Question 11

Let $$n!=123* ...n$$ for integer $$n \geq 1$$.

If $$p = 1!+(22!)+(33!)+... +(1010!)$$, then $$p+2$$ when divided by 11! leaves a remainder of

Solution

According to given condiiton we have p = (1 × 1!) + (2 × 2!) + (3 × 3!) + (4 × 4!) + … + (10 × 10!) . So n × n! = [(n + 1) - 1] × n! = (n + 1)! - n!. So equation becomes p = 2! - 1! + 3! - 2! + 4! - 3! + 5! - 4! +… + 11! - 10!. So p = 11! - 1! = 11! - 1. p + 2 = 11! + 1 .So when it is divided by 11! gives a remainder of 1. Hence, option 4.

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Let $$n!=123* ...n$$ for integer $$n \geq 1$$.

If $$p = 1!+(22!)+(33!)+... +(1010!)$$, then $$p+2$$ when divided by 11! leaves a remainder of