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Remainder of factorial sum

CAT 2005 Slot 1 · Quantitative Ability · Medium · Number System

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Let n!=1×2×3×⋯×nn! = 1 \times 2 \times 3 \times \dots \times n for integer n≥1n \ge 1. If p=1!+(2×2!)+(3×3!)+⋯+(10×10!)p = 1! + (2 \times 2!) + (3 \times 3!) + \dots + (10 \times 10!), then p+2p + 2 when divided by 11!11! leaves a remainder of

  1. A.

    10

  2. B.

    0

  3. C.

    7

  4. D.

    1

Answer

D

Explanation

Notice that k×k!=(k+1−1)k!=(k+1)!−k!k \times k! = (k + 1 - 1)k! = (k + 1)! - k!. Sum p=(2!−1!)+(3!−2!)+(4!−3!)+⋯+(11!−10!)=11!−1!=11!−1p = (2! - 1!) + (3! - 2!) + (4! - 3!) + \dots + (11! - 10!) = 11! - 1! = 11! - 1. So p+2=(11!−1)+2=11!+1p + 2 = (11! - 1) + 2 = 11! + 1. When 11!+111! + 1 is divided by 11!11!, the remainder is 11.

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