[Calvin] Which solution will you feature? (3)

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) (a,b) are there such that 2a+3b=100 2a + 3b = 100 ?

You may try the problem by clicking on the above link.

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

Remarks from Calvin \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 bb must be even, and hence there are no solutions for bb odd. It then explains that with b=2cb=2c, we have 1 solution for each cc 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 aa, 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 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.

#FeaturedSolutions #Math

Note by Calvin Lin
8 years, 5 months ago

No vote yet
4 votes

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Comments

Solution A - We first see that bb must be even, so we set b=2c b=2c. Dividing by 22 we obtain a+3c=50a+3c=50, that is a=503ca=50-3c Since a0a\ge0 and c0c\ge0, cc can assume all the values between 00 and 1616, i.e. 1717 different values.

Calvin Lin Staff - 8 years, 5 months ago

Solution C - 3b3b must be even since 2a=3b2a=3b is even and 2a2a for all integers aa. looking at the sequence 3×0,3x2,...,3x30,3x323 \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+(n1)dy(n)= a+(n-1)*d where d=2a=0d=2 a=0 and y(n)=32y(n)=32 on the sequence 0,2,4,6...,30,320,2,4,6...,30,32 to find the number of terms n. n=17n=17

Calvin Lin Staff - 8 years, 5 months ago

[Solution edited out - Calvin]

Gurpreet Singh - 8 years, 5 months ago

[Solution edited out - Calvin]

Shivang Jindal - 8 years, 5 months ago

[Solution edited out - Calvin]

Zi Song Yeoh - 8 years, 5 months ago

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

Calvin Lin Staff - 8 years, 5 months ago
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