply

nothing, it is impossible. Because, square root of 4 is 2, and 2 square is 4.

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Nothing, because the square root of 4 is 2 and 2 square is 4.

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What are the last three digits of \(2003^{2002^{2001}}\)?

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Reply to Calvin, 1. In a 400-year cycle, there are 58 years was the first of January celebrated on a Sunday.

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For the first question, are we assuming that leap years happen every 4 years, or are we following the leap year rules and omitting the leap year on 1700, 1800, and 1900?

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Ah, you got me. I was hoping that would fly beneath the radar.

To answer this question well, you have to accurately account for all of the leap years, and their effects on the calendar.

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But an even more trick question: are we supposed to accommodate for the 1752 calendar change?

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Not my fault that Britain (and her colonies) were late to the party.

FYI: The gregorian calendar was adopted at different points in time. Note that not everyone use the gregorian calendar.

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