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Chapter 1: Q25E (page 49)

Calculate 2125mod127using any method you choose. (Hint: 127 is prime.)

Short Answer

Expert verified

2125mod127is equal to

Step by step solution

01

Given condition

To find the modulo of any large number we can divide that number into small parts and find the modulo of the smaller part and recognized the pattern to get the answer of the large integer.

02

Calculate 2125mod127

For,2125mod127, we can write2125as

2125=2717+6

Where27mod127 can be written as,

27mod127=128mod127=1

Then, for 2125mod127, we have,

2125mod127=2717+6mod127=2717.26mod127=117.26mod127=64mod127=64

So,2125mod127 is equal to 64.

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Most popular questions from this chapter

A positive integer N is a power if it is of the formqk , where q,role="math" localid="1658399000008" k are positive integers and k>1.

(a) Give an efficient algorithm that takes as input a number and determines whether it is a square, that is, whether it can be written asq2 for some positive integer q. What is the running time of your algorithm?

(b) Show that if N=qk (with role="math" localid="1658399171717" N,q , andk all positive integers), then either role="math" localid="1658399158890" k≤logNorN=1.

(c) Give an efficient algorithm for determining whether a positive integerN is a power. Analyze its running time.

On page 38, we claimed that since about a 1nfraction of n-bit numbers are prime, on average it is sufficient to draw O(n)random n -bit numbers before hitting a prime. We now justify this rigorously. Suppose a particular coin has a probability p of coming up heads. How many times must you toss it, on average, before it comes up heads? (Hint: Method 1: start by showing that the correct expression is∑i=1∞i(1-p)i-1p . Method 2: if E is the average number of coin tosses, show that E=1+(1-p)E).

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.

In the RSA cryptosystem, Alice’s public key (N,e)is available to everyone. Suppose that her private key d is compromised and becomes known to Eve. Show that e=3if (a common choice) then Eve can efficiently factor N.

Show that if a≡b(modN)and if M Divides Nthena≡b(modM)

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