Suppose $a_2,a_3,a_4,a_5,a_6,a_7$ are integers such that $$\frac57=\frac{a_2}{2!}+\frac{a_3}{3!}+\frac{a_4}{4!}+\frac{a_5}{5!}+\frac{a_6}{6!}+\frac{a_7}{7!},$$ where $0\leq a_j\lt j$ for $j=2,3,4,5,6,7.$ The sum $a_2+a_3+a_4+a_5+a_6+a_7$ is

A. $8$
B. $9$
C. $10$
D. $11$

I tried simplifying things but couldn't move further .when i checked the answer it used hit and trial.So is there a more general way (without using hit and trial) to do these questions. Any help will be appreciated

  • 1
    $\begingroup$ Please don't use pictures. What have you tried? $\endgroup$ – Dietrich Burde Oct 15 '19 at 14:51

Putting everything over $7!$ gives:


Then use a greedy algorithm to get $a_2=1, a_3=1, a_4=1, a_5=0, a_6=4, a_7=2$

gives $2520+840+210+28+2=3600$

so the answer is B (9).



Multiply both sides by $2!$

to find $10/7=a_2+$ some proper fraction as $a_j<j$

$\implies a_2=1$

$3(10/7-1)=a_3+$ some proper fraction

$\implies a_3=?$

and so on


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