Question

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

• A. 10
• B. 7
• C. 1

Answer 1

The correct option is C 1

If $P=1!=1$

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

If $P=1!+2\times 2!=5$

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

Hence, $P=1!+(2\times 2!)+(3\times 3!)+.....+(10\times 10!)$

when divided by $11!$ leaves remainder 1.

Alternative method:

$P=1+2.2!+3.3!+....10.10!$
$=(2-1)1!+(3-1)2!+(4-1)3!+.....+(11-1)10!$
$=2!-1!+3!-2!+...+11!-10!$
$=11!-1$
Hence, (p+2) will have a remainder of 1 when divided by 11.