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Math127121 assignment 2solutions-35071

Course

Discrete Mathematics  (Mathematics 1271)

23 Documents
Students shared 23 documents in this course
Academic year: 2021/2022
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Lakehead University

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Math 1271 Assignment 2 Solutions

  1. Compute the logarithms: log 42 ,log 264 ,and log 84.

SolutionIf log 4 2 =y,we have 4y= 2,soy= 1/ 2 .If log 2 64 =x,we have 2x= 64, sox= 6 log 8 4 =z,we have 8z= 4,soz= 2/ 3.

  1. Find the prime factor decompositions of the numbers 300 and 180, and use these to find their greatest common divisor and least common multiple.

SolutionWe have 300 = 3·100 = 3· 4 ·25 = 2 2 · 3 · 52 and 180 = 18·10 = 3 2 · 2 · 2 ·5 = 22 · 32 · 5 .Their greatest common divisor is 2 2 · 3 ·5 = 60,and their least common multiple is 2 2 · 32 · 52 = 900.

  1. Write out the termst 1 tot 10 in the sequencetn=b(n−1)/ 3 c.

SolutionWe haveb 0 / 3 c = 0,b 1 / 3 c = 0,b 2 / 3 c = 0,b 3 / 3 c = 1,b 4 / 3 c = 1, b 5 / 3 c= 1,b 6 / 3 c= 2,b 7 / 3 c= 2,b 8 / 3 c= 2,andb 9 / 3 c= 3,so our sequence is 0 , 0 , 0 , 1 , 1 , 1 , 2 , 2 , 2 , 3.

  1. Find the value of each of the sums. ∑ 50

n=

(−1)n

∑ 10

n=

5 n

SolutionWe have

∑ 50

n=0(−1) n= 1−1+1−1+1· · ·−1+1 = (1−1)+· · ·+(1−1)+1 =

1 .Also,

∑ 10

n=1 5

n= 5(∑ 9 n=0 5

n) = 5((1− 510 )/(1−5)) = 5(1−9765625)/(−4) =

12207030 ,where we have used the formula for the sum of a geometric progression given in class.

  1. Show that

∑k n=1(tn−tn− 1 ) =tk−t 0 .Use this to compute

∑ 100

n=

1 (n+1)(n+2).

SolutionExpanding,

∑k n=1(tn−tn− 1 ) = (t 1 −t 0 ) + (t 2 −t 1 ) +· · ·+ (tk−tk− 1 ) = tk−t 0 ,since all the other terms cancel in pairs. We have

∑ 100

n=

1 ∑ (n+1)(n+2) = 100 n=

( 1

n+1−

1 n+

)

.Lettn=n 1 +2, using the formula above, our sum becomes −(1/ 102 − 1 /2) = 50/ 102.

  1. Convert the following integers from decimal notation to binary notation: 103, 412 ,and 3123.

SolutionWe have 2 6 = 64 and 2 7 = 128,so we write 103 = 64+39 = 64+32+7 = 64 + 32 + 4 + 2 + 1 = 2 6 + 2 5 + 2 2 + 2 1 + 2 0 = (1100111) 2 .We have 2 8 = 256 and 29 = 512,so we write 412 = 256 + 156 = 256 + 128 + 28 = 256 + 128 + 16 + 12 = 256+128+16+8+4 = 2 8 +2 7 +2 4 +2 3 +2 2 = (110011100) 2 .We have 2 10 = 1024, 211 = 2048,and 2 12 = 4096,so we write 3123 = 2048 + 1075 = 2048 + 1024 + 51 = 2048+1024+32+19 = 2048+1024+32+16+2+1 = 2 11 +2 10 +2 5 +2 4 +2 1 +2 0 = (110000110011) 2.

  1. Convert the following integers from binary notation to decimal notation: 1001, 10101 ,and 10110. 1

2

SolutionWe have (1001) 2 = 2 0 + 2 3 = 1 + 8 = 9,(10101) 2 = 2 0 + 2 2 + 2 4 = 1 + 4 + 16 = 21,and (10110) 2 = 2 1 + 2 2 + 2 4 = 2 + 4 + 16 = 22.

  1. Given the following two numbers in binary notation, find their sumand product, in binary notation: 1101 and 1011.

Solution:We perform the same operations as with decimal notation. We get:

1101 + 11000

and

1101 × 1011 1101 11010 000000 1101000 10001111

  1. Convert the following number from hexadecimal notation to binary notation: 5B736.

SolutionWe have 16 = 2 4 = (10000) 2 ,5 = 4 + 1 = (101) 2 , B= 11 = 8 + 2 + 1 = (1011) 2 ,3 = 2 + 1 = (11) 2 ,7 = 4 + 2 + 1 = (111) 2 ,and 6 = 4 + 2 = (110) 2 .We get 5 B376 = 5· 164 + 11· 163 + 3· 162 + 7·16 + 6 = (101) 2 (10000) 42 + (1011) 2 (10000) 32 + (11) 2 (10000) 22 + (111) 2 (10000) 2 + (110) 2 = (1011011001101110110) 2.

  1. Convert the following number from decimal notation to hexadecimal notation:

SolutionWe have 16 2 = 256 and 16 3 = 4096,so we write 4368 = 4096 + 272 = 4096 + 256 + 16 = (1110) 16.

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Math127121 assignment 2solutions-35071

Course: Discrete Mathematics  (Mathematics 1271)

23 Documents
Students shared 23 documents in this course

University: Lakehead University

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Math 1271 Assignment 2
Solutions
1. Compute the logarithms: log42,log264,and log84.
Solution If log42 = y, we have 4y= 2,so y= 1/2.If log264 = x, we have 2x= 64,
so x= 6.If log84 = z, we have 8z= 4,so z= 2/3.
2. Find the prime factor decompositions of the numbers 300 and 180, and use these
to find their greatest common divisor and least common multiple.
Solution We have 300 = 3·100 = 3·4·25 = 22·3·52and 180 = 18·10 = 32·2·2·5 =
22·32·5.Their greatest common divisor is 22·3·5 = 60,and their least common
multiple is 22·32·52= 900.
3. Write out the terms t1to t10 in the sequence tn=b(n−1)/3c.
Solution We have b0/3c= 0,b1/3c= 0,b2/3c= 0,b3/3c= 1,b4/3c= 1,
b5/3c= 1,b6/3c= 2,b7/3c= 2,b8/3c= 2,and b9/3c= 3,so our sequence is
0,0,0,1,1,1,2,2,2,3.
4. Find the value of each of the sums.
50
X
n=0
(−1)n
10
X
n=1
5n
Solution We have P50
n=0(−1)n= 1−1+1−1+1 · · ·−1+1 = (1−1)+· · ·+(1−1)+1 =
1.Also, P10
n=1 5n= 5(P9
n=0 5n) = 5((1 −510)/(1 −5)) = 5(1 −9765625)/(−4) =
12207030,where we have used the formula for the sum of a geometric progression
given in class.
5. Show that Pk
n=1(tn−tn−1) = tk−t0.Use this to compute P100
n=1 1
(n+1)(n+2) .
Solution Expanding, Pk
n=1(tn−tn−1) = (t1−t0) + (t2−t1) + · · · + (tk−tk−1) =
tk−t0,since all the other terms cancel in pairs. We have P100
n=1 1
(n+1)(n+2) =
P100
n=1 1
n+1 −1
n+2 .Let tn=1
n+2 .Then, using the formula above, our sum becomes
−(1/102 −1/2) = 50/102.
6. Convert the following integers from decimal notation to binary notation: 103,
412,and 3123.
Solution We have 26= 64 and 27= 128,so we write 103 = 64+ 39 = 64+32 +7 =
64 + 32 + 4 + 2 + 1 = 26+ 25+ 22+ 21+ 20= (1100111)2.We have 28= 256 and
29= 512,so we write 412 = 256 + 156 = 256 + 128 + 28 = 256 + 128 + 16 + 12 =
256 + 128 + 16 + 8 + 4 = 28+ 27+ 24+ 23+ 22= (110011100)2.We have 210 = 1024,
211 = 2048,and 212 = 4096,so we write 3123 = 2048 + 1075 = 2048 + 1024 + 51 =
2048+1024+32+19 = 2048+1024+32+16+2+1 = 211 + 210 +25+24+21+20=
(110000110011)2.
7. Convert the following integers from binary notation to decimal notation: 1001,
10101,and 10110.
1

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