Suppose a_2, a_3, a_4, a_5, a_6, a_7 are inte | vedclass

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(B) Given $\frac{5}{7} = \frac{a_2}{2!} + \frac{a_3}{3!} + \frac{a_4}{4!} + \frac{a_5}{5!} + \frac{a_6}{6!} + \frac{a_7}{7!}$.Multiplying both sides b

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$9$

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(B) Given $\frac{5}{7} = \frac{a_2}{2!} + \frac{a_3}{3!} + \frac{a_4}{4!} + \frac{a_5}{5!} + \frac{a_6}{6!} + \frac{a_7}{7!}$.Multiplying both sides by $7! = 5040$,we get:$5 	imes \frac{5040}{7} = 5 	imes 720 = 3600$.So,$3600 = a_2 	imes \frac{5040}{2} + a_3 	imes \frac{5040}{6} + a_4 	imes \frac{5040}{24} + a_5 	imes \frac{5040}{120} + a_6 	imes \frac{5040}{720} + a_7$.$3600 = 2520 a_2 + 840 a_3 + 210 a_4 + 42 a_5 + 7 a_6 + a_7$.Since $0 \leq a_j $a_2 = 1$ ($2520 	imes 1 = 2520  3600$).$3600 - 2520 = 1080$.$a_3 = 1$ ($840 	imes 1 = 840  1080$).$1080 - 840 = 240$.$a_4 = 1$ ($210 	imes 1 = 210  240$).$240 - 210 = 30$.$a_5 = 0$ ($42 	imes 0 = 0  30$).$30 = 7 a_6 + a_7$.$a_6 = 4$ ($7 	imes 4 = 28  30$).$a_7 = 30 - 28 = 2$.Sum $= 1 + 1 + 1 + 0 + 4 + 2 = 9$.

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

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