2
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Please don't use pictures. What have you tried? $\endgroup$ – Dietrich Burde Oct 15 '19 at 14:51
3
$\begingroup$

Putting everything over $7!$ gives:

$a_7+7a_6+42a_5+210a_4+840a_3+2520a_2=5\cdot720=3600$

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).

$\endgroup$
1
$\begingroup$

Hint:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.