Bascules Registres
Mémoires •Circuit asynchrone : les
3 Congruence
We read this as “a is congruent to b modulo (or mod) n. 2 mod 7 and 103 ? 3 mod 10. ... is based on the simple congruence 10 =? ?1 mod 11.
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10:00-11:30 AM Introduction to Women's Day the academic 10-ago. WEDNESDAY. MODULE 1-. INTROPM. 4 Week. Break. MODULE 1 -. INTROPM. 31-ago. 01-sept.
Math 127: Chinese Remainder Theorem
Moreover ?(10) = 4
livre-algorithmes EXo7.pdf
Il ne reste plus qu'à calculer le reste modulo 10 (par exemple @x2GG2IHA272IH module de™im—l le calcul de un pour n = 11 donne 1000 décimales de 10 :.
Cours darithmétique
Exercice 5 Montrer que 2x + 3 est un multiple de 11 si et seulement si 5x + 2 l'est que 100 ? 0 (mod 4) ou que 10 ? ?1 (mod 11)
HRS 2012 Module 10 – Page 1 Final Version – Last Modified 11/05
5 nov. 2012 Final Version – Last Modified 11/05/12 ... Z267_RanValueMod10 "PREASSIGNED RANDOM VALUE - 2012 Mod 10" ... MODULE 10 SCENARIO ASSIGNMENT.
Congruences et théorème chinois des restes
u 1(1) 2(8)
Handout 11: Module 10 Slide 197
https://www.nextgenscience.org/sites/default/files/Handout%2011-Module%2010%2C%20Culminating%20Task%20Debrief%20Questions.pdf
3 Congruence
We read this as “a is congruent to b modulo (or mod) n. 2 mod 7 and 103 ? 3 mod 10. ... is based on the simple congruence 10 =? ?1 mod 11.
[PDF] chapitre 3 : congruences et arithmétique modulaire
Congruences Définition 1 1 Soit m a b entiers On dit que a est congru à b modulo m si m divise a ? b (On dit aussi que “a et b sont congrus modulo m”
[PDF] Les compteurs : (modulo 8 10 et 16) ? Les décompteurs - PDF4PRO
Avec 2 bascules on peut avoir jusqu'à 4 états différents : 00 01 10 et 11 ce qui permet de compter de 0 à 3 en binaire naturel Avec 3 bascules on a 8 états
[PDF] Compteur œ Décompteur - D DUBOIS ESTIT
Remarque: Pour réaliser un compteur modulo 32 il faut 5 bascules J- K 00 01 11 10 Exemple 2: Réalisation d'un décompteur synchrone modulo 10 :
[PDF] FONCTIONS SEQUENTIELLES - AlloSchool
Les compteur 7490 (modulo 10) 7492 (modulo 12) et 7493 (modulo 16) sont des compteurs asynchrones (figure 13) composés de 4 bascules dont les connexions
[PDF] Compteur Intégré
Compteur Modulo 100 Afin de réaliser un compteur supérieur au Modulo 10 on a besoins d'utiliser plus d'un circuit intégré 7490 Pour l'exemple d'un compteur
[PDF] CHAPITRE III : LES COMPTEURS - Technologue pro
10 1 0 1 0 11 Compteur modulo 10 : (Avec front descendant) On à 2 3< 10 < 2 Décompteur asynchrone modulo 8 : (Avec front montant) On à 2
[PDF] Compteur/Décompteur Synchrone
1 2 Réalisation d'un circuit compteur/décompteur synchrone modulo 4 dans le code GRAY Réponse 1 : Un compteur binaire Modulo 6 (état initial (2)10) peut
[PDF] CHAPITRE 6 COMPTEURS SYNCHRONES
10 Réaliser la logique combinatoire de stimulation à partir des Ce sont des compteurs de type 9---0---9---0---9 : 10 états ? compteur à 11) 1
[PDF] Division modulo et clefs de contrôle
o`u le résultat est soit un nombre entre 0 et 9 soit le symbole x si le reste modulo 11 est 10 Vérifier que le code ci-dessus est correct Montrer que la clef
Math 127: Chinese Remainder Theorem
Mary Radclie
1 Chinese Remainder Theorem
Using the techniques of the previous section, we have the necessary tools to solve congruences of the form
axb(modn). The Chinese Remainder Theorem gives us a tool to consider multiple such congruences simultaneously.First, let's just ensure that we understand how to solveaxb(modn).Example 1.Findxsuch that 3x7 (mod10)
Solution.Based on our previous work, we know that 3 has a multiplicative inverse modulo 10, namely 3 '(10)1. Moreover,'(10) = 4, so the inverse of 3 modulo 10 is 33277 (mod10). Hence, multiplying both sides of the above equation by 7, we obtain3x7 (mod10)
,73x77 (mod10) ,x499 (mod10) Hence, the solution isx9 (mod10).Example 2.Findxsuch that 3x6 (mod12). Solution.Uh oh. This time we don't have a multiplicative inverse to work with. So what to do? Well, let's take a look at what this would mean. If 3x6 (mod12), that means 3x6 is divisible by 12, so there is somek2Zsuch that 3x6 = 12k. Now that we're working in the integers, we can happily divide by 3, and we thus obtain thatx2 = 4k. Hence, we have thatx2 (mod4)solves the desired congruence.Of course, the strategy outlined here will not always work. Imagine, if instead of 3x6 (mod12), we
wanted 3x7 (mod12). Obviously that wouldn't be possible, as writing out the corresponding integer equation yields 3x7 = 12k, and there are no integersx;ksuch that 3x12k= 7, by Bezout's Lemma. In general, we have thataxb=nyfor somey2Z, and henceaxny=b. This implies that we can nd a solution to this congruence if and only if gcd(a;n)jb, again by Bezout's Lemma. Proposition 1.Letn2N, and leta;b2Z. The congruenceaxb(modn) has a solution forxif and only if gcd(a;n)jb. Moreover, the strategy we employed in Example 2 will in general work. Suppose that we haveax b(modn), and we have that gcd(a;n) =d. Then in order that this has a solution, we know thatbis divisible byd. In particular, there exist integersa0;b0;n0such thata=a0d;b=b0d;n=n0d. We can then work as we did in Example 2 to rewrite this equation asa0xb0(modn0). 1Example 3.Findx, if possible, such that
2x5 (mod7);
and 3x4 (mod8) Solution.First note that 2 has an inverse modulo 7, namely 4. So we can write the rst equiva- lence asx456 (mod7). Hence, we have thatx= 6 + 7kfor somek2Z. Now we can substitute this in for the second equivalence:3x4 (mod8)
3(6 + 7k)4 (mod8)
18 + 21k4 (mod8)
2 + 5k4 (mod8)
5k2 (mod8):
Recalling that 5 has an inverse modulo 8, namely 5, we thus obtain k102 (mod8):Hence, we have thatk= 2 + 8jfor somej2Z.
Plugging this back in forx, we have thatx= 6 + 7k= 6 + 7(2 + 8j) = 20 + 56jfor somej2Z. In fact, any choice ofjwill work here. Hence, we have thatxis a solution to the system of congruences if and only ifx20 (mod56).Example 4.Findx, if possible, such that x3 (mod4); andx0 (mod6): Solution.Let's work as we did above. From the rst equivalence, we have thatx= 3 + 4k for somek2Z. Then, the second equivalence implies that 3 + 4k0 (mod6), and hence4k 33 (mod6). However, this is impossible, since we know that gcd(4;6) = 2 and 26 j3.Ok, so not every system of congruences will have a solution, but our strategy of trying to solve them
will reveal when there is no solution also. Notice the problem that occurred here: when we considered the rst equivalence, we ended up witha coecient of 4 in front of thek. Since 4 is not relatively prime to 6, there was a chance that the next
equivalence would not have a solution, and indeed that is what happened. In general this will be the case:
if we consider two equivalences of the form xb1(modn1) xb2(modn2); then the method we developed above will take the following approach: rst, writex=b1+kn1. Plug that in to the second equation to obtainkn1b2b1(modn2). Ifn1andn2share factors, then we may not be able to solve this equivalence, per Proposition 1. Hence, we can demand thatn1andn2are relatively prime, and this should solve that problem. Continuing, then, if we assume thatn1andn2are relatively prime, we have reduced this system to kn1b2b1(modn2). Then we obtainkn1b2+b1=jn2for somej2Z. Rearranging, we have
kn1jn2=b2b1. Sincen1andn2are relatively prime, we know from Bezout's Lemma that we will be
2 able to solve this equation forkandj. Once we knowkandj, we can then backsolve to give us a solution forx.This strategy of considering relatively prime moduli, in general, will yield a solution to this problem.
The general form is given by the following theorem. Theorem 1.Letn1;n2;:::;nkbe a set of pairwise relatively prime natural numbers, and letb1;b2;:::;bk2 Z. PutN=n1n2:::nk, the product of the moduli. Then there is a uniquex(modN)such thatx b i(modni)for all1ik. Note that working modNshould be unsurprising; this is how we ended up in the rst example as well. You can see that the method of backsolving forxwill end up multiplying the moduli together.Proof.For eachiwith 1ik, putmi=Nn
i. Notice that since the moduli are relatively prime, and m iis the product of all the moduli other thanni, we have thatni?mi, and hencemihas a multiplicative inverse moduloni, sayyi. Moreover, note thatmiis a multiple ofnjfor allj6=i.Putx=y1b1m1+y2b2m2++ykbkmk:
Notice that for eachiwith 1ik, we obtain
xy1b1m1+y2b2m2++ykbkmk(modni) yibimi(modni) (since eachmjwithj6=iis a multiple ofni) bi(modni) (sinceyiis an inverse tomimoduloni):Therefore, we have thatxbi(modni) for all 1ik.
Finally, we wish to show uniqueness of the solution (modN). Suppose thatxandyboth solve the congruences. Then we have that for eachi,niis a divisor ofxy. Since theniare relatively prime, thismeans thatNis a divisor ofxy, and hencexyare congruent moduloN.Example 5.Use the Chinese Remainder Theorem to nd anxsuch that
x2 (mod5) x3 (mod7) x10 (mod11) Solution.SetN= 5711 = 385. Following the notation of the theorem, we havem1=N=5 = 77,m2=N=7 = 55, andm3=N=11 = 35.
We now seek a multiplicative inverse for eachmimoduloni. First:m1772 (mod5), and hence an inverse tom1modn1isy1= 3. Second:m2556 (mod7), and hence an inverse tom2modn2isy2= 6. Third:m3352 (mod11), and hence an inverse tom3modn3isy3= 6. Therefore, the theorem states that a solution takes the form: x=y1b1m1+y2b2m2+y3b3m3= 3277 + 6355 + 61035 = 3552: Since we may take the solution moduloN= 385, we can reduce this to 87, since 285287 (mod385).3
Example 6.Find all solutionsx, if they exist, to the system of equivalences:2x6 (mod14)
3x9 (mod15)
5x20 (mod60)
Solution.As in Example 2, we rst wish to reduce this, where possible, using the strategy outlined following the statement of Proposition 1. Since gcd2;14 = 2, we can cancel a 2 from all terms in the rst equivalence to writex3 (mod7). Likewise, we simplify the other two equivalences to reduce the entire system to x3 (mod7) x3 (mod5) x4 (mod12): We can now follow the strategy of the Chinese Remainder Theorem. Following the notation in the theorem, we have m1= 512 = 604 (mod7);y14510242 (mod7)
m2= 712 = 844 (mod5);y243644 (mod5)
m3= 75 = 3511 (mod12);y3113(1)3 111 (mod12):
Hence, we havex=y1m1b1+y2m2b2+y3m3b3= 2603 + 4843 + 11354 = 2908.Hence, we have any solutionx2908388 (mod420).4
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