Problem 115
Question
Let \(t_{n}=\underbrace{1.1 \ldots 1}_{n \text { times }}\), then (A) \(t_{912}\) is not prime (B) \(t_{951}\) is not prime (C) \(t_{480}\) is not prime (D) \(t_{91}\) is not prime
Step-by-Step Solution
Verified Answer
(D) \( t_{91} \) is not prime.
1Step 1: Understanding the Problem
We are given a sequence where the term \( t_n \) is represented by the number 111...1, repeated \( n \) times. We need to determine the primality of the number for the given choices. Each option corresponds to a different \( n \): 912, 951, 480, and 91.
2Step 2: Checking Prime Number Definition
A prime number is only divisible by 1 and itself. To check if \( t_n \) is prime, we need to see if there are any divisors other than 1 and \( t_n \).
3Step 3: Converting Numbers to Formula
The number \( t_n \) can be expressed as \( t_n = \frac{10^n - 1}{9} \). This formula helps in analyzing its divisibility properties.
4Step 4: Checking Divisibility for Each Option
Since \( t_n = \frac{10^n - 1}{9} \), we check if \( 10^n - 1 \) is divisible by \( 9k \) for some integer \( k \), hence making \( t_n \) composite. We specifically consider divisibility by smaller primes or forms like \( 10^k - 1 \) that factorize it.
5Step 5: Analyzing Options
1. For \( t_{912} \), notice that 912 is divisible by 3 (since 9+1+2=12, and 12 is divisible by 3), and thus \( 10^{912} - 1 \) will be divisible by \( 10^3 - 1 = 999 \), making \( t_{912} \) composite.2. For \( t_{951} \), 951 is divisible by 3 (9+5+1=15, and 15 is divisible by 3), so \( 10^{951} - 1 \) is divisible by \( 10^3 - 1 \).3. For \( t_{480} \), 480 is divisible by 3 (4+8+0=12, and 12 is divisible by 3), repeating the reasoning as before.4. Check divisibility of \( 10^{91} - 1 \) by simpler factors (consider divisibility by the form \( 10^p - 1 \) for any smaller \( p \), e.g., using Mersenne-like factorization if any).
6Step 6: Identifying the Prime Option
For exactly one of these cases (\( t_{n} \)), the number turns out not to have such factorization properties suitable for making it non-prime. By identifying the outlier among \(91, 912, 951, 480\), we find that \( t_{91} \) doesn't immediately present obvious small divisibility, hinting toward it possibly being a prime number.
Key Concepts
DivisibilityComposite NumbersNumber Representation
Divisibility
Divisibility is a central concept in understanding numbers. It refers to the ability of a number to be divided by another number without leaving a remainder. For instance, 10 is divisible by 2 because when 10 is divided by 2, the result is an integer, 5, without any remainder.
In the context of prime numbers, divisibility is crucial. A prime number is defined as one that can only be divided evenly by 1 and itself. To check if a number, like the ones in our problem, is prime, we examine if it has any divisors other than 1 and the number itself.
For the sequence given by the formula \( t_n = \frac{10^n - 1}{9} \), determining primality involves checking if the resulting number is divisible by smaller primes. For instance:
In the context of prime numbers, divisibility is crucial. A prime number is defined as one that can only be divided evenly by 1 and itself. To check if a number, like the ones in our problem, is prime, we examine if it has any divisors other than 1 and the number itself.
For the sequence given by the formula \( t_n = \frac{10^n - 1}{9} \), determining primality involves checking if the resulting number is divisible by smaller primes. For instance:
- If \( 10^n - 1 \) is divisible by any factor other than itself, then \( t_n \) is not prime.
- Understanding divisibility helps identify properties such as composite numbers which have additional divisors.
Composite Numbers
Composite numbers are the opposite of prime numbers. A composite number is any integer greater than one that is not prime, meaning it has divisors other than 1 and itself.
To identify composite numbers we look for number representations that allow factors and divisors.
For our task, where each number \( t_n \) is represented by repeating '1' for \( n \) times, checking divisibility by smaller primes, like 3 or 5, helps to quickly determine composite nature.
To identify composite numbers we look for number representations that allow factors and divisors.
For our task, where each number \( t_n \) is represented by repeating '1' for \( n \) times, checking divisibility by smaller primes, like 3 or 5, helps to quickly determine composite nature.
- If \( n \) is divisible by a smaller number like 3, the resulting \( 10^n - 1 \) often becomes divisible too. For example, 912 and 951 being divisible by 3 suggests their composite status.
- Composite numbers reveal themselves when the formula \( \frac{10^n - 1}{9} \) yields non-prime through factorization.
Number Representation
Number representation refers to how numbers are expressed or formatted. It plays a vital role in problems dealing with sequences and patterns.
In our problem, the number \( t_n = \underbrace{111\ldots1}_{n\text{ times}} \) is efficiently represented by the formula \( t_n = \frac{10^n - 1}{9} \). This concise representation aids in exploring its properties, such as divisibility and primality.
In our problem, the number \( t_n = \underbrace{111\ldots1}_{n\text{ times}} \) is efficiently represented by the formula \( t_n = \frac{10^n - 1}{9} \). This concise representation aids in exploring its properties, such as divisibility and primality.
- The representation \( \frac{10^n - 1}{9} \) splits a repetitive number into a numeric expression, simplifying complex calculations.
- This representation uncovers the structure and factors of numbers, allowing easier evaluation of their property, whether prime or composite.
- Understanding the representation aids in tackling diverse mathematical problems efficiently.
Other exercises in this chapter
Problem 113
Let \(\left(1+x^{2}\right)^{2}(1+x)^{n}=\sum_{k=0}^{n+4} a_{k} x^{k}\). If \(a_{1}, a_{2}, a_{3}\), are in A.P., then \(n\) is equal to (A) 1 (B) 2 (C) 3 (D) 4
View solution Problem 114
If \(a, b, c\) are non-zero real numbers such that 3 \(\left(a^{2}+b^{2}+c^{2}+1\right)=2(a+b+c+a b+b c+c a)\), then, \(a, b, c\) are in (A) A.P. (B) G. P. (C)
View solution Problem 118
Sum to \(n\) terms of the series \(\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\ldots .\) is (A) \(\frac{n}{2 n+1}\) (B) \(\frac{n}{2 n-1}\) (C)
View solution Problem 119
Sum to infinite terms of the series \(\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\ldots .\) is (A) \(\frac{1}{4}\) (B) \(\frac{1}{3}\) (C) \(\f
View solution