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ASU 248 All Soviet Union MO 1977 (x_1+...+x_n)=(y_1+...+y_m)<mn

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July 6, 2019
combinatoricsalgebraSum

Problem Statement

Given natural numbers x1,x2,...,xn,y1,y2,...,ymx_1,x_2,...,x_n,y_1,y_2,...,y_m. The following condition is valid: (x1+x2+...+xn)=(y1+y2+...+ym)<mn    (∗)(x_1+x_2+...+x_n)=(y_1+y_2+...+y_m)<mn \,\,\,\, (*) Prove that it is possible to delete some terms from (*) (not all and at least one) and to obtain another valid condition.
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