   Chapter 8.4, Problem 13ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
1 views

# a. Prove that for every integer n ≥ 1 , 10 n ≡ ( − 1 ) n ( mod   11 ) . b. Use part (a) to prove that a positive integer is divisible by 11 if, and only if, the alternating sum of its digits is divisible by 11. (For instance, the alternating sum of the digits of 82,379 is 8 − 2 + 3 − 7 + 9 = 11 and 82 , 379 = 11.7489. )

To determine

(a)

To prove:

The statement 10n(1)n(mod11) for all integers n1 ,

Explanation

Given information:

The given statement is 10n(1)n(mod11).

Formula used:

Division algorithm:

Let a be an integer and d be a positive integer. Then, there are unique integers qandr with 0r<d such that a=dq+r, here, q is the quotient and r is the remainder.

When m divides ab then a is congruent to b modulo m. The notation is ab(modm).

The base b representation of n is aka2a1a0, then n=akbk+ak1bk1++a1b+a0.

a divides b if there exists an integer c such that b=ac. The notation is a|b.

Proof:

To prove the statement 10n(1)n(mod11), by using mathematical induction on n.

Let P(n) be 10n(1)n(mod11).

For n=1 ,

By the definition, 11 divides 101(1)1.

101(1)1=10+1=11

11|(101( 1)1)

Here, 1 is an integer.

Thus, P(1) is true.

Let the statement 10n(1)n(mod11) is true for m=k.

So, by the definition of divides, there exists an integer l such that, 10k(1)k=11l

To determine

(b)

To prove:

A positive integer is divisible by 11 if and only if the alternating sum of its digits is divisible by 11.

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 