China Mathematical Competition (Extra Test)

Suppose that $x_i$ ($i=1,2,\dots ,7$) minutes is the time for $i$-th player on the field. Now, the problem is to find the number of solution groups of positive integers for the following equation: $$x_1+x_2+\cdots +x_7=270$$ when the conditions $7\mid x_i$ ($i=1,2,3,4$) and $13\mid x_j$ ($j=5,6,7$) are satisfied.

Suppose $x_1+x_2+x_3+x_4=7m$ and $x_5+x_6+x_7=13n$. Then $$7m+13n=270,$$ and $m,n\in {\mathbb{N}}_{+},m\ge 4$ and $n\ge 3$.

When $(m,n)=(33,3)$, $x_5=x_6=x_7=13$. Let $x_i=7y_i$ ($i=1,2,3,4$), then $$y_1+y_2+y_3+y_4=33.$$ We get $C_{33-1}^4=C_{32}^3=4960$ solution groups of positive integers $(y_1,y_2,y_3,y_4)$ and in this case, we have $4960$ solution groups of positive integers satisfying the conditions.

When $(m,n)=(20,10)$, let $x_i=7y_i$ ($i=1,2,3,4$) and $x_j=13y_j$ ($j=5,6,7$). Hence $$y_1+y_2+y_3+y_4=20\text{ and }{y}_5+y_6+y_7=10.$$ In this case, we have $C_{19}^3\times C_9^3=34884$ solution groups of positive integers satisfying the conditions.

When $(m,n)=(7,17)$, set $x_i=7y_i$ ($i=1,2,3,4$) and $x_j=13y_j$ ($j=5,6,7$). Hence $$y_1+y_2+y_3+y_4=7\text{ and }{y}_5+y_6+y_7=17.$$ In this case, we have $C_6^3\times C_{16}^3=2400$ solution groups of positive integers satisfying the conditions.

Consequently, for $(1)$, there are $$4960+34884+2400=42244$$ solution groups of positive integers satisfying the conditions.