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  • Exam Code: C1000-065
  • Exam Name: IBM Cognos Analytics Developer V11.1.x
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NEW QUESTION: 1
The RSA Algorithm uses which mathematical concept as the basis of its encryption?
A. PI (3.14159...)
B. Two large prime numbers
C. 16-round ciphers
D. Geometry
Answer: B
Explanation:
Explanation/Reference:
Source: TIPTON, et. al, Official (ISC)2 Guide to the CISSP CBK, 2007 edition, page 254.
And from the RSA web site, http://www.rsa.com/rsalabs/node.asp?id=2214 :
The RSA cryptosystem is a public-key cryptosystem that offers both encryption and digital signatures (authentication). Ronald Rivest, Adi Shamir, and Leonard Adleman developed the RSA system in 1977
[RSA78]; RSA stands for the first letter in each of its inventors' last names.
The RSA algorithm works as follows: take two large primes, p and q, and compute their product n = pq; n is called the modulus. Choose a number, e, less than n and relatively prime to (p-1)(q-1), which means e and (p-1)(q-1) have no common factors except 1. Find another number d such that (ed - 1) is divisible by (p-1)(q-1). The values e and d are called the public and private exponents, respectively. The public key is the pair (n, e); the private key is (n, d). The factors p and q may be destroyed or kept with the private key.
It is currently difficult to obtain the private key d from the public key (n, e). However if one could factor n into p and q, then one could obtain the private key d. Thus the security of the RSA system is based on the assumption that factoring is difficult. The discovery of an easy method of factoring would "break" RSA (see Question 3.1.3 and Question 2.3.3).
Here is how the RSA system can be used for encryption and digital signatures (in practice, the actual use is slightly different; see Questions 3.1.7 and 3.1.8):
Encryption
Suppose Alice wants to send a message m to Bob. Alice creates the ciphertext c by exponentiating: c = me mod n, where e and n are Bob's public key. She sends c to Bob. To decrypt, Bob also exponentiates:
m = cd mod n; the relationship between e and d ensures that Bob correctly recovers m. Since only Bob knows d, only Bob can decrypt this message.
Digital Signature
Suppose Alice wants to send a message m to Bob in such a way that Bob is assured the message is both authentic, has not been tampered with, and from Alice. Alice creates a digital signature s by exponentiating: s = md mod n, where d and n are Alice's private key. She sends m and s to Bob. To verify the signature, Bob exponentiates and checks that the message m is recovered: m = se mod n, where e and n are Alice's public key.
Thus encryption and authentication take place without any sharing of private keys: each person uses only another's public key or their own private key. Anyone can send an encrypted message or verify a signed message, but only someone in possession of the correct private key can decrypt or sign a message.

NEW QUESTION: 2
ポイントツーポイントipv6接続の最も効率的なサブネットマスクは何ですか?
A. / 32
B. / 64
C. / 128
D. / 127
E. / 48
Answer: C
Explanation:
Explanation
ref : https://tools.ietf.org/html/rfc6164

NEW QUESTION: 3
ユーザーはhttps://myapps.microsoft.comにサインインしようとすると、次のメッセージを受け取ります。
「あなたのサインインはブロックされました。このサインインについて何か異常なものを検出しました。たとえば、新しい位置情報デバイスまたはアプリからサインインしている可能性があります。続行する前に、本人確認が必要です。管理者に連絡してください。」ユーザーがサインインできないのはどの構成ですか?
A. セキュリティとコンプライアンスのデータ損失防止(DIP)ポリシー
B. セキュリティとコンプライアンスの監督ポリシー
C. Microsoft Azure Active Directory(Azure AD)ID保護ポリシー
D. Microsoft Azure Active Directory(Azure AD)条件付きアクセスポリシー
Answer: D
Explanation:
参照:
https://docs.microsoft.com/en-us/azure/active-directory/conditional-access/overview

NEW QUESTION: 4
A customer wants to automate some tasks within a private cloud. The customer previously deployed
VMfocused infrastructures but now prefer a solution that meets these requirements.
* Enables better performance and lower overhead than VMs
* Enables environment abstraction at the OS level
* Boots in milliseconds
* Offers stable, maximum performance independent of workloads
Which technology satisfies these requirements?
A. container-based virtualization
B. higher-performance VMs
C. Vblock
D. KVM
E. FlexPod
Answer: A

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