Grade 12 - Exp. / Annals on arithmetic 5 exercises (100% corrected)

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1. Unclassified financial years E.3597 The aim of the exercise is to show that there exists a natural number n whose decimal writing of the cube ends in 2009 , i.e. such that n 3 ≡ 2009 ( mod. 10 000) . Part A 1 Determine the remainder of the Euclidean division of 2 009 2 by 16 . 2 Deduce that : 2 009 8 001 ≡ 2 009 ( mod. 16) Part B Consider the sequence u n defined on N by u 0 =2 009 2 − 1 and, for any natural number n : u n +1 = u n +1 5 − 1 . 1 a Demonstrate that u 0 is divisible by 5 . b Demonstrate, using Newton’s binomial formula, that for any natural number n : u n +1 = u n · u 4 n + 5 · u 3 n + 2 · u 2 n + 2 · u n + 1 c Demonstrate by recurrence that, for any natural num-ber n , u n is divisible by 5 n +1 . 2 a Check that u 3 =2 009 250 − 1 then deduce that 2 009 250 ≡ 1 ( mod. 625) . b Then demonstrate that : 2 009 8 001 ≡ 2 009 ( mod. 625) Part C 1 Using Gauss’s theorem and the results established in the previous questions, show that : 2 009 8 001 − 2 009 is divisible by 10 000 . 2 Conclude, i.e., determine a natural number whose deci-mal writing of the cube ends in 2 009 . E.3631 The three parts I , II , III can be treated independently of each other. Part I Let: E = 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9 ; 10 . Determine the pairs a ; b of integers distinct from E such that the remainder of the Euclidean division of ab by 11 is 1 . Part II Let n be a natural number greater than or equal to 3 . 1 Is the integer ( n − 1)! even? 2 Is the integer ( n − 1)!+1 divisible by an even natural num-ber? 3 Prove that the integer (15 − 1)!+1 is not divisible by 15 . 4 Is the integer (11 − 1)!+1 divisible by 11 ? Part III Let p be a non-prime natural number ( p 2 ) . 1 Prove that p admits a divisor q ( 1 <q<p ) that divides ( p − 1)! 2 Does the integer q divide the integer ( p − 1)!+1 ? 3 Does the integer p divide the integer ( p − 1)!+1 ? E.3208 For each of the following five propositions, indicate whether it is true or false and give a demonstration of the chosen answer. An unproven answer scores no points : Proposition 1: ˇfor any natural integer n , 3 divides the integer 2 2 n − 1 ı. Proposition 2: ˇsif a relative integer x is a solution of the equation x 2 + x ≡ 0 ( mod. 6) then x ≡ 0 ( mod. 3) ı. Proposition 3: ˇthe set of pairs of relative integers ( x ; y ) solutions of the equation 12 x − 5 y =3 is the set of pairs : (4+10 k ; 9+24 k ) où k ∈ Z ı. Proposition 4: ˇThere exists a single pair ( a ; b ) of natural numbers, such that : a<b ; PPCM ( a;b ) − PGCD ( a;b )=1 ı. Two natural numbers M and N are such that M is written abc in base ten and N is written bca in base ten. Proposition 5: ˇSi the integer M is divisible by 27 then the integer M − N is also divisible by 27ı. E.3321 1 Consider the set : A 7 = 1 ; 2 ; 3 ; 4 ; 5 ; 6 a For any element a of A 7 write in the table below the unique element y of A 7 such that : a · y ≡ 1 ( mod. 7) . a 1 2 3 4 5 6 y 6 b For x relative integer, show that the equation 3 · x ≡ 5 ( mod. 7) equals x ≡ 4 ( mod. 7) . c If a is an element of A 7 , show that the only relative integers x solutions of the equation a · x ≡ 0 ( mod. 7) are multiples of 7 . 2 Throughout this question, p is a prime integer greater than or equal to 3 . Consider the set A p = 1 ; 2 ; ::: ; p − 1 of non-zero natural integers strictly less than p . Let a be an element of A p . a Verify that a p − 2 is a solution of the equation : a · x ≡ 1 ( mod. p ) . b Let r be the remainder in the Euclidean division of a p − 2 by p . Show that r is the unique solution x in A p , of the equation a · x ≡ 1 ( mod. p ) . c Let x and y be two relative integers. Show that x · y ≡ 0 ( mod. p ) if, and only if, x is a multiple of p or y is a multiple of p . d Application : p =31 . Solve in A 31 the equations : 2 x ≡ 1 ( mod. 31) ; 3 x ≡ 1 ( mod. 31) Using the previous results, solve in Z the equation : 6 x 2 − 5 x +1 ≡ 0 ( mod. 31) . https://chingmath.fr chapExoCorrec/3597 sacados/3597 Liban Juin 2009 chapExoCorrec/3631 sacados/3631 chapExoCorrec/3208 sacados/3208 chapExoCorrec/3321 sacados/3321
E.3626 1 In this question, we propose to determine all the relative integers N such that : N ≡ 5 ( mod. 13) N ≡ 1 ( mod. 17) a Verify that 239 is solution of this system. b Let N be a relative integer solution of this system. Show that N can be written as : N = 1 + 17 · x = 5 + 13 · y où x and y are two relative integers verifying the rela-tionship 17 x − 13 y =4 . c Solve the equation 17 · x − 13 · y =4 où x and y are rela-tive integers. d Deduce that there exists a relative integer k such that : N = 18 + 221 · k . e Demonstrate the equivalence between : N ≡ 18 ( mod. 221) and N ≡ 5 ( mod. 13) N ≡ 1 ( mod. 17) 2 In this question, any trace of research, even if incomplete, or initiative, even if unsuccessful, will be taken into ac-count in the assessment. a Is there a non-zero natural number k such that : 10 k ≡ 1 ( mod. 17) ? b Is there a natural number ‘ such that : 10 ≡ 18 ( mod. 221) ? https://chingmath.fr chapExoCorrec/3626 sacados/3626 Asie Juin 2009