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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

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

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鈮�0(modpq), we proved the formula role="math" localid="1658726171933" a(p-1)鈮�1(modpq)in Section 1.4.2.)

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).

Show that if a has a multiplicative inverse modulo N, then this inverse is unique (modulo N).

1.38. To see if a number, say 562437487, is divisible by 3, you just add up the digits of its decimalrepresentation, and see if the result is divisible by role="math" localid="1658402816137" 3.

( 5+6+2+4+3+7+4+8+7=46, so it is not divisible by 3).

To see if the same number is divisible by 11, you can do this: subdivide the number into pairs ofdigits, from the right-hand end(87,74,43,62,5) , add these numbers and see if the sum is divisible by11 (if it's too big, repeat).

How about 37? To see if the number is divisible by 37, subdivide it into triples from the end(487,437,562) add these up, and see if the sum is divisible by37 .


This is true for any prime pother than2 and 5. That is, for any prime 辫鈮�2,5, there is an integer rsuch that in order to see ifp divides a decimal number n, we breakn into r-tuples of decimal digits (starting from the right-hand end), add up these r-tuples, and check if the sum is divisible by p.

(a) What is the smallest rsuch for p=13? Forp=13 ?

(b) Show thatr is a divisor of p-1.
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