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Chapter 1: Q26E (page 49)

What is the least significant decimal digit of (1717)17? (Hint: For distinct primesp,q, and any a is not equal to role="math" localid="1658726105638" a≢0(modpq), we proved the formula role="math" localid="1658726171933" a(p-1)≡1(modpq)in Section 1.4.2.)

Short Answer

Expert verified

The least significant decimal digit can be categorized as the rightmost digit of the decimal number written in scientific notation having mod of 10 .

Step by step solution

01

Calculate (1717)17 mod 1

We have to find least significant digit of (1717)17.

For that we need to find out modulo of (1717)17mod10.

Factors of 10 are,1,2,5,10 .

Prime number between the factor of 10 are 2 and 5.10=2,5

Formula we need to use is,

ap-1q-1=1modpq

02

Calculations

Given that,

a=17,p=2 , and q=5.

By putting these value in the formula, we get

172-15-1=1mod10174=mod10

1717can also be written as

1717=42+117

So, 171717mod10can be written as,

174×cmod10×17mod10=7

Where, C is constant.

The least significant decimal digit of 171717 is 7.

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Most popular questions from this chapter

Digital signatures, continued.Consider the signature scheme of Exercise 1.45.

(a) Signing involves decryption, and is therefore risky. Show that if Bob agrees to sign anything he is asked to, Eve can take advantage of this and decrypt any message sent by Alice to Bob.

(b) Suppose that Bob is more careful, and refuses to sign messages if their signatures look suspiciously like text. (We assume that a randomly chosen message−that is, a random number in the range{1,...,N-1} is very unlikely to look like text.) Describe a way in which Eve can nevertheless still decrypt messages from Alice to Bob, by getting Bob to sign messages whose signatures look random.

Unlike a decreasing geometric series, the sum of the1,12,13,14,15,..... diverges; that is,∑i=1n1i=∞

It turns out that, for large n , the sum of the first n terms of this series can be well approximated as

∑i=1n1i≈Inn+y

where is natural logarithm (log base e=2.718...) and y is a particular constant 0.57721...... Show that

∑i=1n1i=θ(logn)

(Hint: To show an upper bound, decrease each denominator to the next power of two. For a lower bound, increase each denominator to the next power of 2 .)

The Fibonacci numbers F0,F1,...are given by the recurrenceFn+1=Fn+Fn-1 ,F0=0,F1=1. Show that for anyn≥1,gcd(Fn+1,Fn)=1.

The grade-school algorithm for multiplying two n-bit binary numbers x and y consist of addingtogethern copies of r, each appropriately left-shifted. Each copy, when shifted, is at most 2n bits long.
In this problem, we will examine a scheme for adding n binary numbers, each m bits long, using a circuit or a parallel architecture. The main parameter of interest in this question is therefore the depth of the circuit or the longest path from the input to the output of the circuit. This determines the total time taken for computing the function.
To add two m-bit binary numbers naively, we must wait for the carry bit from position i-1before we can figure out the ith bit of the answer. This leads to a circuit of depthΟ(m). However, carry-lookahead circuits (seewikipedia.comif you want to know more about this) can add inΟ(logn)depth.

  1. Assuming you have carry-lookahead circuits for addition, show how to add n numbers eachm bits long using a circuit of depth Ο(lognlogm).
  2. When adding three m-bit binary numbers x+y+z, there is a trick we can use to parallelize the process. Instead of carrying out the addition completely, we can re-express the result as the sum of just two binary numbersr+s, such that the ith bits of r and s can be computedindependently of the other bits. Show how this can be done. (Hint: One of the numbers represents carry bits.)
  3. Show how to use the trick from the previous part to design a circuit of depthΟ(logn)for multiplying two n-bit numbers.

Wilson's theorem says that a numberis prime if and only if
(N-1)!=-1(modN).

(a) If is prime, then we know every number1≤x<p is invertible modulo . Which of thesenumbers is their own inverse?
(b) By pairing up multiplicative inverses, show thatrole="math" localid="1658725109805" (p-1)!=-1(modp) for prime p.
(c) Show that if N is not prime, then(N-1)!≢(modN) .(Hint: Considerd=gcd(N,(N-1)!.)
(d) Unlike Fermat's Little Theorem, Wilson's theorem is an if-and-only-if condition for primality. Why can't we immediately base a primality test on this rule?
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