A four number puzzle (Posted on 2019-06-15)

	Submitted by Brian Smith
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Let X=A+C and Y=B+D. Then A*B+B*C+C*D = X*Y-A*D and X+Y=2019, X>A, and Y>D. To maximize X*Y-A*D, the product A*D needs to be at a minimum while X*Y is at a maximum. With positive integers this happens when X=1009, Y=1010, A=1, D=1; implying B=1009, C=1008, and the value of A*B+B*C+C*D equals 1019089. To minimize X*Y-A*D, the product A*D needs to be as close as possible to X*Y. This occurs when A and D are one less than X and Y respectively. This implies C=1 and B=1, reducing A*B+B*C+C*D to A+1+B. Since A+B+C+D=2019 and C=1 and B=1 then the value of A*B+B*C+C*D equals 2018. In summary, the largest and smallest values of A*B+B*C+C*D are 1019089 and 2018.

	Subject	Author	Date
	re(2): proposed solution	xdog	2019-06-17 10:37:49
	re(3): proposed solution -- higher value	broll	2019-06-16 22:41:56
	re(2): proposed solution -- higher value	Charlie	2019-06-16 21:33:26
	re: proposed solution	Brian Smith	2019-06-16 21:22:59
	proposed solution	xdog	2019-06-16 08:46:45