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Question

Let n!=1×2×3×...........×n for integer n>1. If p=1!+(2×2!)+(3×3!)+...........+(10×10!), then (p+2) when divided by 11! leaves a remainder of: (CAT 2005)
  1. 10
  2. 1
  3. 7
  4. 0


Solution

The correct option is B 1

If P=1!=1

Then P+2=3, when divided by 2!, remainder will be 1.

If P=1!+2×2!=5

Then, P+2=7 when divided by 3!, remainder is still 1.

Hence, P=1!+(2×2!)+(3×3!)+.....+(10×10!)

when divided by 11! leaves remainder 1.

Alternative method:

P=1+2.2!+3.3!+....10.10!
=(21)1!+(31)2!+(41)3!+.....+(111)10!
=2!1!+3!2!+...+11!10!
=11!1
Hence, (p+2) will have a remainder of 1 when divided by 11.

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