CS 1050 B: Constructing Proofs
April 12, 2008
Solutions to Homework 7
Lecturer: Sasha Boldyreva
Problem 7.1, 12 points.
Problem 18 from Section 6.2 of Rosen’s textbook.
a) 1/7.
b) Following the Birthday bound reasoning:
Pr(at least 2 are born on the same day)
=
1

Pr(all born on different days)
=
1

6
7
·
5
7
·
. . .
·
8

n
7
.
c) Plug in n= 2, 3 and so on, and see that the first time it exceeds 1/2 is when n=4, so
this is the answer. You can use the Birthday bound too.
Problem 7.2, 5 points.
Problem 20 from Section 6.2 of Rosen’s textbook.
The probability that none of
n
people has a birthday today is (366
/
366)
n
. We need to
find the smallest
n
such that this is less than 1/2. We can find that the answer is 254 by
trialanderror or by using logarithms.
Problem 7.3, 5 points.
Problem 24 (d) and (e) from Section 3.6 of Rosen’s textbook.
Show the intermediate results.
d) gcd(1529,14039)=gcd(1529,278)=gcd(278,139)=gcd(139,0)=139.
e) gcd(1529,14038)=gcd(1529,277)=gcd(277,144)=gcd(144,133)=gcd(133,11)=gcd(11,1)=gcd(1,0)=1.
Problem 7.4, 5 points.
Run the Extended GCD algorithm by hand on input (1529,14039)
from the previous problem and show the intermediate and final results.
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 Spring '05
 HUANG
 Inductive Reasoning, Probability theory, Negative and nonnegative numbers, Euclidean algorithm, Mathematical proof

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