1. NO. 11 HAL. 221
<div id="noscript-padding"></div>
Using Fermat's Little Theorem, find the least positive residue of
3 999999999 mod7 3999999999mod7
up vote4down
votefavorite
2
I'm going through the problems in Rosen's Elementary Number Theory
and am having some trouble with the this problem,
Find the least positive residue of 3 999999999 mod7 3999999999mod7 .
elementary-number-theory modular-arithmetic
share|cite|improve
this question
edited Jan 26 '15 at 6:46
Andrés Caicedo
50.7k5113172
asked Jun 13 '13 at 0:30
crf
2,7051035
add a comment |
3 Answers
active oldest votes
up
vote4
down
votea
ccept
ed
Fermat's Little theorem says that
If p p is a prime and a∤p a∤p , then
a p−1 ≡1modp. ap−1≡1modp.
We can use this theorem to reduce exponents, such as in the problem at hand. First,
we should observe that the hypotheses of Fermat's Little Theorem are met. Namely,
7 is prime and 3∤7. 7 is prime and 3∤7.
So we have that,
3 6 ≡1mod7. 36≡1mod7.
The exponent can therefore be reduced in the following manner,
3 999999999 ≡3 999999996 ⋅3 3 ≡(3 6 ) 166666666 ⋅3 3 ≡1 166666666 ⋅3 3 ≡1⋅27≡6
mod7. 3999999999≡3999999996⋅33≡(36)166666666⋅33≡1166666666⋅33≡1⋅27≡6m
od7.
share|cite|improve this
answer
edited Jun 13 '13 at 9:59 answered Jun 13 '13 at 0:40
2. Gamma Function
2,49041543
1
Why used the generalied case when in this cases n n is prime? – Thomas
Andrews Jun 13 '13 at 0:49
@ThomasAndrews Yeah, I'm not sure why I went immediately to the
generalization. – Gamma Function Jun 13 '13 at 1:00
add a comment |
up
vote4do
wn vote
From FLT, 3 6 ≡1(mod7). 36≡1(mod7). Since
999999999=166666666×6+3 999999999=166666666×6+3 we have
3 999999999 ≡3 3 ≡6. 3999999999≡33≡6.
share|cite|improve this answer answered Jun 13 '13 at 0:37
Ragib Zaman
25.3k33580
FLT≠ ≠ Fermat's last theorem, but the little one. – vadim123 Jun 13 '13 at
1:01
add a comment |
up
vote4do
wn vote
Hint:
mod 7: 3 3 ≡−1⇒(3 3 ) 333333333 ≡(−1) 333333333 ≡−1 mod 7: 33≡−1⇒(33)333
333333≡(−1)333333333≡−1
No. 6 hal 236
1. (a) (5 points) Find the last digit of the decimal expansion of 7999,999.
(b) (5 points) Find the least positive residue of 21,000,000 modulo 17.
Solution: a) This is problem 6.3.6 from the book/Homework
We have to study 7 (mod 10). Notice that φ(10) = 4, then, by Euler’s theorem, 74 ≡ 1
(mod 10). On the other hand, 1, 000, 000 is multiplo of 4m, that means that 999, 999 =
4k + 3 for some k ∈ Z. then
7999,999 ≡ 74k+3 ≡ 73 ≡ 9 · 7 ≡ 3 (mod 10).
The last digit is 3.
This part can be done without using Euler’s theorem, just by checking that 74 ≡ 1 (mod
10).
b) This is problem 6.1.12 from the book/Homework
3. By Fermat’s little theorem, 216 ≡ 1 (mod 17). Since 16 | 1, 000, 000, we have that
21,000,000 ≡ 1 (mod 17).
Just learning based on question page 253 no 6d
2. (a) Find the form of all positive integers satisfying ¿(n) = 10: What is the
smallest positive integer for which this is true?
(b) Show that there are no positive integers satisfying ¾(n) = 10:
Solution:
(a) If n = p1
e1 ¢¢¢ pr
er
; then ¿(n) = (e1 + 1) ¢¢¢ (er + 1) = 10: Thus9e1 + 14 = 2 and
e2 + 1 = 5 (or vice-versa), or e1 + 1 = 10: Thusn has the form p or pq for some
primes p and q: The smallest such is 3 ¢ 24
= 48:
Since ¾(n) ¸ n + 1 > n for n > 1; it su±ces to compute ¾(n) for n < 10: None equal 10
