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Question Number 185728 by aba last updated on 26/Jan/23
(x−2013)!=6!7!  x=??
$$\left(\mathrm{x}−\mathrm{2013}\right)!=\mathrm{6}!\mathrm{7}! \\ $$$$\mathrm{x}=?? \\ $$
Answered by mr W last updated on 26/Jan/23
6!7!  =1×2×3×4×5×6×7!  =1×3×5×6×8!  =1×5×2×9!  =10!  (x−2013)!=6!7!=10!  x−2013=10  ⇒x=2023
$$\mathrm{6}!\mathrm{7}! \\ $$$$=\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}×\mathrm{5}×\mathrm{6}×\mathrm{7}! \\ $$$$=\mathrm{1}×\mathrm{3}×\mathrm{5}×\mathrm{6}×\mathrm{8}! \\ $$$$=\mathrm{1}×\mathrm{5}×\mathrm{2}×\mathrm{9}! \\ $$$$=\mathrm{10}! \\ $$$$\left({x}−\mathrm{2013}\right)!=\mathrm{6}!\mathrm{7}!=\mathrm{10}! \\ $$$${x}−\mathrm{2013}=\mathrm{10} \\ $$$$\Rightarrow{x}=\mathrm{2023} \\ $$
Commented by Frix last updated on 27/Jan/23
Is there another pair (m, n) with  n!(n+1)!=m!  except the trivial ones?  Or maybe n!(n+k)!=m!  [m, n, k ∈N]
$$\mathrm{Is}\:\mathrm{there}\:\mathrm{another}\:\mathrm{pair}\:\left({m},\:{n}\right)\:\mathrm{with} \\ $$$${n}!\left({n}+\mathrm{1}\right)!={m}! \\ $$$$\mathrm{except}\:\mathrm{the}\:\mathrm{trivial}\:\mathrm{ones}? \\ $$$$\mathrm{Or}\:\mathrm{maybe}\:{n}!\left({n}+{k}\right)!={m}! \\ $$$$\left[{m},\:{n},\:{k}\:\in\mathbb{N}\right] \\ $$

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