/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 For which elements \(n \in\\{0,1... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For which elements \(n \in\\{0,1,2,3,4,5\\}\) does 6 divide \(n^{3}+5 n ?\)

Short Answer

Expert verified
The elements are \(n = 0, 2, 4\).

Step by step solution

01

Understand the Problem

We need to find elements \(n\) within the set \(\{0,1,2,3,4,5\}\) such that the expression \(n^3 + 5n\) is divisible by 6. This means that \(n^3 + 5n\) should equal \(6k\) for some integer \(k\).
02

Setup Expression for Divisibility

A number is divisible by 6 if it is divisible by both 2 and 3. Therefore, \(n^3 + 5n\) must be divisible by 2 and by 3. Let's test each element for these conditions.
03

Test Divisibility by 2

A number is divisible by 2 if it is even. The expression \(n^3 + 5n\) can be simplified to \(n(n^2 + 5)\). Note that \(n\) is multiplied by \((n^2 + 5)\). If \(n\) is even, \(n^3 + 5n\) is even. Test each \(n\): - \(n = 0\): Even - \(n = 1\): Odd - \(n = 2\): Even - \(n = 3\): Odd - \(n = 4\): Even - \(n = 5\): Odd.
04

Test Divisibility by 3

A number is divisible by 3 if the sum of its digits is divisible by 3. We check \(n^3 + 5n\) modulo 3 for each \(n\): - \(n = 0\): \(0^3 + 5(0) = 0 \equiv 0 \mod 3\) - \(n = 1\): \(1^3 + 5(1) = 6 \equiv 0 \mod 3\) - \(n = 2\): \(2^3 + 5(2) = 18 \equiv 0 \mod 3\) - \(n = 3\): \(3^3 + 5(3) = 54 \equiv 0 \mod 3\) - \(n = 4\): \(4^3 + 5(4) = 84 \equiv 0 \mod 3\) - \(n = 5\): \(5^3 + 5(5) = 150 \equiv 0 \mod 3\).
05

Combine Conditions for n to be divisible by 6

We combine the results of divisibility by 2 and by 3. Those \(n\) which satisfy both conditions are: - \(n = 0\): Divisible by both 2 and 3 - \(n = 2\): Divisible by both 2 and 3 - \(n = 4\): Divisible by both 2 and 3.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Divisibility Rules
Understanding divisibility rules makes checking whether one number divides another much simpler. These rules are like shortcuts that save you time.
  • Divisibility by 2: A number is divisible by 2 if it is even. This means the last digit of the number is 0, 2, 4, 6, or 8.
  • Divisibility by 3: A number is divisible by 3 if the sum of all its digits is divisible by 3.
  • Divisibility by 6: A number is divisible by 6 only if it is divisible by both 2 and 3. So, a number must be even and the sum of its digits should be divisible by 3 at the same time.
In our problem, we need to check both conditions for divisibility by 6. Once you practice these rules with a few examples, you'll be able to identify divisible numbers more quickly. They can be handy in solving many mathematical problems. Simple arithmetic with these rules can also give insights before using more involved methods.
Modulo Operation
The modulo operation helps determine the remainder after division. It's a handy tool for checking divisibility and simplifies complex calculations greatly. When we say "a modulo b" (written as \(a \equiv c \pmod{b}\)), we are interested in the remainder when dividing a by b.
For example, if you calculate \(18 \div 3\), the remainder is 0 because 18 divides by 3 evenly. Therefore, \(18 \equiv 0 \pmod{3}\). If you instead take \(17 \div 3\), the remainder is 2 thus, \(17 \equiv 2 \pmod{3}\).
In our exercise, we used modulo 2 and modulo 3 to examine if the polynomial \(n^3 + 5n\) is divisible by these numbers:
  • If \(n^3 + 5n \equiv 0 \pmod{2}\), it means the result is even, satisfying one of the divisibility conditions for 6.
  • Likewise, if \(n^3 + 5n \equiv 0 \pmod{3}\), the result is divisible by 3.
By combining conditions with modulo operation, it allowed us to test each candidate \(n\) straightforwardly.
Problem Solving Steps
Solving mathematical problems systematically can help break down what seems complex into more straightforward tasks. Let's take an example from our exercise to see how structuring your approach can lead to solutions:
  • Understand the Problem: Begin by clearly understanding what you're supposed to find. In our case, it was elements \(n\) such that \(n^3 + 5n\) is divisible by 6.
  • Break Down the Requirements: Recognize divisibility by 6 requires divisibility by both 2 and 3, prompting us to check these separately.
  • Test Each Condition: Methodically test all possible values in the given set for these conditions. For example, see which values of \(n\) make \(n^3 + 5n\) even, and which make it divisible by 3.
  • Combine Findings: Find those values satisfying all criteria. Here, only \(n = 0, 2,\) and \(4\) met both divisibility conditions.
Following these steps doesn't just answer the exercise; it builds confidence and understanding in approaching any similar problem. Each step serves as leverage to reach the next, and before you know it, a seemingly complicated problem has been tackled. \( \)

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A common use of induction is to prove various facts that seem to be fairly obvious but are otherwise awkward or impossible to prove. These frequently involve expressions with ellipses. Use induction to show that: (a) \(X \cup\left(X_{1} \cap X_{2} \cap X_{3} \cap \cdots \cap X_{3}^{n}\right)=\left(X \cup X_{1}\right) \cap\left(X \cup X_{2}\right) \cap \cdots \cap\left(X \cup X_{n}\right)\) (b) \(X \cap\left(X_{1} \cup X_{2} \cup X_{3} \cup \ldots \cup X_{n}\right)=\left(X \cap X_{1}\right) \cup\left(X \cap X_{2}\right) \cup \ldots \cup\left(X \cap X_{n}\right)\) (c) \(\overline{\left(X_{1} \cap X_{2} \cap \cdots \cap X_{n}\right)}=\overline{X_{1}} \cup \overline{X_{2}} \cup \ldots \cup \overline{X_{n}}\) (d) \(\overline{\left(X_{1} \cup X_{2} \cup \ldots \cup X_{n}\right)}=\overline{X_{1}} \cap \overline{X_{2}} \cap \ldots \cap \overline{X_{n}}\)

Let \(A=\\{1,2,3, \ldots, 10\\}, B=12,3,6,8\\},\) and \(C=(3,5,4,8,2\\} .\) Find the following: (a) \(B \cup C\) (b) \(B \cap C\) (c) \(B-C\) (d) \(A-B\) (e) \(A-C\)

Prove that in a boolean algebra, DeMorgan's Laws hold; that is, $$ \begin{array}{l} \neg(x \vee y)=\neg x \wedge \neg y \\ \neg(x \wedge y)=\neg x \vee \neg y \end{array} $$

Let \(p\) denote the proposition "Sue is a computer science major" and \(q\) denote the proposition "Sam is a physics major." Write out what the following propositions mean: (a) \(\neg q\) (b) \(q \wedge p\) (c) \(p \vee q\) (d) \(\neg q \wedge p\) (e) \(q \rightarrow p\) (f) \(p \leftrightarrow q\) (g) \(\neg q \rightarrow p\)

The enrollment for the four courses Biol212, Poli115, Econ313, and Fina215 is 108 . \(203,315,\) and \(212,\) respectively. No student is in all four of these courses. No student is in the three courses Biology 212 , Fina 215 , and Poli 115 . No student takes \(\mathrm{E} \operatorname{con} 313\) and Fina 215 in the same semester. Polit 15 and Fina 215 are not allowed in the same term. There are 39 students in both Biol212 and Poli115, and 48 students in both Polit 15 and Econ313 as well as in the two courses Biol2 12 and Econ313. Biol212, Polit 15 . and \(\mathrm{F} \operatorname{con} 313\) have a common enrollment of \(73 .\) Biol 212 and Fina 215 have a common enrollment of \(67 .\) How many different students are enrolled in these four courses?

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.