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Q) There are unique integers $a_2, a_3, a_4, a_5, a_6, a_7$ such that $$\frac{a_2}{2!}+\frac{a_3}{3!}+\frac{a_4}{4!}+\frac{a_5}{5!}+\frac{a_6}{6!}+\frac{a_7}{7!}=\frac 57$$,where $0\le a_i < i$. Then the value of $a_2+a_3+a_4+a_5+a_6+a_7$
(1)$8$
(2)$9$
(3)$10$
(4)$11$

Answer: $(2)$

My workout

all these are unique integers hence all are having different values.
now multiplying the big one equation by $7!$

$$7\left\{6!\frac{a_2}{2!}+6!\frac{a_3}{3!}+6!\frac{a_4}{4!}+6!\frac{a_5}{5!}+6!\frac{a_6}{6!}\right\}+a_7= 6!.5$$ which is like the remainder theorem which implies, $6!.5=3600$, when divided by 7 leaves remainder $a_7$ so dividing 3600 by 7 gives 2 hence $a_7=2$ . Now from inequality equation we can say that

$a_2\in\{0,1\}$
$a_3\in\{0,1,2\}$
$a_4\in\{0,1,2,3\}$
$a_5\in\{0,1,2,3,4\}$
$a_6\in\{0,1,2,3,4,5\}$
$a_7\in\{0,1,2,3,4,5,6\}$

so from all the data above i can conclude that $a_2=0, a_3=1, a_4=3, a_5=4, a_6=5, a_7=2$ hence summing them up $0+1+3+4+5+2=15$ which is none of the above option

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    $\begingroup$ The integers don't need to have different values. The word "unique" only means that there is only one solution to the equation where all the numbers are integers. If each had a different values, the word distinct would be used. In fact, since the smallest possible sum of $6$ distinct integers is $0+1+2+3+4+5 =15$, you obviously need to duplicate some of them. $\endgroup$ – 5xum Apr 11 '16 at 8:40
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    $\begingroup$ ok, so it is a misinterpretation mistake $\endgroup$ – anni saini Apr 11 '16 at 8:47
  • $\begingroup$ Yes. Also, there are $6$, not $7$, variables in your equation. $\endgroup$ – 5xum Apr 11 '16 at 8:57
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5xum has already pointed out your error.


Multiplying the both sides by $6!$ gives $$\text{(integer)}+\frac{a_7}{7}=\frac{5\cdot 6!}{7}\quad\Rightarrow\quad a_7\equiv 5\cdot 6!\pmod 7\quad\Rightarrow\quad a_7=2$$

Multiplying the both sides by $7!/6$ gives $$\text{(integer)}+\frac{7a_6+a_7}{6}=5!\times 5\quad\Rightarrow\quad 7a_6+a_7\equiv 0\pmod 6\quad\Rightarrow \quad a_6=4$$

Multiplying the both sides by $7!/5$ gives $$\small\text{(integer)}+\frac{42a_5+7a_6+a_7}{5}=6\times 4!\times 5\quad\Rightarrow\quad 42a_5+7a_6+a_7\equiv 0\pmod 5\quad\Rightarrow\quad a_5=0$$

I think you can continue from here.

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It is easy to see that $(a_2,\ldots,a_7)=(1,1,1,0,4,2)$ is a solution. Indeed, we have $$ \frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{4}{720}+\frac{2}{5040}=\frac{5}{7}. $$ In which case we have $a_2+\cdots +a_7=9$.

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