 Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum            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.   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.   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.   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!   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.   Problem D6 MDQH GRH is one possible solution.   Problem D7 Wrap around: Y + 3 = B. This works partially because B has been vacated by moving the rest of the alphabet forward.   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.   Problem D9 An example: CODES becomes KUNQG.   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.   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     Session 9: Index | Notes | Solutions | Video