As such, which of the following are equivalent to (15) (15) (15) mod 23? -17 -6 17 270 3375

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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To compute (15)(15)(15) mod 23, what can you do?

Select all that apply.

Group of answer choices
- Take the product (15)(15)(15) then find its remainder when dividing by 23.
- Take the result from the previous question, multiply it by 15, then compute the remainder when divided by 23.
- Perhaps inefficiently, add or subtract 23 (or any multiple of 23) from (15)(15)(15) repeatedly until we arrive at a - - number between 0 and 22.
As such, which of the following are equivalent to (15) (15) (15) mod 23?
U
-17
-6
17
270
3375
Transcribed Image Text:As such, which of the following are equivalent to (15) (15) (15) mod 23? U -17 -6 17 270 3375
For p = 7, q11, e = 13, and n = pq, find the multiplicative inverse of e, reduced modulo
☀ (n) using the extended Euclidean algorithm.
Remark: The textbook presents this algorithm as Exercises 28 and 29 in section 8.4 but gives a
different presentation of the table using unindexed variables and works left to right instead of
down. The same thing is essentially going on with the example from my lecture slides but with
less keeping track of redundant information. The exam question will have the same format as the
lecture slides. Back substitution is messy and more likely to result in errors.
Transcribed Image Text:For p = 7, q11, e = 13, and n = pq, find the multiplicative inverse of e, reduced modulo ☀ (n) using the extended Euclidean algorithm. Remark: The textbook presents this algorithm as Exercises 28 and 29 in section 8.4 but gives a different presentation of the table using unindexed variables and works left to right instead of down. The same thing is essentially going on with the example from my lecture slides but with less keeping track of redundant information. The exam question will have the same format as the lecture slides. Back substitution is messy and more likely to result in errors.
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