MHB 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 $$7!\;,$$ 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 $$\log_{3}\left(a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}\right)=\log_{3}(9)=2$$

 

[kaliprasad]

[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]

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