Direct proof divisibility: Suppose $x$ is an integer such that $2 \cdot 3 \cdot 4 \cdot 5 \cdot x = 59 \cdot 58 \cdot 57 \cdot 56 \cdot 55$

Question

Suppose $\,x\,$ is an integer such that $\,2 \cdot 3 \cdot 4 \cdot 5 \cdot x = 59 \cdot 58 \cdot 57 \cdot 56 \cdot 55.\,$ Does $\,59 \mid x$? Does $\,29 \mid x$? Does $\,118 \mid x$?

Answer 1

$ \require{cancel} \require{color} %\let\oldcdot\cdot \let\oldcancel\cancel %\renewcommand{\cdot}{\!\oldcdot\!} \renewcommand{\cancel}[1]{\!\color{blue}\oldcancel{\,\color{black}#1\,}} $ The answer is yes, numbers $\,59,$ $\,29,$ and $\,118\,$ all divide $\,x$.

From the equation $\,2 \cdot 3 \cdot 4 \cdot 5 \cdot x = 59 \cdot 58 \cdot 57 \cdot 56 \cdot 55.\,$ we conclude that $$ \begin{aligned} x &= \dfrac{ 59 \cdot 58 \cdot 57 \cdot 56 \cdot 55}{2 \cdot 3 \cdot 4 \cdot 5} = \dfrac{ 59 \cdot \left( 29\cdot \! \cancel{2} \right) \cdot \left( 19\cdot \! \cancel{3} \right) \cdot \left( 14\cdot \! \cancel{4} \right) \cdot \left( 11\cdot \! \cancel{5} \right) } {\cancel{2} \cdot \cancel{3} \cdot \cancel{4} \cdot \cancel{5}} =\\ &= 59 \cdot 29 \cdot 19 \cdot 14 \cdot 11 = 59 \cdot 29 \cdot 19 \cdot \left(7\cdot2\right) \cdot 11 \end{aligned} $$ Thus we have $$ \bbox[5pt, border:2.5pt solid #FF0000]{ x = 59 \cdot 29 \cdot 19 \cdot 11 \cdot 7 \cdot 2} $$ Since $\,118 = 59\cdot2,\,$ we see right away that $\,59 \mid x,\,$ $\,29 \mid x,\,$ and $\,118 \mid x.\,$

Answer 2

59 is a prime number. Since the right hand side is divisible by 59, so must the left hand side. And 2, 3, 4, 5 don't provide a factor divisible by 59. So x must be divisible by 59.

The other parts to the question follow on similarly by noticing that $58 = 2 × 29$, $118 = 2 × 59$, etc.