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The number of ways in which we can arrange the digits 1.2,3, \ldots .9 such that the product of five digits at any of the five consecutive positions is divisible by 7 is

Option: 1

7 !


Option: 2

{ }^{9} P_{7}


Option: 3

8 !


Option: 4

5(7 !)


Answers (1)

 Let an arrangement of a digit number be \mathrm{x_{1} x_{2}x_{3} x_{4} x_{5} x_{6} x_{7} x_{8} x_{9}}. Note that we require product of each of \mathrm{\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right):\left(x_{2}, x_{3}, x_{4}, x_{5}, x_{6}\right) ; \ldots:\left(x_{5}, x_{6}, x_{7}, x_{8}, x_{9}\right)} is divisible by 7 .

This is possible if the 5th digit is 7 .

Therefore, we can arrange the 9 digits in desired number of ways in 8 ! ways.

 

Posted by

Ramraj Saini

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