0 votes 0 votes Let $n! = 1 \times 2 \times 3 \times \dots \times n$ for integer $n \geq 1$. If $p = 1! (2 \times 2!) + (3 \times 3!) + \dots + (10 \times 10!)$, then $p+2$ when divided by $11!$ leaves a remainder of $10$ $0$ $7$ $1$ Quantitative Aptitude cat2005 quantitative-aptitude number-systems + – go_editor asked Dec 29, 2015 • edited Mar 29, 2022 by Lakshman Bhaiya go_editor 13.9k points 351 views answer comment Share See all 0 reply Please log in or register to add a comment.