Below, we present a problem from the 12/31 Algebra and Number Theory set, along with 3 student submitted solution. You may vote up for the solutions that you think should be featured, and should vote down for those solutions that you think are wrong (voting is anonymous!). Also, feel free to make remarks about these solutions.

Diophantine Solutions How many ordered pairs of non-negative integers \( (a,b) \) are there such that \( 2a + 3b = 100 \)?

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All solutions may have LaTeX edits to make the math appear properly. The exposition is presented as is, and have not been edited.

\[ \mbox{Remarks from Calvin} \]

This is a simple problem at first glance. However, many solutions had difficult justifying that they obtained all the possible solutions (short of being told that their numerical answer was determined to be correct).

I agree that Solution A by Riccardo is the best written. It explains why \(b\) must be even, and hence there are no solutions for \(b\) odd. It then explains that with \(b=2c\), we have 1 solution for each \(c\) from 0 to 16.

Looking at Solution B, it doesn't provide any explanation of why the solution sets must follow an Arithmetic Progression, or why no other solutions are possible.

Solution C is extremely similar to A. However, it is not as clear, because it doesn't explain what the corresponding value of \(a\), or why those are the only possible solutions. The other issue I had with this, was the lack of a punctuation mark between \( d=2, a=0\). While it was typed in as < d = 2 a = 0 >, I can only read what is written, and not what I think he's trying to get at.

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## Comments

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TopNewestSolution A - We first see that \(b\) must be even, so we set \( b=2c\). Dividing by \(2\) we obtain \(a+3c=50\), that is \[a=50-3c\] Since \(a\ge0\) and \(c\ge0\), \(c\) can assume all the values between \(0\) and \(16\), i.e. \(17\) different values.

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[Solution edited out - Calvin]

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[Solution edited out - Calvin]

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[Solution edited out - Calvin]

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Solution C - \(3b\) must be even since \(2a=3b\) is even and \(2a\) for all integers \(a\). looking at the sequence \(3 \times 0, 3x2, ... , 3x30, 3x32\) we can show there are accompanying even numbers that will sum to 100. We can then use \(y(n)= a+(n-1)*d\) where \(d=2 a=0\) and \(y(n)=32\) on the sequence \(0,2,4,6...,30,32\) to find the number of terms n. \(n=17\)

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Solution B - \(a=2,b=32; a=5,b=30, a=8,b=28\) which proceed for \(a\) up to 50 maintaining A.P and for \(b=32, 30, 28, ....\) up to 0 maintaining A.P with common difference 2 in descending order. As \(a,b\) are non-negative integer we take smallest value of \(b\) is zero(0) and that of \(a=2\)

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