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Solutions for Session 9, Part D
See solutions for Problems: D1 | D2 | D3 | D4 | D5 | D6| D7| D8| D9| D10| D11
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Problem D1 | |
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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
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Problem D2 | |
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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
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Problem D3 | |
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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.
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Problem D4 | |
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Since 4 does not have a reciprocal, we need to use the table to find all solutions to 4x = 7. No such solution exists!
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Problem D5 | |
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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.
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Problem D7 | |
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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
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Problem D8 | |
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"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.
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Problem D10 | |
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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
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Problem D11 | |
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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
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