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8/8/2019 Criptografie si securitate
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Criptografie si securitate
Number: 000-000Passing Score: 500Time Limit: 45 minFile Version: 1.0
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Exam A
QUESTION 1
A.
B.
C.
D.
Answer: CSection: (none)
Explanation/Reference:C
QUESTION 2
A. MAZIB. MAXIC. LAZH
D. LAZIAnswer: DSection: (none)
Explanation/Reference:D
QUESTION 3
A. CASAB. CASEC. SACAD. CASE
Answer: BSection: (none)
Explanation/Reference:
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B
QUESTION 4
A. VXBQB. WXBQC. VXCRD. VXBP
Answer: DSection: (none)
Explanation/Reference:D
QUESTION 5
A. ZAREB. AZURC. ZARID. RAZA
Answer: BSection: (none)
Explanation/Reference:B
QUESTION 6
A.
B.
C.
D.
Answer: DSection: (none)
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Explanation/Reference:
QUESTION 7
A. EYUGB. EEUYC. YUGED. UEYG
Answer: ASection: (none)
Explanation/Reference:A
QUESTION 8Considerati alfabetul latin din care eliminati litera de frecventa redusa Q. Folosind sistemul de codificarePolybios, codificati textul clar STIRPIRE. Alegeti varianta corecta din cele de mai jos.
A. DDDEBDDBDABDDBAEB. DCDDBDDBDABDDBAEC. DCDEBDDBDABDDBAED. DCDDBDDBDABDDDAE
Answer: BSection: (none)
Explanation/Reference:B
QUESTION 9Considerati alfabetul latin din care eliminati litera de frecventa redusa W. Folosind sistemul de codificarePolybios, decodificati textul criptat AEECACCBAACCAADCAE. Alegeti varianta corecta din cele de mai jos.
A. EXCLUDEREB. EXCLAMAREC. EXCHANGED. INCLUDERE
Answer: BSection: (none)
Explanation/Reference:B
QUESTION 10Considerati alfabetul latin din care eliminati litera de frecventa redusa Q. Folosind sistemul de codificare
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Polybios, decodificati textul criptat AEECACCBAACCAADCAE. Alegeti varianta corecta din cele de mai jos.
A. EXCLUDEREB. EXCLAMAREC. INCLUDERED. EXCLAMASE
Answer: DSection: (none)
Explanation/Reference:D
QUESTION 11Considerati alfabetul latin din care eliminati litera de frecventa redusa W. Folosind sistemul de codificarePolybios, codificati textul clar UNIVERSITATE. Alegeti varianta corecta din cele de mai jos.
A. EACDBDEBAEDCDDBDDEAADEAD
B. EACDBDEBAEDCDDBDDEAADEAEC. AECDBDEBAEDCDDBDDEAADEAED. AECDBDEBAEDCDDBDDEAADEAD
Answer: BSection: (none)
Explanation/Reference:B
QUESTION 12
A. AHFUTB. BNFZTC. BNFUTD. ANFVT
Answer: ASection: (none)
Explanation/Reference:A
QUESTION 13
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A. YFOZGB. WDMXEC. XENYFD. VCLWD
Answer: C
Section: (none)Explanation/Reference:C
QUESTION 14
A. DJNST
B. BHLQRC. CIMRSD. EKOTU
Answer: ASection: (none)
Explanation/Reference:A
QUESTION 15
A. EZGAWB. ZEGXWC. ZEGWXD. EZGAX
Answer: DSection: (none)
Explanation/Reference:D
QUESTION 16
A. UVLBO
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B. UVLBRC. VULBOD. VULCO
Answer: CSection: (none)
Explanation/Reference:C
QUESTION 17Codificati textul clar USCAT folosind sistemul de criptare Vigenere cu cheia secreta LABIL. Solutia este...
A. FSDIFB. ESDIFC. FSDIED. ESDIE
Answer: CSection: (none)
Explanation/Reference:C
QUESTION 18Codificati textul clar CRIPTOGRAFIE folosind sistemul de criptare Vigenere cu cheia secreta GRUPA. Solutiaeste...
A. IIIHTUXRSFOVB. IICETUXIPFOVC. IICETUXLPFOVD. IICHTUXLSFOV
Answer: CSection: (none)
Explanation/Reference:C
QUESTION 19Codificati textul clar INCAS folosind sistemul de criptare Vigenere cu cheia secreta GARA. Solutia este...
A. UNTAYB. NMTAYC. ONTAYD. ONTAZ
Answer: CSection: (none)
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Explanation/Reference:C
QUESTION 20Codificati textul clar ZIMBRU folosind sistemul de criptare Vigenere cu cheia secreta BILA. Solutia este...
A. ARYBSCB. AQXCSBC. AQXBSCD. AQYBSC
Answer: CSection: (none)
Explanation/Reference:C
QUESTION 21
Codificati textul clar UMBRA folosind sistemul de criptare Vigenere cu cheia secreta CARD. Solutia este...
A. WMSUGB. WMSUCC. KMSUCD. WMSUQ
Answer: BSection: (none)
Explanation/Reference:B
QUESTION 22Codificati textul clar IMPAR folosind sistemul de criptare Vigenere cu cheia secreta VARF. Solutia este...
A. DMGFMB. DUGFMC. DMIAMD. DMGZM
Answer: ASection: (none)
Explanation/Reference:A
QUESTION 23Decodificati textul criptat VUJNIC, codificat cu sistemul de criptare Vigenere cu cheia secreta CUI. Solutiaeste...
A. CABINA
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B. TABLOUC. CANINAD. TABELA
Answer: BSection: (none)
Explanation/Reference:B
QUESTION 24Decodificati textul criptat VUJGFI, codificat cu sistemul de criptare Vigenere cu cheia secreta CUI. Solutiaeste...
A. TABLOUB. CANINAC. TABELAD. CABINA
Answer: CSection: (none)
Explanation/Reference:C
QUESTION 25Decodificati textul criptat EUJKHI, codificat cu sistemul de criptare Vigenere cu cheia secreta CUI. Solutiaeste...
A. CABINA
B. CANINAC. TABELAD. TABLOU
Answer: ASection: (none)
Explanation/Reference:A
QUESTION 26
Decodificati textul criptat EUVKHI, codificat cu sistemul de criptare Vigenere cu cheia secreta CUI. Solutiaeste...
A. TABELAB. TABLOUC. CABINAD. CANINA
Answer: D
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Section: (none)
Explanation/Reference:D
QUESTION 27
Folosind sistemul de criptare asincron cu auto-cheie k=7,codificarea textului clar STRIGAT este ....A. ZSJRXXQB. SSJRXQQC. SZRJQXQD. ZZRJXXQ
Answer: ASection: (none)
Explanation/Reference:A
QUESTION 28Folosind sistemul de criptare asincron cu auto-cheie k=11,codificarea textului clar GRADINA este ....
A. RIILTGGB. RIJLTGGC. RIILSHHD. RIILSHI
Answer: ASection: (none)
Explanation/Reference:A
QUESTION 29Folosind sistemul de criptare asincron cu auto-cheie k=7,codificarea textului clar GRADINA este ....
A. LEHHPCDB. MEHHPCDC. NEEHPCCD. OEEHPCC
Answer: CSection: (none)
Explanation/Reference:C
QUESTION 30Folosind sistemul de criptare asincron cu auto-cheie k=8,codificarea textului clar CORIDA este ....
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A. LXRQAAB. KYPXAAC. LYRXAAD. KYPQAA
Answer: B
Section: (none)Explanation/Reference:B
QUESTION 31Folosind sistemul de criptare asincron cu auto-cheie pentru k = 11, decodificarea textului criptat RIILTGGeste ...
A. GRADINIB. GRANINIC. GRADINA
D. GRINDINA
Answer: CSection: (none)
Explanation/Reference:C
QUESTION 32Folosind sistemul de criptare asincron cu auto-cheie pentru k = 8, decodificarea textului criptat ZZEEPP este ...
A. RAVENAB. RAFALAC. RAFALED. RAMURA
Answer: BSection: (none)
Explanation/Reference:B
QUESTION 33
Folosind sistemul de criptare asincron cu auto-cheie pentru k = 9, decodificarea textului criptat AAVZMMeste ...
A. RAVENAB. RAMURAC. RAFALED. RAFALA
Answer: A
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Section: (none)
Explanation/Reference:A
QUESTION 34
Folosind sistemul de criptare asincron cu auto-cheie pentru k = 10, decodificarea textului criptat BBNHYYeste ...
A. RAFALAB. RAVENAC. RAFALED. RAMURA
Answer: DSection: (none)
Explanation/Reference:D
QUESTION 35Folosind sistemul de criptare asincron cu auto-cheie pentru k = 11, decodificarea textului criptat CCHHSW este...
A. RAFALEB. RAMURAC. RAVENAD. RAFALA
Answer: ASection: (none)
Explanation/Reference:A
QUESTION 36Folosind sistemul aditiv fluid binar de criptare sa se codifice x=101011, avand cheia k=1010
A. 101010B. 101011C. 000001
D. 100000
Answer: CSection: (none)
Explanation/Reference:C
QUESTION 37
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Se da secventa binara de text clar 110100.Codificati aceasta secventa folosind cheia fluida 1011, folosind unsistem aditiv fluid binar de criptare.
A. 001010B. 011010C. 010010D. 010110
Answer: BSection: (none)
Explanation/Reference:B
QUESTION 38Folosind un sistem aditiv fluid binar de criptare se cere criptarea secvenei de text clar x = 101101,cunoscnd cheia fluid z = 1101.
A. 101101B. 101010C. 010010D. 011010
Answer: DSection: (none)
Explanation/Reference:D
QUESTION 39
A. 1001 1101 1000 1111B. 0011 0111 1000 1111C. 1100 1110 1000 1111D. 0011 1011 0010 1111
Answer: DSection: (none)
Explanation/Reference:D
QUESTION 40
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A. 1011 1000 0100 0110B. 1011 0100 0100 1010C. 0111 0010 1000 0101D. 0110 1001 1000 1010
Answer: C
Section: (none)Explanation/Reference:C
QUESTION 41
A. 0111 0001 1000 0110B. 0110 1010 1000 0110C. 1011 0001 1000 1001D. 0111 0000 1000 0110
Answer: ASection: (none)
Explanation/Reference:A
QUESTION 42
A. 1100 0110 1000 0010B. 1100 0110 1000 0011C. 1010 0110 0100 0011D. 1100 1010 1000 0011
Answer: BSection: (none)
Explanation/Reference:
B
QUESTION 43
A. 1111 0101 1010 0111
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B. 1111 0101 1100 0111C. 1111 1010 0101 1011D. 1111 0110 1001 0111
Answer: CSection: (none)
Explanation/Reference:C
QUESTION 44
A. MODELAREB. SOFTWARE
C. ALGORITMD. GESTIUNE
Answer: ASection: (none)
Explanation/Reference:A
QUESTION 45
A. MODELAREB. SOFTWAREC. ALGORITMD. GESTIUNE
Answer: BSection: (none)
Explanation/Reference:B
QUESTION 46
A. MODELAREB. SOFTWARE
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C. ALGORITMD. GESTIUNE
Answer: CSection: (none)
Explanation/Reference:
C
QUESTION 47
A. MODELAREB. SOFTWAREC. ALGORITM
D. GESTIUNEAnswer: DSection: (none)
Explanation/Reference:D
QUESTION 48
A. 0010 0110 0100 1101B. 1110 1011 0001 0000C. 1110 1100 1101 0001D. 0010 0100 0110 1100
Answer: ASection: (none)
Explanation/Reference:A
QUESTION 49
A. 0010 0100 0110 1100
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B. 1110 1100 1101 0001C. 0010 0110 0100 1101D. 1110 1011 0001 0000
Answer: DSection: (none)
Explanation/Reference:D
QUESTION 50
A. 0010 0110 0100 1101B. 1110 1100 1101 0001
C. 1110 1011 0001 0000D. 0010 0100 0110 1100
Answer: BSection: (none)
Explanation/Reference:B
QUESTION 51
A. 1101 0001 1001 0011B. 1000 1111 1100 1111C. 0100 0011 0000 1001D. 1000 1000 1010 0110
Answer: CSection: (none)
Explanation/Reference:C
QUESTION 52
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A. 0100 0011 0000 1001B. 1000 1111 1100 1111C. 1000 1000 1010 0110D. 1101 0001 1001 0011
Answer: C
Section: (none)Explanation/Reference:C
QUESTION 53
A. 1101 0001 1001 0011
B. 1000 1000 1010 0110C. 1000 1111 1100 1111D. 0100 0011 0000 1001
Answer: CSection: (none)
Explanation/Reference:C
QUESTION 54
A. 1110 0111 0010 0111B. 1111 0000 0011 1100C. 1110 1000 1101 1011D. 1110 0111 1101 0111
Answer: BSection: (none)
Explanation/Reference:B
QUESTION 55
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A. 1110 0111 0010 0111B. 1111 0000 0011 1100C. 1110 1000 1101 1011D. 1110 0111 1101 0111
Answer: DSection: (none)
Explanation/Reference:D
QUESTION 56
A. 1110 0111 0010 0111B. 1111 0000 0011 1100C. 1110 1000 1101 1011D. 1110 0111 1101 0111
Answer: CSection: (none)
Explanation/Reference:C
QUESTION 57
A. 1110 0111 0010 0111B. 1111 0000 0011 1100C. 1110 1000 1101 1011D. 1110 0111 1101 0111
Answer: ASection: (none)
Explanation/Reference:A
QUESTION 58
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A. 1100 1100 1100 0011B. 1001 1001 1001 0110
C. 1111 0101 1100 0000D. 1111 1100 0110 0000
Answer: BSection: (none)
Explanation/Reference:B
QUESTION 59
A. 1100 1100 1100 0011B. 1001 1001 1001 0110C. 1111 0101 1100 0000D. 1111 1100 0110 0000
Answer: ASection: (none)
Explanation/Reference:
A
QUESTION 60
A. 1100 1100 1100 0011B. 1001 1001 1001 0110C. 1111 0101 1100 0000D. 1111 1100 0110 0000
Answer: DSection: (none)
Explanation/Reference:D
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QUESTION 61
A. 1100 1100 1100 0011B. 1001 1001 1001 0110C. 1111 0101 1100 0000D. 1111 1100 0110 0000
Answer: CSection: (none)
Explanation/Reference:C
QUESTION 62
A. 0001 1000 1111 1010B. 0010 0000 1101 1010C. 0010 1101 1100 1111D. 1101 1010 0111 0101
Answer: DSection: (none)
Explanation/Reference:D
QUESTION 63
A. 0001 1000 1111 1010B. 0010 0000 1101 1010
C. 0010 1101 1100 1111D. 1101 1010 0111 0101
Answer: ASection: (none)
Explanation/Reference:A
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QUESTION 64
A. 0001 1000 1111 1010B. 0010 0000 1101 1010C. 0010 1101 1100 1111D. 1101 1010 0111 0101
Answer: BSection: (none)
Explanation/Reference:B
QUESTION 65
A. 0001 1000 1111 1010B. 0010 0000 1101 1010C. 0010 1101 1100 1111D. 1101 1010 0111 0101
Answer: CSection: (none)
Explanation/Reference:C
QUESTION 66
A. 1001 0000 0110 1011
B. 0001 1101 0111 1100C. 1001 0001 1111 1010D. 1010 1110 1100 0110
Answer: BSection: (none)
Explanation/Reference:B
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QUESTION 67
A. 1001 0000 0110 1011B. 0001 1101 0111 1100C. 1001 0001 1111 1010D. 1010 1110 1100 0110
Answer: DSection: (none)
Explanation/Reference:D
QUESTION 68
A. 1001 0000 0110 1011B. 0001 1101 0111 1100C. 1001 0001 1111 1010D. 1010 1110 1100 0110
Answer: ASection: (none)
Explanation/Reference:A
QUESTION 69
A. 1001 0000 0110 1011B. 0001 1101 0111 1100C. 1001 0001 1111 1010D. 1010 1110 1100 0110
Answer: CSection: (none)
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Explanation/Reference:C
QUESTION 70Sistemul de criptare El-Gamal se bazeaza pe...
A. dificultatea calculului logaritmului discret intr-un corp finitB. dificultatea descompunerii in factori primi a numerelor mari (de sute de cifre)C. teoria algebrica a codurilorD. problema {0, 1} a rucsacului
Answer: ASection: (none)
Explanation/Reference:A
QUESTION 71
Sistemul de criptare Merkle-Hellman se bazeaza pe...
A. dificultatea calculului logaritmului discret intr-un corp finitB. dificultatea descompunerii in factori primi a numerelor mari (de sute de cifre)C. teoria algebrica a codurilorD. problema {0, 1} a rucsacului
Answer: DSection: (none)
Explanation/Reference:D
QUESTION 72Sistemul de criptare RSA se bazeaza pe...
A. dificultatea calculului logaritmului discret intr-un corp finitB. dificultatea descompunerii in factori primi a numerelor mari (de sute de cifre)C. teoria algebrica a codurilorD. problema {0, 1} a rucsacului
Answer: BSection: (none)
Explanation/Reference:B
QUESTION 73Sistemul de criptare McEliece se bazeaza pe...
A. dificultatea calculului logaritmului discret intr-un corp finitB. dificultatea descompunerii in factori primi a numerelor mari (de sute de cifre)
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Answer: BSection: (none)
Explanation/Reference:B
QUESTION 77Fie d = 11 exponentul de decriptare al sistemului de criptare RSA construit cu numerele prime p = 7, q = 11.Determinati exponentul de criptare e.
A. 7B. 3C. 11D. 13
Answer: CSection: (none)
Explanation/Reference:C
QUESTION 78Consideram sistemul de criptare RSA construit cu numerele prime p = 3, q = 5.Daca exponentul de criptare este e = 4 si se doreste codificarea textului clar m = 11, determinati textul criptat c.
A. 1B. 3C. 5D. 7
Answer: ASection: (none)
Explanation/Reference:A
QUESTION 79Un utilizator al sistemului de criptare RSA are ca cheie publica (n, e) = (35, 5) si cheia secreta d = 5. Dacaprimeste textul criptat c = 3, atunci textul clar decodificat de utilizator este...
A. 20B. 7C. 19D. 33
Answer: DSection: (none)
Explanation/Reference:D
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QUESTION 80Un utilizator al cifrului RSA are ca cheie publica (n, e) = (35, 5) si ca cheie secreta d = 5. Daca primeste textulcifrat c = 33 atunci textul in clar m corespunzator este :
A. m = 3
B. m = 15C. m = 13D. m = 11
Answer: ASection: (none)
Explanation/Reference:A
QUESTION 81Fie d exponentul de decriptare al cifrului RSA construit cu numerele prime p = 3 si q = 5. Daca exponentul decriptare este e = 7, atunci
A. d = 5B. d = 7C. d = 11D. d = 3
Answer: BSection: (none)
Explanation/Reference:B
QUESTION 82Alice alege numerele prime p=5, q=11 i exponentul de criptare e=27. Semntura RSA pentru documentulm=24 emis de Alice este s = 24d mod 55 , unde d este exponentul de decriptare corespun- ztor lui e. Avem:
A. s=13B. s=19C. s=43 sau BD. s=31
Answer: CSection: (none)
Explanation/Reference:
QUESTION 83Alice alege numerele prime p=5, q=11 i exponentul de criptare e=27. Semntura RSA pentru documentulm=15 emis de Alice este s = md mod 55 , unde d este exponentul de decriptare corespunz- tor lui e. Avem:
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A. s = 20 sau C
B. s = 13
C. s = 41D. s = 31
Answer: A
Section: (none)Explanation/Reference:
QUESTION 84Concepte de baz ce apar n criptografie :K este o multime numit spaiul cheilor . Un element al lui K este nu-mit cheie. Fiecare element e din K determin n mod unic o bijeciede la M (spaiul mesajelor) la C (spaiul textelor cifrate), notatEe, numit:
A. funcie (transformare) de criptare
B. funcie transformare
C. funcie de criptare
Answer: ASection: (none)
Explanation/Reference:
QUESTION 85Concepte de baz ce apar n criptografie :Pentru fiecare element d din spaiul cheilor K, Dd este o bijecie de la C (spaiul textelor cifrate), la M (spaiulmesajelor), numit:
A. funcie transformareB. funcie (transformare) de decriptareC. funcie de decriptare
Answer: BSection: (none)
Explanation/Reference:
QUESTION 86Concepte de baz ce apar n criptografie:A este o multime finit, numit alfabetul de definiie.M reprezint spaiul mesajelor i contine iruri de simboluri dintr-un alfabet de definiie.Elementele din M se numesc:
A. mesaje n clar
B. funcie transformare
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C. funcie (transformare) de decriptare
D. funcie de decriptare
Answer: ASection: (none)
Explanation/Reference:
QUESTION 87Concepte de baz ce apar n criptografie:Multimea C contine iruri de simboluri dintr-un alfabet de definiie care poate diferi de alfabetul lui M (spaiulmesajelor). Un element din C este numit text cifrat. Multimea C se numeste:
A. spaiul textelor cifrateB. scheme de criptare
C. mesaje n clarD. funcie (transformare) de decriptare
Answer: ASection: (none)
Explanation/Reference:
QUESTION 88Concepte de baz ce apar n criptografie:
Multimea {Ee |e din K } a functiilor de criptare i o multime cores- punzatoare D d K a functiilor de
decriptare, cu proprietatea ca pentru fiecare e din K exista o cheie unica din K astfel incat= e D E , formeaz:
A. scheme de criptare
B. spaiul textelor cifrate
Answer: ASection: (none)
Explanation/Reference:
QUESTION 89Considerai alfabetul latin din care eliminai litera de frecven redu- s W.Folosind sistemul de codificare POLYBIOS, codificai textul clar UNIVERSITATE. Alegei varianta corect dincele de mai jos
A. EACD BDEB AEDC DDBD DEAA DEAE
B. EACD BDEB AEDC DDBD DEAA DEAAC. EACD BDEB AEDD DDBD DEAA DEAE
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Explanation/Reference:
QUESTION 93Considerai cifrul de permutare al lui HILL. Pentru numrul ntreg fixat d = 2 i cheia de criptare
M = (33 22) codificarea textului clar CLAR este..(73 75)
A. NHZHB. HIWK
Answer: ASection: (none)
Explanation/Reference:
QUESTION 94Considerai sistemul de criptare bloc DES. Folosind modul de utili- zare CBC al acestuia, codificai secvena detext clar
x=1101 1101 1010 1001, folosind cheia de criptare cul initial este IV=1100.
A. 1000 1010 0000 1100
B. 1000 1010 0000 1101
C. 1000 1010 0000 1110D. 1000 1010 0000 1111
Answer: ASection: (none)
Explanation/Reference:
QUESTION 95Considerai sistemul de criptare bloc DES. Folosind modul de utilizareOFB al acestuia, codificai secvena de text clar x=1011 1101 0011 0101 folosind cheia de criptare (1 2 3 4)
( 2 3 1 4). Blocul initial este IV=1001
A. 1000 1000 1010 0111B. 1000 1000 1010 0110
C. 1000 1000 1010 1000
D. 1000 1000 1010 0111
Answer: BSection: (none)
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Explanation/Reference:
QUESTION 96Considerai sistemul de criptare bloc DES. Folosind modul de utilizare
CBC al acestuia, codificai secvena de text clar x = 1010 0101 1000 0001, folosind cheia de criptare
Blocul iniial este IV=1111.
A. 1100 0110 1011 0010
B. 1100 0110 1011 0011C. 1100 0110 1011 1011
D. 1100 0110 1011 0111
Answer: B
Section: (none)
Explanation/Reference:
QUESTION 97Considerai sistemul de criptare bloc DES. Folosind modul de utilizareECB al acestuia, codificai secvena de text clar x = 1111 0101 1010 1101, folosind cheia de criptare(1 2 3 4)(3 2 1 4)
A. 1111 0101 1010 0111
B. 1111 0101 1010 0110C. 1111 0101 1010 0100
D. 1111 0101 1010 0000
Answer: ASection: (none)
Explanation/Reference:
QUESTION 98
Considerai sistemul de criptare bloc DES. Folosind modul de utili- zare CFB al acestuia, codificai secvena detext clarx = 0001 1000 1101 1011, folosind cheia de criptare
(1 2 3 4)(4 1 2 3)
Blocul iniial este IV = 1101.
A. 1111 0111 0110 1000
B. 1111 0111 0110 1001C. 1111 0111 0110 1011
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Blocul iniial IV=0010
A. 0111 1101 1110 1000
B. 0111 1101 1110 1001C. 0111 1101 1110 1011D. 0111 1101 1110 1010
Answer: ASection: (none)
Explanation/Reference:
QUESTION 102Considerai sistemul de criptare bloc DES. Folosind modul de utili- zare OFB al acestuia, codificai secvena de
text clarx=1001 1100 0100 0001, folosind cheia de criptare
si blocul initial IV=0011
A. 0011 0000 0001 0010
B. 0011 0000 0001 0011C. 0011 0000 0001 0110
D. 0011 0000 0001 1110
Answer: ASection: (none)
Explanation/Reference:
QUESTION 103Considerai sistemul de criptare bloc DES. Folosind modul de utili- zare CFB al acestuia, codificai secvena detext clarx=0001 1000 1101 1011, folosind cheia de criptare
si blocul initial IV=1101
A. 1111 0111 0110 1000B. 1111 0111 0110 1001C. 1111 0111 0110 1010
D. 1111 0111 0110 1100
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A. 0011 0111 0101 1010B. 0011 0111 0101 1111
C. 0011 0111 0101 1011D. 0011 0111 0101 0011
Answer: CSection: (none)
Explanation/Reference:
QUESTION 107Considerai sistemul de criptare bloc DES. Folosind modul de utili- zare CBC al acestuia, codificai secvena detext clar
x = 1010 0101 1000 0001, folosind cheia de criptare
blocul initial este IV = 1111.
A. 1100 0110 1011 1011
B. 1100 0110 1011 0111C. 1100 0110 1011 0011
D. 1100 0110 1011 0010
Answer: CSection: (none)
Explanation/Reference:
QUESTION 108Considerai sistemul de criptare bloc DES. Folosind modul de utilizare OFB al acestuia, codificai secvena detext clar x=1001 1100 0100 0001, folosind cheia de criptare
(1 2 3 4 )( 3 1 4 2 )
i blocul iniial IV=0011
A. 0011 0000 0001 0010
B. 0011 0000 0001 0011C. 0011 0000 0001 1010D. 0011 0000 0001 1110
Answer: ASection: (none)
Explanation/Reference:
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QUESTION 109Considerm d exponentul de decriptare, e exponentul de criptare i p i q dou numere prime impare. Pentrup=7, q=5 i d=11, s se deter- mine e (exponentul de criptare).
A. e = 10B. e = 11C. e = 12
D. e = 13
Answer: BSection: (none)
Explanation/Reference:
QUESTION 110Considerm d exponentul de decriptare, e exponentul de criptare i p i q doua numere prime impare. Pentrup=5, q=11 i e=3 , s se determine d (exponentul de decriptare).
A. d=13B. d=14
C. d=15D. d=16
Answer: ASection: (none)
Explanation/Reference:
QUESTION 111Considerm d exponentul de decriptare, e exponentul de criptare i p i q doua numere prime impare. Pentrup=5, q=7 i e=5 , s se determine d (exponentul de decriptare).
A. d=4B. d=5C. d=6D. d=7
Answer: BSection: (none)
Explanation/Reference:
QUESTION 112Considerm e exponentul de criptare, p i q dou numere prime im- pare i n=pq. (n,e) = (15, 4) -> cheiepublic. Fiind dat n=15, e=4 i textul clar m=11, determinai textul codificat c .
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A. c=0B. c=1C. c=3D. c=4
Answer: BSection: (none)
Explanation/Reference:
QUESTION 113Considerm modul de utilizare CBC al cifrului bloc DES care aplic permutri biilor unui vector de bii delungime 4. Criptai textul clarx = 1011 0001 0100 1010 folosind cheia
( 1 2 3 4 )( 2 3 4 1 )
si blocul iniial IV = 1010
A. 0010 0110 0100 1100B. 0010 0110 0100 1101
C. 0010 0110 0100 1111
D. 0010 0110 0100 1000
Answer: BSection: (none)
Explanation/Reference:
QUESTION 114Considerm modul de utilizare ECB al cifrului bloc DES care aplic permutri biilor unui vector de bii delungime 4.Criptai textul clar x = 1011 0001 0100 1010 folosind cheia
1 2 3 4
2 3 4 1
A. 0111 0010 1000 0101
B. 0111 0010 1000 0111
C. 0111 0010 1000 1101
D. 0111 0010 1000 0100
Answer: ASection: (none)
Explanation/Reference:
QUESTION 115
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QUESTION 118Dac d este exponentul de decriptare al unui cifru RSA construit cu numerele prime p=5 i q=17 i avndexponentul de criptare e=3, atunci
A. d = 41
B. d = 42
C. d = 43D. d = 44
Answer: CSection: (none)
Explanation/Reference:
QUESTION 119Dac N este numrul tuturor cheilor cifrului DES, atunci
A. N = 256
B. N = 257
C. N = 258
D. N = 259
Answer: ASection: (none)
Explanation/Reference:
QUESTION 120Dac r este numrul rundelor cifrului DES ca cifru Feistel, atunci
A. r = 13B. r = 18C. r = 5D. r = 16 ?
Answer: DSection: (none)
Explanation/Reference:
QUESTION 121Din alfabet se scoate o liter mai putin folosit, i anume litera Q. Folosind POLYBIOS, decodificai textulcriptat DCDDBDDBDABDDBAE. Care este textul n clar?
A. STIRPIRE
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B. STIRPIRR
Answer: ASection: (none)
Explanation/Reference:
QUESTION 122El-Gamal
p = 23q = 7a = 6b = 3
B->A (B,C)=(21,11)
A. m = 5
B. m = 6
C. m = 7D. m = 8
Answer: CSection: (none)
Explanation/Reference:
QUESTION 123Fie cifrul El-Gamal asociat numrului prim p=11 i rdcinii primiti- ve modulo7, g=5. Cheia secret a lui Alice
este a=4, iar cea a lui Bob este b=7. Alice primete de la Bob textul criptat (3,7) pe care l deco- dificigsete mesajul clar ....
A. x = m(textul clar) = 10B. x = m(textul clar) = 11
C. x = m(textul clar) = 12D. x = m(textul clar) = 13
Answer: ASection: (none)
Explanation/Reference:
QUESTION 124Fie cifrul El-Gamal asociat numrului prim p=11 i rdcinii primiti- ve modulo11 g=2. Cheia secret a lui Aliceeste a=4 i a lui Bob b=7. Alice primete de la Bob textul criptat (3,7). Decodificnd se obine mesajul clar:
A. 9B. 10C. 11
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D. 12
Answer: BSection: (none)
Explanation/Reference:
QUESTION 125Fie cifrul El-Gamal asociat numrului prim p=11 i rdcinii primiti- ve modulo11 g=2. Cheia secret a lui Aliceeste a=3, iar cea a luiBob b=4. Dac Bob cripteaz mesajul n clar m=9 pentru a fi trans- mis lui Alice obine (B,C). Avem:(B,C) = (7,2)(B,C) = (5,3)(B,C) = (9,6)
A. 4B. 3C. 2
D. nici un raspuns corect
Answer: DSection: (none)
Explanation/Reference:
QUESTION 126Fie cifrul El-Gamal asociat numrului prim p=23 i rdcinii primiti- ve modulo23, g=7. Cheia secret a lui Aliceeste a=6, iar cea a lui Bob b=3. Bob cripteaza textul clar m=7 i obine (B,C). Avem:
A. (B, C)=(21, 10)
B. (B, C)=(21, 11)
C. (B, C)=(21, 12)D. (B, C)=(21, 13)
Answer: BSection: (none)
Explanation/Reference:
QUESTION 127Fie cifrul El-Gamal asociat numrului prim p=7 i rdcinii primiti- ve, modulo7, g=5. Cheia secret a lui Aliceeste a=3, iar cea a lui Bob este b=4. Dac Bob codific textul clar x=11 i l transmite lui Alice, atunci aceastaprimete codificarea ....
A. c = (c1,c2) = (2,4)
B. c = (c1,c2) = (2,5)
C. c = (c1,c2) = (2,6)D. c = (c1,c2) = (2,7)
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Answer: ASection: (none)
Explanation/Reference:
QUESTION 128Fie cifrul RSA construit cu numerele prime p, q i fie n = pq. Dac eeste exponentul de criptare, atunci:
A. gcd(e, (p - 1)(q - 1)) = 1B. gcd(e, (p - 1)(q - 1)) = 2
C. gcd(e, (p - 1)(q - 1)) = 3
D. gcd(e, (p - 1)(q - 1)) = 4
Answer: ASection: (none)
Explanation/Reference:
QUESTION 129
Fie f transformarea afin f7,12 = (7 x + 12) (mod 26) . Atunci
f -1 = ay + b . S se afle ai b .
A. a=15, b=3B. a=15, b=-3
C. a=15, b=4
D. a=15, b=-4
Answer: ASection: (none)
Explanation/Reference:
QUESTION 130
Fie m = 4 x 5 x 7 = 140, a aparine a N, 0
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Explanation/Reference:
QUESTION 131
Fie m = 4 x 9 x 5 = 180, a aparine a N, 0
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Fie S={0,1,2,,25} i n numrul functiilor de criptare afin fa,b:S- S,fa,b(x)(ax+b)mod26 unde a,b apartin S igcd(a, 26) = 1. Avem:
A. n=312B. n=676C. n=100D. n=250
Answer:Section: (none)
Explanation/Reference:
QUESTION 135Fie sistemul de criptare bloc DES, modul ECB. S se afle codarea secvenei n clar: 1001 1101 1000 1111, nbaza cheii
(1 2 3 4)( 2 3 1 4 )
A. 0011 1011 0010 1111
B. 0011 1011 0010 1110
C. 0011 1011 0010 1100D. 0011 1011 0010 1000
Answer: ASection: (none)
Explanation/Reference:
QUESTION 136Fie sistemul de criptare bloc DES, modul ECB. S se afle codarea secvenei n clar: 0001 1000 11101 1011, nbaza cheii(1 2 3 4)(4 1 2 3)i y=1101
A. 1111 0111 0110 1001
B. 1111 0111 0110 1011
C. 1111 0111 0110 1111
D. 1111 0111 0110 1000
Answer: DSection: (none)
Explanation/Reference:
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QUESTION 137Fie transformarea afin ek(x) = (7x+3)(mod 26). S se codifice cu- vntul AMURG
A. c = DJNST
B. c = DJNSS
C. c = DJSSTD. c = DJNTT
Answer: ASection: (none)
Explanation/Reference:
QUESTION 138Fie un cifru RSA cu cheia publica (n,e)=(35,5) i cheia secret d=5. Dac textul cifrat este c=3, atunci textul n
clar m=?A. m = 33
B. m = 34
C. m = 35D. m = 36
Answer: ASection: (none)
Explanation/Reference:
QUESTION 139Folosind criptarea VIGENERE, s se afle care din urmatoarele vari- ante reprezint criptarea textului clarINCAS, pe baza cheii GARA
A. ONTAA
B. ONTAYC. ONTAT
D. ONTAO
Answer: B
Section: (none)
Explanation/Reference:
QUESTION 140Folosind protocolul Diffie Hellman, Alice i Bob aleg p=7 i pe g=3 ca rdcina primitiv modulo7. Dac cheiasecret a lui Alice este a=4, cheia secret a lui Bob este b=2, iar k este cheia secret co- mun, atunci
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A. k = 1
B. k = 2
C. k = 3D. k = 4
Answer: B
Section: (none)Explanation/Reference:
QUESTION 141Folosind protocolul Diffie Hellman, Alice i Bob aleg p=7 i pe g=3 ca rdcina primitiv modulo7. Dac cheiasecret a lui Alice este a=7, cheia secret a lui Bob este b=4, iar k este cheia secret comuna, atunci:
A. k=3B. k=4
C. k=5
D. k=6
Answer: CSection: (none)
Explanation/Reference:
QUESTION 142Folosind sistemul de criptare ASINCRON cu auto-cheie pentru k=7, codificarea textului clar GRADINAeste ........
A. NEEHPCCB. NEEHCCCC. NEEHPCP
D. NEEHPCH
Answer: ASection: (none)
Explanation/Reference:
QUESTION 143Folosind sistemul de criptare ASINCRON cu auto-cheie, pentru k=7, decodificarea textului criptat ZSJRXXQeste:
A. STRIGASB. STRIGAAC. STRIGAT
D. STRIGAG
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Answer: CSection: (none)
Explanation/Reference:
QUESTION 144Folosind un sistem ADITIV FLUID BINAR de criptare se cere crip- tarea secvenei de text clar x = 101101,cunoscnd cheia fluid z=1101.
A. 011011
B. 011010C. 011000D. 011111
Answer: BSection: (none)
Explanation/Reference:
QUESTION 145Folosind un sistem de criptre asincron cu auto-cheie, pentru k = 11, codificai /decodificai textul clar SPIRUHARET.
A. DSARLSSJNNB. DSARLSSJNG
C. DSARLSSJNJ
D. DSARLSSJNS
Answer: BSection: (none)
Explanation/Reference:
QUESTION 146n sistemul de criptare Vigenere, s consideram cuvntul cheie GRUPA. Criptm cu aceast cheie textul clarCRIPTOGRAFIE i se obine textul criptat ...
A. IICETUXLPFOV
B. IICETUXLPFOOC. IICETUXLPFOA
D. IICETUXLPFOI
Answer: ASection: (none)
Explanation/Reference:
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QUESTION 147Considerai sistemul de criptare bloc DES. Folosind modul de utili- zare ECB al acestuia, codificai secvena detext clarx = 1100 1101 1001 1110, folosind cheia de criptare
(1 2 3 4 )(3 4 2 1 )
A. 0011 0111 0101 1010
B. 0011 0111 0101 1011C. 0011 0111 0101 1111
D. 0011 0111 0101 0011
Answer: BSection: (none)
Explanation/Reference:
QUESTION 148Considerai sistemul de criptare bloc DES. Folosind modul de utili- zare ECB al acestuia, codificai secvena detext clarx = 1111 0101 1010 1101, folosind cheia de criptare
(1 2 3 4)(3 2 1 4)
A. 1111 0101 1010 0111
B. 1111 0101 1010 0110C. 1111 0101 1010 0011
D. 1111 0101 1010 0101
Answer: ASection: (none)
Explanation/Reference:
QUESTION 149VIGENERE text IMPAR cheie VARF
A. DMGFD
B. DMGFM
C. DMGFFD. DMGMM
Answer: BSection: (none)
Explanation/Reference:
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QUESTION 150Sistemul Aditiv Fluid Binar x = 111001 cheia fluid z = 1001
A. 011110
B. 011100
C. 011000D.
011111
Answer: DSection: (none)
Explanation/Reference:
QUESTION 151Se d secvena binar de text clar 101011. Codificai aceasta secven- a folosind cheia fluid 1010, folosind unsistem aditiv fluid binar de criptare.
A. 000001B. 000011C. 000111
D. 000000
Answer: ASection: (none)
Explanation/Reference:
QUESTION 152Se consider secvena binara de text criptat 111001. Folosind intr-un sistem aditiv fluid binar de criptare cheiafluid secret 1001, se cere decriptarea secvenei.
A. 011000
B. 011100
C. 011110D. 011111
Answer: DSection: (none)
Explanation/Reference:
QUESTION 153Se consider p=7 (p numr prim) =5 , 3 Cheia secret a lui Alice4 Cheia secret a lui Bobm=11 textul clar de criptat de ctre Bob pentru a-l trimite lui Alice. Care este codificarea textului clar?
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A. M=10
B. M=11
C. M=12D. M=13
Answer: A
Section: (none)Explanation/Reference:
QUESTION 154Se consider p = 11 (p numr prim), = 2 (a rdcina primitiv).3 Cheia secret a lui Alice4 Cheia secret a lui Bobm = 9 textul clar de criptat de catre Bob pentru a-l trimite lui Alice. Care este codificarea textului clar?
A. (y1 , y2) = (5 ,4)
B. (y1 , y2) = (5 ,5)C. (y1 , y2) = (5 ,6)
D. (y1 , y2) = (5 ,7)
Answer: ASection: (none)
Explanation/Reference:
QUESTION 155
Se consider p = 11 , = 2.3 Cheia secret a lui Alice4 Cheia secret a lui BobAlice primete de la Bob (y1,y2) = (5,3) S se determine textul clar x.
A. X=9
B. X=8
C. X=7D. X=6
Answer: ASection: (none)
Explanation/Reference:
QUESTION 156Se consider p = 11, = 2.4 Cheia secret a lui Alice7 Cheia secret a lui BobS se determine textul clar x pentru (y1,y2) = (3,7).
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A. X=11
B. X=12
C. X=10D. X=9
Answer: C
Section: (none)Explanation/Reference:
QUESTION 157
A.B.C.D.
Answer:Section: (none)
Explanation/Reference: