In the Sieve of Eratosthenes for numbers less than 100, explain why, after crossing out all the multiples of 2, 3, 5, and 7, the remaining numbers are prime.

Choose the correct answer below.

AThe Sieve of Eratosthenes is meant to find prime numbers; it will not identify composite numbers.
BThe next prime number after 7 is 11. The smallest composite number which does not have $2,3,5$, or 7 as a factor is $11^2$, which is greater than 100.
CThe product $2 \cdot 3 \cdot 5 \cdot 7$ is greater than 100, so the Sieve will have crossed out all of the composite numbers less than 100.
DThe Sieve of Eratosthenes is used to find large prime numbers; its results will be inconsistent when used on small numbers, such as for the numbers less than 100.

Question

addisonlong04 addisonlong04

31-10-2026
Mathematics

Answered

In the Sieve of Eratosthenes for numbers less than 100, explain why, after crossing out all the multiples of 2, 3, 5, and 7, the remaining numbers are prime.

Choose the correct answer below.

AThe Sieve of Eratosthenes is meant to find prime numbers; it will not identify composite numbers.
BThe next prime number after 7 is 11. The smallest composite number which does not have $2,3,5$, or 7 as a factor is $11^2$, which is greater than 100.
CThe product $2 \cdot 3 \cdot 5 \cdot 7$ is greater than 100, so the Sieve will have crossed out all of the composite numbers less than 100.
DThe Sieve of Eratosthenes is used to find large prime numbers; its results will be inconsistent when used on small numbers, such as for the numbers less than 100.

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-10-31T12:03:08-04:00

A composite number $n$ must have a prime factor $p \le \sqrt{n}$.
For $n \le 100$, all composite numbers must have a prime factor $\le \sqrt{100} = 10$.
The prime numbers $\le 10$ are 2, 3, 5, and 7. Crossing out their multiples eliminates all composite numbers $\le 100$.
The smallest composite number without these factors is $11^2 = 121$, which is greater than 100, confirming the remaining numbers must be prime.
The correct option is B. $\boxed{B}$

Explanation

Understanding Composite Numbers and Their Factors

To determine why crossing out multiples of 2, 3, 5, and 7 leaves only prime numbers for values less than 100, we must examine the properties of composite numbers. A composite number $n$ is a positive integer greater than 1 that is not prime. By definition, any composite number $n$ must have at least one prime factor $p$ such that $p \le \sqrt{n}$.

Applying the Prime Factor Bound

We are interested in identifying all composite numbers up to 100. According to the property mentioned above, any composite number $n \le 100$ must have a prime factor $p \le \sqrt{100} = 10$. Therefore, to eliminate all composite numbers up to 100, we only need to remove the multiples of all prime numbers less than or equal to 10. The prime numbers less than or equal to 10 are 2, 3, 5, and 7. Once we have removed all multiples of these numbers, any remaining number (greater than 1) cannot have any prime factors less than or equal to 10.

Conclusion and Identification of the Correct Option

If a remaining number $n \le 100$ had a prime factor, that prime factor would have to be greater than 10. The smallest prime number greater than 10 is 11. If a number $n$ were composite and had no prime factors $\le 10$, its smallest prime factor would be at least 11. Consequently, the smallest such composite number would be the product of its smallest prime factors, which is $11 \times 11 = 121$. Since 121 is greater than 100, there are no composite numbers less than or equal to 100 that do not have 2, 3, 5, or 7 as a prime factor. Thus, all numbers remaining after crossing out the multiples of 2, 3, 5, and 7 must be prime. This aligns perfectly with Option B.

Examples

The Sieve of Eratosthenes is a fundamental algorithm used in computer science for generating prime numbers efficiently. For example, if you are developing a cybersecurity application that requires generating large prime numbers for encryption (like in RSA algorithm), you would use optimized variations of this sieve to quickly identify candidates. Understanding why the sieve stops at the square root of the limit is essential for optimizing the algorithm's performance, as it avoids unnecessary computations for larger potential factors.