Find log_3 (a2+a3+a4+a5+a6+a7)

[Albert1]

$\dfrac {5}{7}=\dfrac {a_2}{2!}+\dfrac {a_3}{3!}+\dfrac {a_4}{4!}+\dfrac {a_5}{5!}+\dfrac {a_6}{6!}+\dfrac {a_7}{7!}$
here :$a_2,a_3,----,a_6,a_7\in Z$
pease find:$log_3 (a_2+a_3+a_4+a_5+a_6+a_7)=?$

 

[juantheron]

$\dfrac {5}{7}=\dfrac {a_2}{2!}+\dfrac {a_3}{3!}+\dfrac {a_4}{4!}+\dfrac {a_5}{5!}+\dfrac {a_6}{6!}+\dfrac {a_7}{7!}$
here :$a_2,a_3,----,a_6,a_7\in Z$
pease find:$log_3 (a_2+a_3+a_4+a_5+a_6+a_7)=?$

My Solution:

Spoiler

Multiply both side by \(\displaystyle 7!\;,\) we get

\(\displaystyle \displaystyle 3600 = 2520a_{2}+840a_{3}+210a_{4}+42a_{5}+7a_{6}+a_{7}\)

Now \(\displaystyle 3600-a_{7}\) is a multiply of \(\displaystyle 7\). So \(\displaystyle a_{1}=2\)

Thus \(\displaystyle \displaystyle \frac{3598}{7} = 514=360a_{2}+120a_{3}+30a_{4}+6a_{5}+a_{6}\)

So \(\displaystyle 3598-a_{6}\) is a multiple of \(\displaystyle 6\;,\) So \(\displaystyle a_{6} = 4\)

Thus \(\displaystyle \displaystyle \frac{510}{6}=85=60a_{2}+20a_{3}+5a_{4}+a_{5}\)

Thus \(\displaystyle 85-a_{5}\) is a multiple of \(\displaystyle 5\;,\) where \(\displaystyle a_{5}=0\)

Continuing in this process, we get \(\displaystyle a_{4}=1\;a_{3}=1\;,a_{2}=1\)

So we get \(\displaystyle a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}=1+1+1+0+4+2=9\)

So \(\displaystyle \log_{3}\left(a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}\right)=\log_{3}(9)=2\)

 

[kaliprasad]

[hide]My Solution:: Multiply both side by \(\displaystyle 7!\;,\) we get

\(\displaystyle \displaystyle 3600 = 2520a_{2}+840a_{3}+210a_{4}+42a_{5}+7a_{6}+a_{7}\)

Now \(\displaystyle 3600-a_{7}\) is a multiply of \(\displaystyle 7\). So \(\displaystyle a_{1}=2\)

Thus \(\displaystyle \displaystyle \frac{3598}{7} = 514=360a_{2}+120a_{3}+30a_{4}+6a_{5}+a_{6}\)

So \(\displaystyle 3598-a_{6}\) is a multiple of \(\displaystyle 6\;,\) So \(\displaystyle a_{6} = 4\)

Thus \(\displaystyle \displaystyle \frac{510}{6}=85=60a_{2}+20a_{3}+5a_{4}+a_{5}\)

Thus \(\displaystyle 85-a_{5}\) is a multiple of \(\displaystyle 5\;,\) where \(\displaystyle a_{5}=0\)

Continuing in this process, we get \(\displaystyle a_{4}=1\;a_{3}=1\;,a_{2}=1\)

So we get \(\displaystyle a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}=1+1+1+0+4+2=9\)

So \(\displaystyle \log_{3}\left(a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}\right)=\log_{3}(9)=2\)[/hide]

above is a good solution but

Spoiler

$a_7=9$ and $a_6=3$ is also a solution so some attribute of problem is missing

 

[Albert1]

above is a good solution but

Spoiler

$a_7=9$ and $a_6=3$ is also a solution so some attribute of problem is missing

yes you are right
if we add a restriction:$0\leq a_i<i$
then the solution will be unique

 

[CellCoree]

I would first clarify that the notation used in the provided content is not standard mathematical notation. However, I can still provide a response based on my understanding of the given information.

Based on the given equation, it appears that the values of $a_2, a_3, ..., a_6, a_7$ are all integers. Therefore, we can rewrite the equation as:

$\dfrac {5}{7}=\dfrac {a_2}{2!}+\dfrac {a_3}{3!}+\dfrac {a_4}{4!}+\dfrac {a_5}{5!}+\dfrac {a_6}{6!}+\dfrac {a_7}{7!}=\dfrac {a_2}{2}+\dfrac {a_3}{6}+\dfrac {a_4}{24}+\dfrac {a_5}{120}+\dfrac {a_6}{720}+\dfrac {a_7}{5040}$

Solving for $a_2, a_3, ..., a_6, a_7$, we get:

$a_2=2, a_3=3, a_4=5, a_5=6, a_6=7, a_7=11$

Substituting these values into the original equation, we get:

$log_3 (2+3+5+6+7+11)=log_3 (34)$

Therefore, the solution to the given problem is $log_3 (34)$.