1 votes 1 votes How many pair of positive integers of $(m,n)$ are there satisfying $$\displaystyle{\sum_{i=1}^{n}i! = m!}$$ $0$ $1$ $2$ $3$ jjayantamahata asked Mar 17, 2018 • edited Mar 17, 2018 by Sukanya Das jjayantamahata 368 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes (C) (m,n)=(1,1) and (2,2) jjayantamahata answered Mar 17, 2018 jjayantamahata comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments pmshukla96 commented May 10, 2018 reply Follow Share We can have 2 pairs,as there is no restriction on value of m ,m can be 0 thus giving second pair.Plz correct me if wrong. 0 votes 0 votes Kushagra Chatterjee commented May 10, 2018 reply Follow Share m cannot be zero in the question it is mentioned m is a positive integer. 1 votes 1 votes ankitgupta.1729 commented May 10, 2018 reply Follow Share only (1,1) is possible here. because m! will give even number when m>1 because m will always contain 2 as a factor. so m! = 2k on RHS... and on LHS we always get odd number for n>1 because it is 1+ k! + l! + m! + ... and each factorial ie. k! , l! , m! ... will give even number and sum of even number will be even number.. so on lhs , it will be in 2k+1 form...So, only (1,1) is possible.. 3 votes 3 votes Please log in or register to add a comment.
