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Math problem of calendar ~! ! ! ! About the calendar ~ @ # Hungry
Three dates may be vertically adjacent, so let the first one be X, then the second one is x+7 and the third one is x+ 14.

x+(x+7)+(x+ 14)= 3x+2 1。

Because x+ 14

X>= 1, then 3x+21> =24.

So there are only 24, 40, 56, 72 possibilities.

Also: when 3x+2 1=40, x= 19/3 is not an integer, so.

3x+2 1 = 56, and x = 35/3 is not an integer.

Therefore, only 24,72 people are eligible.