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Chapter 1: Q21E (page 49)

How many integers modulo113 have inverses?(Note:113=1331)

Short Answer

Expert verified

The number of integers coprime to113 are 1210

Step by step solution

01

Explain Inverse modulo

The modular multiplicative inverse can be defined as the GCD(a,b)must be equal to 1 and a, b are relatively prime.

02

Calculate the number of integers

Consider that the numbers1,11,121,1331 are the factors of 113. It is given that 113=1331, the number of integers can be calculated as follows,

113-1-112-1=1331-1-121-1=1330-120=1210

Therefore, The number of integers coprime to 113 are 1210

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Most popular questions from this chapter

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鈭�i=1鈭�i(1-p)i-1p . Method 2: if E is the average number of coin tosses, show that E=1+(1-p)E).

Determine necessary and sufficient conditions on xandc so that the following holds: for anya,b, if ax鈮�bxmodc, thena鈮�bmodc .

Show that any binary integer is at most four times as long as the corresponding decimal integer. For very large numbers, what is the ratio of these two lengths, approximately?

1.36. Square roots. In this problem, we'll see that it is easy to compute square roots modulo a prime pwith p鈮�3(mod4).
(a) Suppose p鈮�3(mod4). Show that(p+1)/4 is an integer.
(b) We say x is a square root of a modulo p if a=x2(modp). Show that if p鈮�3(mod4)and if a has a square root modulo p, thena(p+1)/4 is such a square root.

Prove or disprove: If a has an inverse modulo b, then b has an inverse modulo a.
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