Section 6.3 177 9. (a) Make a box for each person and put into that box a checkmark for each acquaintance with that person. There are 208 checkmarks to be entered into these boxes, so some box must contain at least r208/30l = 7 checkmarks. (b) If this is wrong, then each person knows at least seven others. This would mean at least 30(7) = 210 ordered pairs of people who know each other; that is, in the 210, Herb and Edgar are counted twice, once because Herb knows Edgar and once because Edgar knows Herb. So we get at least !(21O) = 105 pairs of acquaintances, a contradiction. 10. [BB] Mimicking the solution to Problem 14, we let ai be the number of sets Martina plays on day i. Then we have 1 :::; al < al + a2 < . .. < al + a2 + . .. + a77 :::; 132. Now it is not true that the only integer in the range 1-77 which is divisible by 21 is 21 itself. Thus, while two of these sums must leave the same remainder upon division by 21, we can conclude this time only that the difference of these sums is divisible by 21, not that the difference
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