# Number theory problem involving factorials

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

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

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).

Hint:

Multiply both sides by $$2!$$

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

$$\implies a_2=1$$

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

$$\implies a_3=?$$

and so on