Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum

Monthly Update sign up
Mailing List signup
Search
Follow The Annenberg Learner on LinkedIn Follow The Annenberg Learner on Facebook Follow Annenberg Learner on Twitter
MENU
Learning Math Home
Patterns, Functions, and Algebra
 
Session 9 Part A Part B Part C Part D Part E Homework
 
Glossary
Algebra Site Map
Session 9 Materials:
Notes
Solutions
 

A B C D
Homework

Video

Solutions for Session 9, Part D

See solutions for Problems: D1 | D2 | D3 | D4 | D5 | D6| D7| D8| D9| D10| D11


Problem D1

First, add the opposite of 5 to both sides. This is 5, because 5 + 5 = 0. Then:

 

7x + 5 = 9

 

7x + 5 + 5 = 9 + 5

 

7x + 0 = 4

 

7x = 4

Now, we need to multiply by the reciprocal of 7. This is 3, because 7 * 3 = 1. Then:

 

7x = 4

 

3(7x) = 3 * 4

 

(3 * 7)x = 2

 

x = 2

This method will be effective, so long as the opposite of our added number exists, and so long as the inverse of our multiplied number exists.

<< back to Problem D1


 

Problem D2

Follow the same procedure as in Problem D1. The opposite of 7 is 3, and the reciprocal of 3 is 7. The answer is x = 9.

<< back to Problem D2


 

Problem D3

Since the opposite of 1 exists (it's 9), we get 4x = 8. But 4 doesn't have a reciprocal! The next step is to look through the multiplication table, trying to find any numbers that, when multiplied by 4, produce 8. There are two: x = 2 and x = 7. These are the two solutions. This means that a linear equation in the system of units digit arithmetic can have more than one solution.

<< back to Problem D3


 

Problem D4

Since 4 does not have a reciprocal, we need to use the table to find all solutions to 4x = 7. No such solution exists!

<< back to Problem D4


 

Problem D5

If A has an inverse, then there will be exactly one solution. If A does not have an inverse, then the number of solutions depends on the common factors of A and B. If A has no inverse and A and B do not have a common factor, then there will be no solutions. If A has no inverse and A and B do have a common factor, then there will be more than one solution. Specifically, the number of solutions will be two if A and B have 2 as a common factor and five if A and B have 5 as a common factor.

<< back to Problem D5


 

Problem D6

MDQH GRH is one possible solution.

<< back to Problem D6


 

Problem D7

Wrap around: Y + 3 = B. This works partially because B has been vacated by moving the rest of the alphabet forward.

<< back to Problem D7


 

Problem D8

"Undo" the steps by moving everything backwards 3 letters. The result is "Some people think that mathematics is a serious business that must always be cold and dry; but we think mathematics is fun and we aren't ashamed to admit the fact." The origin of the quote you have deciphered is a wonderful book called Conrete Mathemtics by Graham, Knuth, and Patashnik.

<< back to Problem D8


 

Problem D9

An example: CODES becomes KUNQG.

<< back to Problem D9


 

Problem D10

One way to do this is to build a table of plaintext and ciphertext, then decode using the table (like a decoder ring). For example, this table tells you that C becomes K, so if you are given ciphertext letter K, you know the original letter was C.

<< back to Problem D10


 

Problem D11

The rule requires you to "undo" the operations, solving for the variable P (since P is the original letter).

 

C = 3P + 2, so we'll "undo" the 2 by adding its opposite, which is 24 (2 + 24 = 0 in mod 26)

 

C + 24 = 3P + 2 + 24

 

C + 24 = 3P (mod 26)

Now, we'll "undo" the 3 by finding its reciprocal, a number which makes 3R =1 in mod 26. This number is not too hard to find, since "1" is the same number as 1 + 26 = 27. This means R = 9 is the reciprocal.

 

C + 24 = 3P

 

9(C + 24) = 9(3P)

 

9C + (9 * 24) = (9 * 3)P

9C + 8 = P is the rule. Try it to see if it changes KUNQG into CODES

<< back to Problem D11


 

Learning Math Home | Algebra Home | Glossary | Map | ©

Session 9: Index | Notes | Solutions | Video

© Annenberg Foundation 2014. All rights reserved. Legal Policy