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73) How many rational numbers are in the opening of the binomial?


For the first time I solve such a question. My approach:

$$ \left\lbrace\begin{array}{ccccccl} (\sqrt3)^{2m}×(\sqrt[3]2)^{3n}\in \mathbb{Q} \\[1mm] 2m+3n=100 \end{array}\right.\Longrightarrow 2m+3n=100 \Longrightarrow 2m+6k=100 \Longrightarrow m=50-3k,\left\{k≥0,m≥0 \right\} \Longrightarrow k=0,1,2,3,..,16 \Longrightarrow 16+1=17 $$

Is this the correct way to find the answer?


Each term of $(a+b)^{100}$ is of the form $C(100,k)*a^{k}*b^{(100-k)}$. In this case, $k$ will have to be even and $100-k$ will have to be a multiple of 3.

So, $k = 2r$ and $100-k = 100-2r = 0 (\mod 3) ==> r = 2 (\mod 3)$

As $k$ varies from 0 to 100, $r$ varies from 0 to 50.

There are 17 such $r$ between 0 and 50 $(2 = 3*0+2, 5 = 3*1+2, ..., 50 = 3*16+2)$.

So, 17 of the 101 terms of the expansion are rational.


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