My answer is: Let $X=${$1,2,3,4,5,...,147$} and let for all $x,y\in\mathbb{N}$, $x=y\Leftrightarrow x-y\in 3X$. There are threee classes and such that one class hase size 30, onether 50. 
Can you check my answer?
Find an equivalence relation on $\mathbb{N}$ with exactly 3 classes and such that one class hase size 30, another 50.
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elementary-set-theory
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2that relation isn't an equivalence relation – 2017-01-21
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0Your relation isn't transitive. $1$ is related to $400$ and $400$ to $799$, but $1$ isn't related to $799$. – 2017-01-21
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0Yes. You are right. So, what should be answer? – 2017-01-21
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1Hint: You don't need any arithmetic at all. Just make the two classes you need and ... – 2017-01-21
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2Equivalence relations can be defined by arithmetical formulae and you can try to see what the equivalence classes become... Or you can go the other way and define whatever equivalence classes you want and make an equivalence relation from there. – 2017-01-21
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0@EthanBolker Can you give more clear hint? – 2017-01-21
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0@DavidP Is it eq. relation ? – 2017-01-21
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1Longer hint. Let the first set be $\{1,2, ..., 30\}$. Then let the second set be $\{100, 101, ... , ?\}$. – 2017-01-21
1 Answers
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Let $X_1=\{1,2,\cdots,50\},X_2=\{51,52,\cdots,80\}$ and $X_3=\{81,82,\ldots\}$. If we can make an equivalence relation with these three sets as the equivalence classes, then we're done. And we can: let the relation $\sim$ be defined by $x\sim y$ iff $x$ and $y$ belong to the same $X_i$.