Suppose a2-a3-a4-a5-a6-a7 are intagers such that 57-a22-a33-a44-a55-a66-a77-where 0- aj- j for j-2-3-4-5-6-7- The sum a2-a3-a4-a5-a6-a7 is-

The correct option is B 9 57=a_22!=a_33!=a_44!=a_55!=a_66!=a_77! 5=7a_22!+7a_33!+7a_44!+7a_55!+7a_66!+a_77! 5 × 6! = 2520 a_2 +840 a_3 +210 ...

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Byju's Answer

Standard XII

Mathematics

Definition of Relations

Suppose a2,...

Question

Suppose $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ are intagers 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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Solution

The correct option is B $9$
$\frac{5}{7} = \frac{a_{2}}{2!} = \frac{a_{3}}{3!} = \frac{a_{4}}{4!} = \frac{a_{5}}{5!} = \frac{a_{6}}{6!} = \frac{a_{7}}{7!}$

$5 = \frac{7 a_{2}}{2!} + \frac{7 a_{3}}{3!} + \frac{7 a_{4}}{4!} + \frac{7 a_{5}}{5!} + \frac{7 a_{6}}{6!} + \frac{a_{7}}{7!}$

$5 \times 6! = 2520 a_{2} + 840 a_{3} + 210 a_{4} + 42 a_{5} + 7 a_{6} + a_{7}$
Therefore, $3600 = 2520 a_{2} + 840 a_{3} + 210 a_{4} + 42 a_{5} + 7 a_{6} + a_{7}$
Now $a_{j} \in I$ {from 2 to 7}
So above equation is true if
$a_{2} = a_{3} = a_{4} = 1$
$a_{5} = 1$
$a_{6} = 4, a_{7} = 2$
So the required sum, $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7} = 1 + 1 + 1 + 0 + 4 + 2 = 9$

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Similar questions

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Suppose a2,a3,a4,a5,a6,a7 are intagers such that57=a22!+a33!+a44!+a55!+a66!+a77!where 0≤aj<j for j=2,3,4,5,6,7. The sum a2+a3+a4+a5+a6+a7 is-