The number of ways in which we can arrange the digits $$1$$, $$2$$, $$3$$, …, $$9$$ such that the product of five digits at any of the five consecutive positions is divisible by $$7$$ is

Question

Infinity Learn · Accepted Answer

Let an arrangement of $$9$$ digit number be  $$x1x2x3x4x5x6x7x8x9$$.Note that we require product of each of $$x1$$,$$x2$$,$$x3$$,$$x4$$,$$x5$$;$$x2$$,$$x3$$,$$x4$$,$$x5$$,$$x6$$;…;$$x5$$,$$x6$$,$$x7$$,$$x8$$,$$x9$$ 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.

The number of ways in which we can arrange the digits 1, 2, 3, …, 9 such that the product of five digits at any of the five consecutive positions is divisible by 7 is

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