KVPY SA Exam 2014 Question Paper

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Question : 15 of 80

Marks: +1, -0

Suppose

a_2,a_3,a_4,a_5,a_6,a_7

are integers such that

5/7=a_2/2!+a_3/3!+a^4/4!

+a_5/5!+a_6/6!+a_7/7!,

where

0 ≤ 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

8
9
10
11

Solution:

5 × 6! = 2520a_2 + 840a_3 + 210a_4

+ 42a_5 + 7a_6 + a_7

∵

3600 = 2520a_2 + 840a_3 + 210a_4

+ 42a_5 + 7a_6 + a_7

Now

a_j\ ε\ I

{from 2 to 7}
so above equation is true if

a_2 = a_3 = a_4 = 1

a_5 = 0

a_6 = 4, a_7 = 2

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