Using Fermat's Little Theorem, find the smallest remainder (or least positive residue) of3999,999,999 (mod 7), e21,000,000 (mod 17). e

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Answered: Ising Fermat's Little Theorem, find the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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Problem 1RQ

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1. (a) question: Use Fermat’s Little Theorem to calculate:(i) 23106 (mod 107) (ii) 431 (mod 7) (iii) 316 + 532 + 864 +13128 (mod 17)(iv) the remainder of 21000 divided by 13
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15) Find a remainder after dividing 319⁹⁹ on 7.
Lauren drinks a variety of soft drinks. Over the past month, she has had 15 diet colas, 2 cans of lemonade, and 7 cans of root beer. a. Given this history, what is the probability that her next drink will be a diet cola? Lemonade? Root beer? b. What definition of probability should be used to answer this question? a. The probability that her next drink will be a diet cola is (Round to two decimal places as needed.) Enter your answer in the answer box and then click Check Answer. 3 parts remaining Clear All Copyright © 2021 Pearson Education Inc. All rights reserved. Terms of Use l Privacy Polic ching
1. Show that if pla* + a + a? +a+1 and a -1 mod 5 then p=1 mod 5. Hint: a -1= (a - 1)(a' +a + a? +a+1). What is the period of powers of a when seen modulo p? 2. Show that there are infinite many primes p such that p=1 mod 5. Hint: if pi, P2,. Pn are the first n primes consider A+ A+ A? + A' +1 where 10p, PaPs... Pa -1. A=
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(10) Prove that for any prime number P>3. p² = 1 (mod 24) (Hint: chinese remainder theorem tells us SP² = 1 (mod 3) p² = 1 (mod 8) need to show we only need to p²=1 (mod 8), which we know that P PH two integers. The hard part is to show means S1 p²2-1 81 = (P-1) (PH). must be odd. Hence, P-1, So it is enough to show 2 | P-1 F are PH).
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Proof
Chinese Remainder Theorem Determine FOUR values of x that will satisfy: x ≡ 1 mod 3 & x ≡ 5 mod 8 show your solutions

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Ising Fermat's Little Theorem, find the smallest remainder (or least positive residue) of 3999,999,999 (mod 7), e 21.000,000 (mod 17). e