CAT05 math doubts

Today I took the test and got 10 out of 10 in 1 mark questions but faltered to score anything in 2 marks I 'll be solving all of them one by one . Please help me in understanding how to approach the following two questions. I am not asking for any solution just some sort of hints or how to begin ( though solution and explanations are always welcome) 11. Let n! = 1 x 2 x 3 x ... x n for integer n ? 1. If p = 1! + (2 x 2!) + (3 x 3!) + . . . + (10 x 10!), then p + 2 when divided by 11! leaves a remainder of a. 10 b. 0 c. 7 d. 1 e.Not Attempted 12. Consider a triangle drawn on the X-Y plane with its three vertices at (41, 0), (0, 41) and (0, 0), each vertex being represented by its (X, Y) coordinates. The number of points with integer coordinates inside the triangle (excluding all the points on the boundary) is a. 780 b. 800 c. 820 d. 741 e.Not Attempted

The way I would have solved it

Q11. The way I would have solved it let p be only 1 term i. e 1 + 1! then (p + 2)/2! gives reminder 1 let p be of 2 terms i. e 1! + (2 x 2!) then (p + 2)/3! gives reminder 1 let p be of 3 terms i. e 1! + (2 x 2!) + (3 x 3!) then (p + 2)/4! gives reminder 1 Like so (can be proved with method of deduction) if p be of 10 terms i. e 1! + (2 x 2!) + . . . + (10 x 10!) then (p + 2)/11! 'll give reminder 1. May be not be an elegant method but its logical and correct and 'll fetch 2 mark in CAT which matters most :-))

toatl points on the boundary

toatl points on the boundary = 42*3 - 3 (for the double counting) now x+y = k , where k = 0,1,2,...41 number of total points = summmation of (k+1) from 0 to 41 = 903 hence total number of interior points = 903-123 = 780