Find Sum of 4 Different Natural Numbers with $(7-m)(7-n)(7-p)(7-q)=4$

[Albert1]

$m,n,p,q\in N$
$m\neq n\neq p\neq q$
and $(7-m)(7-n)(7-p)(7-q)=4$
find:$7m+7n+7p+7q=?$

 

[kaliprasad]

$m,n,p,q\in N$
$m\neq n\neq p\neq q$
and $(7-m)(7-n)(7-p)(7-q)=4$
find:$7m+7n+7p+7q=?$

Spoiler

let 7-m < 7-n < 7-p < 7- q without loss of generality

product of 4 different numbers is 4 so they are -2,-1, 1 , 2

so 7-m + 7-n + 7- p + 7- q = 0 or m+n+p+q = 28

so 7m + 7n + 7p + 7q = 196