++as. in Java Integrating QR-Code in Java ++as.

++as. using barcode creation for awt control to generate, create qrcode image in awt applications. Code 9/3 u . \ \ \ A..


\. ~~~~~~~~~~....... ................ o PROBLEM 14. Martina has th awt QR Code ISO/IEC18004 ree weeks to prepare for a tennis tournament. She decides to play at least one set every day but not more than 36 sets in all.

Show that there is a period of consecutive days during which she will play exactly 21 sets. Solution. Suppose Martina plays al sets on day 1, a2 sets on day 2, and so on, a21 sets on the last day of her preparation period.

Consider the 21 natural numbers al, al + a2, ...

, al + a2 +...

+ a21. Since each ai > I and since the. 206 6 Principles of Counting sum al + a2 + + a2l is the jar QR Code total number of sets Martina will play, we have 1 < as < al +-a2 < < al +a2 + +a2l < 36.. Now the only natural numbe r between I and 36 which is divisible by 21 is 21 itself. Therefore, each of these sums is either 21 or it belongs to one of the 20 nonzero congruence classes of integers mod 21. In the first case, we have al + a2 + + ai = 21 for some i; so Martina would play precisely 21 sets on days I through i giving the desired result.

In the second case, by the Pigeon-Hole Principle, two sums lie in the same congruence class and hence have difference divisible by 21. As in Problem 13, we obtain a,+] + a+2 + + a, divisible by 21 for some s and t. Since this is a number between I and 36, it must equal 21 and we conclude that Martina plays 21 games on days t + 1 through s.

U PROBLEM 15. Suppose Martina has 11 weeks to prepare for her tournament and that she intends to play at least one set a day and at most 132 practice sets in all. Draw again the conclusion that during some period of consecutive days, Martina will play precisely 21 sets.

Solution. The approach used before doesn"t work this time (see Exercise 9), so we find a slightly different solution. We let bi be the number of sets Martina plays on days I through i inclusive (b, = al + a2 + + ai in the notation of Problem 14) and consider the 154 numbers bib2,.

. , b7 7 ,bil +-21,b 2 +21, . b 7 7+21.

. The largest number here, b QRCode for Java 77+21, is at most 132+21 = 153. By the Pigeon-Hole Principle, we conclude that two are the same. Since bi > bj if i > j, the only way for two to be equal is for bi = by + 21 for some i and j with i > j.

Therefore, bi- bj = 21 and Martina plays 21 sets on days j + I through i. I We noted earlier that in any group of 13 people, at least two must have birthdays in the same month. The same must hold in any larger group, but surely something stronger is also true.

If at most two people in a group of 30 had birthdays in the same month, we would account for at most 24 people. Thus, in any group of 30, there must be at least three with birthdays in the same month. Recall the definition of the ceilingfunction given in 3.

1.6. For any real number x, [xl means the least integer which is greater than or equal to x.

For example, [3.51 = 4, F0.241 = 1, [-2.

91 = -2 and [3 1 = 3. If n objects are put into m boxes and n > m, then some box must contain at least.
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