Question

Find 4 consecutive even integers where the product of the two smaller numbers is 72 less than the product of the two larger numbers?

Answer 1

`n;\ n+2;\ n+4;\ n+6-numbers\\\\n(n+2)+72=(n+4)(n+6)\\n^2+2n+72=n^2+6n+4n+24\\n^2+2n+72=n^2+10n+24\\n^2-n^2+2n-10n=24-72\\-8n=-48\ \ \ \ /:(-8)\\n=6\\\\Solutions:\fbox6;\ 6+2=\fbox8;\ 8+2=\fbox{10};\ 10+2=\fbox{12}.`