B.Converting a decimal number into binary notation
C.Writing a number as a sum of consecutive integers
D.Writing a number as a product of smaller integers
Correct Answer: Writing a number as a product of smaller integers
Explanation:
Factorization expresses a number as a product of its factors.
Incorrect! Try again.
7After dividing out all prime factors up to , what does a remaining value greater than represent?
Finding prime factors by taking the square root
Easy
A.An additional composite factor that must always be ignored
B.An additional prime factor
C.The number of prime factors
D.The square root of the original number
Correct Answer: An additional prime factor
Explanation:
If the remaining value is greater than , it is a prime factor of the original number.
Incorrect! Try again.
8Which prime factor should be tested first when factorizing by trial division?
Finding prime factors by taking the square root
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Trial division normally begins with the smallest prime, , and is divisible by .
Incorrect! Try again.
9What is the main purpose of binary exponentiation?
Binary Exponentiation
Easy
A.To factor every number efficiently
B.To convert every exponent into a prime number before multiplication
C.To list every prime efficiently
D.To compute powers efficiently
Correct Answer: To compute powers efficiently
Explanation:
Binary exponentiation computes using repeated squaring in time.
Incorrect! Try again.
10Which identity is used when the exponent is even in binary exponentiation?
Binary Exponentiation
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
For an even exponent, the power can be halved and the result squared.
Incorrect! Try again.
11According to Fermat's little theorem, if is prime and is not divisible by , which congruence holds?
Fermat method
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Fermat's little theorem states that under these conditions.
Incorrect! Try again.
12What kind of result does the Fermat primality method generally provide?
Fermat method
Easy
A.A complete prime factorization
B.A probable-prime result
C.A guaranteed proof for every number after testing only one randomly selected base
D.A sorted list of divisors
Correct Answer: A probable-prime result
Explanation:
Some composite numbers can pass Fermat tests, so the method generally identifies probable primes.
Incorrect! Try again.
13What does the Sieve of Eratosthenes find?
Sieve of Eratosthenes
Easy
A.Only the largest prime below a limit
B.Only prime palindromes up to a limit
C.All primes up to a given limit
D.All factors of one given number
Correct Answer: All primes up to a given limit
Explanation:
The sieve marks composite numbers and leaves all prime numbers up to the chosen limit.
Incorrect! Try again.
14When processing a prime in the standard sieve, from which value can its composite multiples be marked?
Sieve of Eratosthenes
Easy
A.
B.
C., because every smaller multiple must be prime
D.
Correct Answer:
Explanation:
Multiples below have already been marked by smaller prime factors.
Incorrect! Try again.
15What is the segmented sieve mainly used for?
Segmented Sieve
Easy
A.Finding the binary form of every number
B.Finding primes within an interval
C.Computing one power with repeated squaring
D.Checking whether every number in memory has exactly two prime factors
Correct Answer: Finding primes within an interval
Explanation:
A segmented sieve efficiently finds primes in a specified range without storing all values up to .
Incorrect! Try again.
16Which primes are needed to mark composites in a segment ending at ?
Segmented Sieve
Easy
A.Primes from to
B.Only primes greater than
C.Every prime up to
D.Primes up to
Correct Answer: Primes up to
Explanation:
Every composite value at most has a prime factor no greater than .
Incorrect! Try again.
17Which technique is most useful when the values in Mansi's series involve repeatedly computing large powers?
Mansi and her series
Easy
A.Bubble sort
B.Linear search
C.A segmented sieve that stores every integer from to the largest possible exponent
D.Binary exponentiation
Correct Answer: Binary exponentiation
Explanation:
Binary exponentiation evaluates large powers efficiently using repeated squaring.
Incorrect! Try again.
18If a term of Mansi's series is defined as , what is its value for ?
Mansi and her series
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
.
Incorrect! Try again.
19If pens are divided into equal collections of pens each, how many collections are formed?
Collections of Pens
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The number of collections is .
Incorrect! Try again.
20What is the smallest prime palindrome greater than ?
Next prime palindrome
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
reads the same in both directions and is also a prime number.
Incorrect! Try again.
21Which statement correctly classifies the number ?
Introduction to Primality Testing
Medium
A. is a prime number
B. is neither prime nor composite
C. is a composite number
D. is both prime and composite
Correct Answer: is neither prime nor composite
Explanation:
A prime number has exactly two positive divisors, while a composite number has more than two. The number has only one positive divisor.
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22To test whether is prime using the method, which divisors are sufficient to check?
O(sqrt(n)) Algorithm for Primality Testing
Medium
A.Only the divisors and
B.
C.
D.Every number from to
Correct Answer:
Explanation:
Since , it is sufficient to test possible divisors up to . Testing the primes and is enough.
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23Why can a primality test stop after checking divisors up to ?
O(sqrt(n)) Algorithm for Primality Testing
Medium
A.The square root contains all prime numbers
B.Every number has a divisor below
C.Numbers above are always prime
D.A factor pair cannot both exceed
Correct Answer: A factor pair cannot both exceed
Explanation:
If and both and were greater than , then , which is impossible. Therefore, any composite number has at least one factor at most .
Incorrect! Try again.
24What is the prime factorization of ?
Factorization of a number
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
.
Incorrect! Try again.
25During trial-division factorization, after all possible factors up to have been removed, what does a remaining value greater than represent?
Factorization of a number
Medium
A.An invalid intermediate value
B.A repeated factor of the original divisor
C.A guaranteed composite factor
D.A remaining prime factor
Correct Answer: A remaining prime factor
Explanation:
After all smaller possible factors are removed, any remaining value greater than cannot have an undiscovered factor below its square root. It is therefore prime.
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26Using trial division up to the square root, what are the prime factors of ?
Finding prime factors by taking the square root
Medium
A. and
B. and
C. and
D. and
Correct Answer: and
Explanation:
. Testing primes up to this value finds .
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27What is the main advantage of dividing out a factor repeatedly during prime factorization?
Finding prime factors by taking the square root
Medium
A.It eliminates the need to test the square root
B.It prevents the same prime factor from being counted repeatedly
C.It guarantees that all divisors are prime
D.It reduces the number before testing later factors
Correct Answer: It reduces the number before testing later factors
Explanation:
Repeatedly dividing by a found factor removes its complete contribution and makes the remaining number smaller, reducing later trial divisions.
Incorrect! Try again.
28Using modular arithmetic, what is the value of ?
Binary Exponentiation
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Since , we have .
Incorrect! Try again.
29What is the time complexity of computing using binary exponentiation?
Binary Exponentiation
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Binary exponentiation repeatedly squares the base and halves the exponent, requiring only logarithmically many iterations.
Incorrect! Try again.
30In binary exponentiation, what operation is performed when the current exponent bit is ?
Binary Exponentiation
Medium
A.The exponent is increased by one
B.The result is divided by the current base
C.The result is multiplied by the current base
D.The base is replaced by its square only
Correct Answer: The result is multiplied by the current base
Explanation:
A set bit indicates that the corresponding power of the base contributes to the final result, so the accumulated result is multiplied by the current base.
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31For a prime number and an integer such that , which congruence is guaranteed by Fermat's Little Theorem?
Fermat method
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Fermat's Little Theorem states that if is prime and is not divisible by , then .
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32Why can a Fermat primality test incorrectly classify some composite numbers as prime?
Fermat method
Medium
A.Some composites satisfy the tested congruence for selected bases
B.Prime numbers never satisfy modular congruences
C.The exponent is always larger than the input number
Correct Answer: Some composites satisfy the tested congruence for selected bases
Explanation:
Certain composite numbers, including Carmichael numbers such as , can satisfy Fermat's congruence for many bases. Such numbers are called pseudoprimes for those bases.
Incorrect! Try again.
33How many prime numbers are there from through ?
Sieve of Eratosthenes
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The primes up to are , giving a total of .
Incorrect! Try again.
34When processing a prime in the Sieve of Eratosthenes, why can marking begin at ?
Sieve of Eratosthenes
Medium
A.The sieve stores only perfect squares
B.Every multiple below is equal to
C.All smaller multiples were already marked by smaller factors
D.Multiples below are always prime
Correct Answer: All smaller multiples were already marked by smaller factors
Explanation:
Any multiple with has a factor smaller than , so it was already marked while processing that smaller factor.
Incorrect! Try again.
35What is the usual time complexity of the Sieve of Eratosthenes for finding all primes up to ?
Sieve of Eratosthenes
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The standard Sieve of Eratosthenes runs in time and uses memory.
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36Which set of primes is required before applying a segmented sieve to the interval ?
Segmented Sieve
Medium
A.Primes up to
B.Primes up to
C.Primes up to
D.Only primes between and
Correct Answer: Primes up to
Explanation:
Every composite number in has a prime factor at most . These base primes are sufficient to mark composites in the segment.
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37Which numbers are prime in the interval ?
Segmented Sieve
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
After marking multiples of the base primes and , the unmarked values in the interval are and .
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38When marking multiples of a base prime in a segment , which starting value is generally correct?
Segmented Sieve
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The first multiple of in the segment is . Starting at also avoids incorrectly marking itself when it lies inside the segment.
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39Suppose Mansi's series is defined by listing consecutive prime numbers: . What is the th term?
Mansi and her series
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The first ten primes are , so the th term is .
Incorrect! Try again.
40If a term in Mansi's prime-based series must be checked for primality and its value is , which approach is sufficient for a single check?
Mansi and her series
Medium
A.Run a sieve up to
B.Test divisors only up to
C.Test prime divisors only up to
D.Check whether is an even number
Correct Answer: Test prime divisors only up to
Explanation:
For a single number, trial division by primes up to is sufficient to determine whether it is prime.
Incorrect! Try again.
41A deterministic primality tester first handles , then tests divisibility only by primes not exceeding . For , which outcome is mathematically correct?
Introduction to Primality Testing
Hard
A. is prime because it has no smaller divisor
B. is composite because it is not prime
C. is neither prime nor composite
D. is prime because it has one positive divisor
Correct Answer: is neither prime nor composite
Explanation:
A prime has exactly two distinct positive divisors. The number has only one, so it is neither prime nor composite.
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42Suppose is composite with integers . Which fact guarantees that trial division need not test candidate divisors larger than ?
Introduction to Primality Testing
Hard
A.Every divisor larger than divides another divisor
B.At least one of and is at most
C.Both and are always below
D.Every composite number has a prime factor equal to
Correct Answer: At least one of and is at most
Explanation:
If both factors exceeded , then , a contradiction. Thus a composite has a nontrivial factor at most .
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43Consider for (long long i = 2; i * i <= n; ++i) with signed 64-bit integers. For very large positive , what is the most serious correctness risk?
O(sqrt(n)) Algorithm for Primality Testing
Hard
A.The loop always performs exactly iterations
B.The loop incorrectly classifies every perfect square
C.The loop skips all even candidate divisors
D.The product can overflow before comparison
Correct Answer: The product can overflow before comparison
Explanation:
Signed overflow can invalidate the loop condition. A safer condition is , assuming positive and .
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44After separately rejecting multiples of and , an implementation tests divisors up to . Which update pattern generates exactly these candidates?
O(sqrt(n)) Algorithm for Primality Testing
Hard
A.Test and , then set
B.Test and , then set
C.Test and , then set
D.Test and , then set
Correct Answer: Test and , then set
Explanation:
Every prime greater than is of the form or . Testing and for covers both forms.
Incorrect! Try again.
45How many positive divisors does have?
Factorization of a number
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
A divisor independently chooses exponents through , through , and through . Hence the count is .
Incorrect! Try again.
46Let , where and are distinct primes. What is the sum of all positive divisors of ?
Factorization of a number
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The divisor-sum function is multiplicative for coprime prime powers: .
Incorrect! Try again.
47A factorization loop repeatedly divides by each candidate while . After the loop, the remaining value is . What must the algorithm do?
Finding prime factors by taking the square root
Hard
A.Append once because the remainder is a prime factor
B.Append twice because the original number may contain
C.Restart trial division because may still be composite
D.Discard because the loop has already passed its square root
Correct Answer: Append once because the remainder is a prime factor
Explanation:
After all factors up to the current square root are removed, any remainder greater than must be prime and occurs once in the remaining value.
Incorrect! Try again.
48Factor by repeated trial division. Which prime factorization is correct?
Finding prime factors by taking the square root
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
. Repeated division is needed to capture each exponent.
Incorrect! Try again.
49Using modular binary exponentiation, what is ?
Binary Exponentiation
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Since and , .
Incorrect! Try again.
50An implementation computes during binary exponentiation, where . Which condition is sufficient for ordinary signed 64-bit multiplication to avoid overflow for every step?
Binary Exponentiation
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Then , so , safely within signed 64-bit range. The larger bounds do not guarantee that every product fits.
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51A Fermat primality test checks whether . Which composite number passes this test for base ?
Fermat method
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
is a base- pseudoprime, so despite being composite.
Incorrect! Try again.
52Why can repeating Fermat tests with many bases still fail to prove that an odd integer is prime?
Fermat method
Hard
A.Some Carmichael numbers pass for every coprime base
B.Every odd composite passes for every nonzero base
C.Binary exponentiation changes the tested congruence
D.Fermat's theorem applies only to primes below
Correct Answer: Some Carmichael numbers pass for every coprime base
Explanation:
A Carmichael number is composite yet satisfies for every coprime to . Fermat testing is therefore probabilistic and vulnerable to these inputs.
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53In a sieve up to , why may marking multiples of a prime begin at ?
Sieve of Eratosthenes
Hard
A.The values below cannot be divisible by
B.Every smaller multiple of is itself a prime number
C.Starting at would omit multiples above
D.Every smaller multiple of has a prime factor below
Correct Answer: Every smaller multiple of has a prime factor below
Explanation:
A multiple has and was already marked when processing a prime factor of .
Incorrect! Try again.
54A sieve stores only odd numbers from through and handles separately. For an odd prime , which marking sequence is correct?
Sieve of Eratosthenes
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Odd multiples of odd differ by . A step of alternates between odd and even multiples, wasting work in an odd-only representation.
Incorrect! Try again.
55To mark multiples of a prime in an inclusive segment , which starting value avoids marking itself as composite while also skipping products already handled conceptually below ?
Segmented Sieve
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
is the first multiple in the segment, while the maximum with prevents marking the prime when it lies in the segment.
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56A segmented sieve processes . Up to what limit must base primes be precomputed to mark every composite in the segment?
Segmented Sieve
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Base primes are needed through . Here , whose square root is slightly greater than , so an integer limit of is sufficient.
Incorrect! Try again.
57Define Mansi's series by . For every , what is ?
Mansi and her series
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Every with ends in . Thus .
Incorrect! Try again.
58Define Mansi's series by and equal to the smallest prime strictly greater than . What is ?
Mansi and her series
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The terms are : after adding , advance to the next strictly larger prime. Therefore .
Incorrect! Try again.
59A collection has pens and must be divided into equal nonempty groups, with neither one group nor groups of one pen allowed. For which is such a division impossible?
Collections of Pens
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Such a division exists exactly when has a nontrivial divisor. Since is prime while , , and , only is impossible.
Incorrect! Try again.
60For an integer , which observation most effectively reduces the search for the smallest prime palindrome greater than ?
Next prime palindrome
Hard
A.Only palindromes ending in an even digit need testing
B.Only odd-length palindromes need to be tested
C.Only palindromes whose digit sum is divisible by need testing
D.Only even-length palindromes need to be tested
Correct Answer: Only odd-length palindromes need to be tested
Explanation:
Every even-length decimal palindrome is divisible by . Since the only even-length prime palindrome is , candidates above must have odd length.
Incorrect! Try again.
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