Prove that $$\frac{(3 a+3 b) !(2 a) !(3 b) !(2 b) !}{(2 a+3 b) !(a+2 b) !(a+b) ! a !(b !)^{2}}$$ is an integer for all pairs of positive integers $a, b$ (American Mathematical Monthly)
My work -
$ v_{p}((3 a+3 b) !(2 a) !(3 b) !(2 b) !)=\sum_{k \geq 1}\left(\left\lfloor\frac{3 a+3 b}{p^{k}}\right\rfloor+\left\lfloor\frac{2 a}{p^{k}}\right\rfloor+\left\lfloor\frac{3 b}{p^{k}}\right\rfloor+\left\lfloor\frac{2 b}{p^{k}}\right\rfloor\right) $
and
$ \begin{array}{l} v_{p}\left((2 a+3 b) !(a+2 b) !(a+b) ! a !(b !)^{2}\right) \\ \quad \quad=\sum_{k \geq 1}\left(\left\lfloor\frac{2 a+3 b}{p^{k}}\right\rfloor+\left\lfloor\frac{a+2 b}{p^{k}}\right\rfloor+\left\lfloor\frac{a+b}{p^{k}}\right\rfloor+\left\lfloor\frac{a}{p^{k}}\right\rfloor+2\left\lfloor\frac{b}{p^{k}}\right\rfloor\right) \end{array} $
now
With the substitution $x=\frac{a}{p^{k}}, y=\frac{b}{p^{k}},$ we have to prove that for any nonnegative real numbers $x, y$ we have $\lfloor 3 x+3 y\rfloor+\lfloor 2 x\rfloor+\lfloor 3 y\rfloor+\lfloor 2 y\rfloor \geq\lfloor 2 x+3 y\rfloor+\lfloor x+2 y\rfloor+\lfloor x+y\rfloor+\lfloor x\rfloor+2\lfloor y\rfloor$
I tried putting $\{x\}+\lfloor x\rfloor=x$ and $\{y\}+\lfloor y\rfloor=y$ and i get things in terms of fractional parts but i am not able to prove after that ....
thankyou
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Ishan
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