Check if Number is Prime in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a prime number?

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

intermediate

Why do we only check divisors up to the square root of a number when testing for primality?

Because if a number has a divisor larger than its square root, it must also have a smaller divisor below the square root. So checking up to the square root is enough to find any divisor.

beginner

What is the simplest way to check if a number n is prime?

Check if n is divisible by any number from 2 up to sqrt(n). If none divide n evenly, then n is prime.

beginner

In Go, which loop structure is commonly used to check divisors for primality?

A for loop is used to iterate from 2 to sqrt(n) to check if n is divisible by any number in that range.

beginner

What should the function return if the input number is less than 2?

It should return false because numbers less than 2 are not prime by definition.

What is the smallest prime number?

A 1

B 3

C 2

D 0

2 is the smallest prime number because it is greater than 1 and divisible only by 1 and itself.

When checking if 29 is prime, up to which number should you check divisors?

A 5

B 6

C 29

D 14

Check divisors up to the square root of 29, which is approximately 5.38, so check up to 5.

If a number n > 3 is divisible by 3, what can you say about n?

A n is prime

B n is not prime

C n is even

D n is less than 3

Since n > 3 and is divisible by 3, it has a divisor other than 1 and itself, so n is not prime.

Which of these numbers is NOT prime?

A 21

B 19

C 17

D 23

21 is divisible by 3 and 7, so it is not prime.

What should a prime checking function return for input 1?

A undefined

B true

C error

D false

1 is not a prime number, so the function should return false.

Explain how to check if a number is prime using a simple algorithm.

Think about dividing the number by smaller numbers to find if it has any divisors.

Describe why checking divisors only up to the square root of a number is enough to determine primality.

Consider how multiplication pairs work for numbers.

Practice

1. What does it mean for a number to be prime?

easy

A. It has more than two divisors.

B. It is divisible by 2 only.

C. It is an even number greater than 2.

D. It has exactly two distinct positive divisors: 1 and itself.

Solution

1. Step 1: Understand the definition of prime number

   A prime number is a number greater than 1 that has no divisors other than 1 and itself.

2. Step 2: Compare options with definition

   It has exactly two distinct positive divisors: 1 and itself. matches this definition exactly, while others describe incorrect properties.

3. Final Answer:

   It has exactly two distinct positive divisors: 1 and itself. -> Option D

4. Quick Check:

   Prime number = two divisors only [OK]

Hint: Prime means only divisible by 1 and itself [OK]

Common Mistakes:

• Thinking even numbers are always prime
• Confusing prime with odd numbers
• Believing 1 is prime

2. Which of the following Go code snippets correctly checks if a number n is prime?

easy

A. for i := 2; i < n; i++ { if n%i == 0 { return false } } return true

B. for i := 2; i <= n; i++ { if n%i == 0 { return false } } return true

C. for i := 1; i < n; i++ { if n%i == 0 { return false } } return true

D. for i := 2; i < n/2; i++ { if n%i == 0 { return true } } return false

Solution

1. Step 1: Analyze loop range for prime check

   We check divisors from 2 up to but not including n to find if n is divisible by any number other than 1 and itself.

2. Step 2: Evaluate each option's correctness

   for i := 2; i < n; i++ { if n%i == 0 { return false } } return true correctly loops from 2 to n-1 and returns false if divisible. for i := 2; i <= n; i++ { if n%i == 0 { return false } } return true incorrectly includes n itself, always returning false. for i := 1; i < n; i++ { if n%i == 0 { return false } } return true starts from 1, which divides all numbers. for i := 2; i < n/2; i++ { if n%i == 0 { return true } } return false returns true on divisor found, which is wrong logic.

3. Final Answer:

   for i := 2; i < n; i++ { if n%i == 0 { return false } } return true -> Option A

4. Quick Check:

   Loop 2 to n-1 and check divisibility [OK]

Hint: Loop from 2 to n-1, return false if divisible [OK]

Common Mistakes:

• Checking divisibility including n itself
• Starting loop from 1 instead of 2
• Returning true when divisor found

3. What will be the output of this Go function call: isPrime(11)?

func isPrime(n int) bool {
 if n <= 1 {
  return false
 }
 for i := 2; i*i <= n; i++ {
  if n%i == 0 {
   return false
  }
 }
 return true
}

medium

A. true

B. false

C. runtime error

D. infinite loop

Solution

1. Step 1: Check input and loop condition

   Input n=11 is greater than 1, so function proceeds. Loop runs from i=2 while i*i <= 11 (i <= 3).

2. Step 2: Test divisibility inside loop

   Check i=2: 11%2 != 0; i=3: 11%3 != 0. No divisor found, loop ends.

3. Final Answer:

   true -> Option A

4. Quick Check:

   11 is prime, function returns true [OK]

Hint: Check divisors up to sqrt(n) for efficiency [OK]

Common Mistakes:

• Not handling n <= 1 correctly
• Looping beyond sqrt(n)
• Assuming 11 is not prime

4. Identify the bug in this Go function that checks if a number is prime:

func isPrime(n int) bool {
 if n <= 1 {
  return false
 }
 for i := 2; i < n/2; i++ {
  if n%i == 0 {
   return false
  }
 }
 return true
}

medium

A. Loop should run until i <= n, not i < n/2

B. Function should return true if n <= 1

C. Loop should run until i*i <= n for efficiency and correctness

D. The modulo operator % is used incorrectly

Solution

1. Step 1: Analyze loop range and condition

   The loop runs from 2 to less than n/2, which is inefficient and may miss divisors for small n.

2. Step 2: Suggest correct loop condition

   Looping until i*i <= n is both correct and efficient because any divisor larger than sqrt(n) pairs with a smaller divisor already checked.

3. Final Answer:

   Loop should run until i*i <= n for efficiency and correctness -> Option C

4. Quick Check:

   Use i*i <= n to check divisors [OK]

Hint: Check divisors up to sqrt(n) to fix bugs [OK]

Common Mistakes:

• Looping only to n/2 misses some divisors
• Returning true for n <= 1
• Misusing modulo operator

5. You want to find all prime numbers up to 20 using Go. Which approach is best to optimize performance?

hard

A. Check divisibility for each number up to n-1 without optimization

B. Check each number from 2 to 20 using a function that tests divisibility up to sqrt(n)

C. Use a loop from 2 to 20 and mark all numbers as prime without checks

D. Only check even numbers for primality since odd numbers are not prime

Solution

1. Step 1: Understand prime checking for multiple numbers

   Checking each number from 2 to 20 individually is necessary to find all primes.

2. Step 2: Optimize divisibility checks

   Testing divisibility only up to sqrt(n) for each number reduces unnecessary checks and improves performance.

3. Final Answer:

   Check each number from 2 to 20 using a function that tests divisibility up to sqrt(n) -> Option B

4. Quick Check:

   Use sqrt(n) limit for each number check [OK]

Hint: Test divisibility up to sqrt(n) for each number [OK]

Common Mistakes:

• Checking all numbers up to n-1 wastes time
• Ignoring odd numbers which can be prime
• Marking all numbers prime without checks