Mat 344
Solution Sketch for Challenge Set 1
1.
a.
The positive integers less than or equal to 100 are the integers
1,2,3,…,100. There are 50 even integers and 50 odd integers.
b.
Let S={1,2,3,…,100}. Let A = {x: x is an even positive integer}, B={x: x
is a positive integer divisible by 5}, and C={x: x is a positive integer
divisible by 11}. Let N(A or B)= {x: x is an integer divisible by 2 or 5}.
The number of integers not divisible by 2 or 5 = 100N(A or B). We need
to determine N(A or B). There is a counting formula for obtaining N(A or
B). It is N (A or B)=N(A)+N(B)N(A and B). N(A and B)={x: x is a
positive integer divisible by 2 and 5} = {x: x is a positive integer divisible
by 10}. N(A) = 50, N(B)=20, and N(A and B)=10. N(A or B)= 50+20
10=60. Hence the number of integers not divisible by 2 or 5 = 10060=40.
c.
N(A or B or C) = {x: x is a positive integer divisible by 2, 5, 11}. The
number of integers not divisible by 2, 5, 11 = 100N(A or B or C). There is
a counting formula for determining N(A or B or C). It is
N(A or B or C) = N(A)+N(B)+N[C] N(A and B)N(A and C)N(B and
C)+N(A and B and C). N(A and C)={x: x is a positive integer divisible by
2 and 11}={x; x is divisible by 22}. N(B and C)={x: x is a positive integer
divisible by 5 and 11}={x: x is divisible by 55}. N(A and B and C)= {x: x
is a positive integer divisible by 2,5,11}={x: x is divisible by 110}.
N(A)=50, N(B)=20, N[C]=9, N(A and B)=10, N(A and C)=4, N(B and
C)=1, N(A and B and C)=0 since no integer is S is divisible by 110.
Hence N(A or B or C) = 50+20+91041=64. The number of integers not
divisible by 2,5, or 11 = 10064=36.
2.
a.
Each pen has a choice of going to person A or person B. By the
fundamental principle of counting, there are 2x2x2x2 =16 ways of
distributing 4 different color pens to 2 different people.
b.
Out of the sixteen cases obtained in part a, there will be two cases where
one person gets all the pens, i.e. person A gets all four pens, or person B
gets all four pens. Therefore, there will be 162=14 cases where each
person will receive at least one pen.
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3.
There are 10x10=100 potential addresses that can be constructed using two
digits, however since 00 is not an address, there are 1001=99 different
addresses on Memory Lane.
4.
a.
The first part is a simple application of the pigeon hole principle. This
principle states that if you try to put n+1 pigeons into n holes, then at least
one hole will have more than one pigeon. If we let the 12 months of the
year be the holes, and the 13 people be the pigeons, then by applying the
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 Fall '06
 miller
 Natural number, Prime number, positive integer

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