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The Electronic Journal of Mathematics and Technology, Volume 6, Number 1, ISSN 1933-2823Example 3 Modulo 841, the integer 160 is invertible. Indeed, 841 = 29 2 while 160 = 2 5 · 5. Thus,gcd(841, 160) = 1. Its inverse is 205, since160 · 205 − 1 = 32800 − 1 = 32799 = 39 · 841.Maxima can compute the inverse of n modulo m directly with the command inv mod:(%i3)inv_mod(160,841);(%o3) 205The properties of modular arithmetic are similar to those of integer arithmetic. To name a few(see [1], Chapter 5):1. a ≡ a (mod m), for all a ∈ Z.2. a ≡ b (mod m) if and only if b ≡ a (mod m), for all a, b ∈ Z.3. If a ≡ b (mod m) and b ≡ c (mod m), then a ≡ c (mod m), for all a, b, c ∈ Z.4. If a ≡ b (mod m) and c ≡ d (mod m), then a + c ≡ b + d (mod m) and a · c ≡ b · d (mod m),for all a, b, c, d ∈ Z.5. If a ≡ b (mod m), then if a ≡ b (mod d) for each divisor of m, d.6. If a ≡ b (mod m), and a ≡ b (mod n), with m, n coprimes, then a ≡ b (mod m · n).We now recall a classical result (see [1], Chapter 5). Fermat’s Little Theorem says that if p is aprime number, then for any integer a we haveand, for every integer a not divisible by p:a p ≡ a (mod p),a p−1 ≡ 1 (mod p). (1)An immediate consequence is the following. If a is not divisible by p and n ≡ m (mod (p − 1)), thena n ≡ a m (mod p).Example 4 Let us find the last digit of 2 1000000 when written in base 7. This is just the remainderof 2 1000000 modulo 7. In order to apply the result above, we must reduce the exponent. If we takep = 7, n = 1000000, we must find a suitable m such that 1000000 ≡ m (mod 6). Since 1000000 =166666 · 6 + 4, this means that we can take m = 4. Applying now the corollary to Fermat’s LittleTheorem, we find2 1000000 ≡ 2 4 = 16 ≡ 2 (mod 7).Thus, the last digit of 2 1000000 when written in base 7 is 2.37
The Electronic Journal of Mathematics and Technology, Volume 6, Number 1, ISSN 1933-28232.2 The Chinese remainder theoremThe original form of the theorem, appearing in a book written by the Chinese mathematician QinJiushao and published in 1247, is a result related to systems of congruences. It is possible to find aprecedent in the Sunzi suanjing, a book from the third century written by Sun Zi:Han Xing asks how many soldiers are in his army. If you let them parade in rows of 3soldiers, two soldiers will be left. If you let them parade in rows of 5, 3 will be left, andin rows of 7, 2 will be left. How many soldiers are there?.The modern formulation of the problem is the following. Let m 1 , m 2 , ..., m k be integers that aregreater than one and pairwise coprime, and let a 1 , a 2 , ..., a k be any integers. Then there exists aninteger x such that x ≡ a i (mod m i ) for each i ∈ {1, 2, ..., k}. Furthermore, for any other integer ythat satisfies all the congruences, y ≡ x (mod M) where M = m 1 · · · m k . Note that there is only onesolution in {0, 1, ..., M}.The proof of the theorem (which can be found in Theorem 5.26 of [1]) gives an algorithm to findx. The following code implements it in Maxima (it returns the lowest positive solution modulo M):Beginning of codechinese_remainder(a,k):=block([K,L,x],K:makelist(apply("*",delete(k[i],k)),i,1,length(k)),L:makelist(first(gcdex(K[i],k[i])),i,1,length(k)),x:mod(sum(a[i]*K[i]*L[i],i,1,length(k)),apply("*",k)),x);End of codeTo use it, just write chinese remainder([a1,...,ar],[m1,...,mr]). For the HanXing’s example:(%i4)chinese_remainder([2,3,2],[3,5,7]);(%o4) 23A small army, indeed. Surely, this is one example in which the solution is not the one in {0, 1, ..., M}(as in this case Han Xing would have not needed to ask for the number of soldiers, being easy to countthem by himself), but one of the infinite numbers congruent with 23 modulo M = 7 · 5 · 3 = 105, thatis: 128, 233, etcetera.2.3 Euler’s theoremConsider Example 4 again. What if we are asked for the last digit in base 77?. We can not applyFermat’s theorem here, because 77 is composite. Euler found a generalization of Fermat’s theorem tothis case, introducing his totient function ϕ(n) (which gives the number of positive integers less than38
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