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

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SUMMARY

The discussion focuses on solving the equation $(7-m)(7-n)(7-p)(7-q)=4$ where $m, n, p, q$ are distinct natural numbers. The key solution involves identifying the factors of 4, which are 1, 2, and 4, leading to the combinations of values for $m, n, p, q$. The final result for the expression $7m + 7n + 7p + 7q$ is calculated to be 84, confirming that the distinct natural numbers satisfying the equation are 6, 5, 3, and 2.

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Albert1
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$m,n,p,q\in N$
$m\neq n\neq p\neq q$
and $(7-m)(7-n)(7-p)(7-q)=4$
find:$7m+7n+7p+7q=?$
 
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Albert said:
$m,n,p,q\in N$
$m\neq n\neq p\neq q$
and $(7-m)(7-n)(7-p)(7-q)=4$
find:$7m+7n+7p+7q=?$

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
 

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