Question Number 180695 by depressiveshrek last updated on 15/Nov/22
$${Find}\:{all}\:{k},\:{m},\:{n}\:\in\:\mathbb{N}\:{such}\:{that} \\ $$$${k}!+\mathrm{3}^{{m}} =\mathrm{3}^{{n}} \\ $$
Answered by nikif99 last updated on 16/Nov/22
![Trying an answer. must be m<n and 3^n mod 3=0 ⇒ (k!+3^m ) mod 3=0 if k≥7 ⇒[6!×7×...+3^m )] mod 3=0 ⇒ (3^m mod 3=0) ∧ [(6!×7×...) mod 3]=0 2nd term is not valid ⇒ k∈{0..6}, so (k, m, n)= (2, 0, 1), (3, 1, 2), (4, 1, 3), (6, 2, 6)](https://www.tinkutara.com/question/Q180716.png)
$${Trying}\:{an}\:{answer}. \\ $$$${must}\:{be}\:{m}<{n}\:{and}\:\mathrm{3}^{{n}} \:{mod}\:\mathrm{3}=\mathrm{0}\:\Rightarrow \\ $$$$\left({k}!+\mathrm{3}^{{m}} \right)\:{mod}\:\mathrm{3}=\mathrm{0} \\ $$$$\left.{if}\:{k}\geqslant\mathrm{7}\:\Rightarrow\left[\mathrm{6}!×\mathrm{7}×…+\mathrm{3}^{{m}} \right)\right]\:{mod}\:\mathrm{3}=\mathrm{0}\:\Rightarrow \\ $$$$\left(\mathrm{3}^{{m}} \:{mod}\:\mathrm{3}=\mathrm{0}\right)\:\wedge\:\left[\left(\mathrm{6}!×\mathrm{7}×…\right)\:{mod}\:\mathrm{3}\right]=\mathrm{0} \\ $$$$\mathrm{2}{nd}\:{term}\:{is}\:{not}\:{valid}\:\Rightarrow \\ $$$${k}\in\left\{\mathrm{0}..\mathrm{6}\right\},\:{so}\:\left({k},\:{m},\:{n}\right)= \\ $$$$\left(\mathrm{2},\:\mathrm{0},\:\mathrm{1}\right),\:\left(\mathrm{3},\:\mathrm{1},\:\mathrm{2}\right),\:\left(\mathrm{4},\:\mathrm{1},\:\mathrm{3}\right),\:\left(\mathrm{6},\:\mathrm{2},\:\mathrm{6}\right) \\ $$
Commented by Frix last updated on 16/Nov/22
$$\mathrm{but}\:{k}!\mathrm{mod3}=\mathrm{0}\:\mathrm{for}\:{k}\geqslant\mathrm{3}\:\mathrm{because}\:\mathrm{obviously} \\ $$$$\mathrm{with}\:{k}\geqslant\mathrm{3}\:\mathrm{we}\:\mathrm{have}\:{k}!=\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}×….=\mathrm{3}{N} \\ $$$$\mathrm{do}\:\mathrm{I}\:\mathrm{get}\:\mathrm{something}\:\mathrm{wrong}? \\ $$
Commented by nikif99 last updated on 16/Nov/22
$${Yes},\:{you}\:{are}\:{right}.\:{It}\:{needs}\:{more} \\ $$$${examination}. \\ $$