{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

challset1sol

# challset1sol - Mat 344 Solution Sketch for Challenge Set 1...

This preview shows pages 1–3. Sign up to view the full content.

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 = 100-N(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 = 100-60=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 = 100-N(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+9-10-4-1=64. The number of integers not divisible by 2,5, or 11 = 100-64=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 16-2=14 cases where each person will receive at least one pen.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
3. There are 10x10=100 potential addresses that can be constructed using two digits, however since 00 is not an address, there are 100-1=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
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

challset1sol - Mat 344 Solution Sketch for Challenge Set 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online