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

Section: Chapter Questions

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
practice for exam. thank you!
Secton 5.1 Excerciser # 18
Number theory problem: part c
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=
I did parts a and b I need help with parts c and d
1. Number Theory a. What is 132035 (mod 71)? (Hint: Fermat's Little Theorem)
(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).
Theorem 12.2. For all odd prime numbers p, ((²¹)!)² = (−1)²+¹ mod p. Remark. Apply Theorem 12.1 to the expression ((²¹)!)² to see that the expression is congruent to (p-1)!(-1). Now apply Wilson's theorem to get the result.
2. From numerical experimentation, it would appear (miraculously) that the value of() for odd primes is determined by the value of q( mod p). What about 2 ? The value of (3) certainly isn't determined by the value of p(mod2) - but maybe it is determined by the value of p(mod4) ? Is Podasip 8 helpful? Make a conjecture. Annexes: Podasip 8. Keep the notation from Podasip 7. If p is an odd prime and a = 0(modp), then... (a) a²¹ = (−1)ª (modp) where N is the number of € that are negative. (b) = (-1)M (modp) 2 by where M = ₁ + €₂ + + €(p-1)/2- Podasip 7. Suppose p is an odd prime and a ‡ 0(modp). For each k = 1,2, ..., Р ak Ekk (modp) where 0
Page 8. 13. Let k be an integer such that k = -1 (mod 5). (a, What is the remainder when 50k10 + 26k³ – 17 is divided by 5? (b), What is the last digit of the base 5 representation of k? Why? Use induction to prove that TD + 22+ + n 14.. 1(1+1) + 3(3+1) n(n+1) for all integers n > 1. п+1 + Drag and drop an image or PDF file or click to browse...

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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