The discussion revolves around finding the value of \( \log_3 (a_2 + a_3 + a_4 + a_5 + a_6 + a_7) \) given a specific equation involving factorials and integer variables \( a_2, a_3, a_4, a_5, a_6, a_7 \). The context includes mathematical reasoning and problem-solving related to the equation.
There is no consensus on the uniqueness of the solution without the proposed restriction. The discussion includes differing views on the conditions necessary for a unique solution.
The discussion highlights the dependence on the values of \( a_i \) and the implications of potential restrictions on these variables, which remain unresolved.
Albert said:$\dfrac {5}{7}=\dfrac {a_2}{2!}+\dfrac {a_3}{3!}+\dfrac {a_4}{4!}+\dfrac {a_5}{5!}+\dfrac {a_6}{6!}+\dfrac {a_7}{7!}$
here :$a_2,a_3,----,a_6,a_7\in Z$
pease find:$log_3 (a_2+a_3+a_4+a_5+a_6+a_7)=?$
jacks said:[hide]My Solution:: Multiply both side by $$7!\;,$$ we get
$$\displaystyle 3600 = 2520a_{2}+840a_{3}+210a_{4}+42a_{5}+7a_{6}+a_{7}$$
Now $$3600-a_{7}$$ is a multiply of $$7$$. So $$a_{1}=2$$
Thus $$\displaystyle \frac{3598}{7} = 514=360a_{2}+120a_{3}+30a_{4}+6a_{5}+a_{6}$$
So $$3598-a_{6}$$ is a multiple of $$6\;,$$ So $$a_{6} = 4$$
Thus $$\displaystyle \frac{510}{6}=85=60a_{2}+20a_{3}+5a_{4}+a_{5}$$
Thus $$85-a_{5}$$ is a multiple of $$5\;,$$ where $$a_{5}=0$$
Continuing in this process, we get $$a_{4}=1\;a_{3}=1\;,a_{2}=1$$
So we get $$a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}=1+1+1+0+4+2=9$$
So $$\log_{3}\left(a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}\right)=\log_{3}(9)=2$$[/hide]
yes you are rightkaliprasad said:above is a good solution but
$a_7=9$ and $a_6=3$ is also a solution so some attribute of problem is missing