Question 1

Directions: For a 5-digit number, without repetition of digits, the following information is available.  (i) The first digit is more than 5 times the last digit. (ii) The two-digit number formed by the last two digits is the product of two prime num…

Accepted Answer

### Concept & Logic To find the largest digit, we must combine the constraint that the first three digits are odd with the constraint that the first digit itself is the largest in the entire number. ### Step-by-Step Solution Let the 5-digit number be $A B C D E$. From statement (iv), $A$ (the first digit) is the largest digit in the entire 5-digit number. From statement (iii), $A, B$, and $C$ are all odd digits. From statement (iv), the digits $3$ and $0$ are not allowed. This means the pool of available odd digits is $1, 5, 7$, and $9$. We know from statement (i) that $A > 5E$. Since $E$ cannot be $0$, $E$ must be at least $1$. If $E = 1$, then $A > 5$. The odd digits greater than $5$ are $7$ and $9$. Since the first three digits must all be distinct odd digits (no repetition), we must choose three digits from the set $\{1, 5, 7, 9\}$. Because $E=1$ (as derived from the inequality $A > 5E$), the digit $1$ is already used for the last position. Therefore, the first three digits $A, B$, and $C$ must exactly be $5, 7$, and $9$ in some order. Since statement (iv) explicitly states that the first digit ($A$) is the largest, and our available digits for $A, B, C$ are $\{5, 7, 9\}$, $A$ must be $9$. ### Exam Strategy & Shortcut You don't need to solve for the whole number. The first three digits are odd and unique. The highest odd digit is $9$. If the first digit is the largest digit overall, it is highly probable to be $9$. Since it must be larger than $5 	imes E$ (where $E \ge 1$), it has to be $>5$. The only available odd digits are $7$ and $9$. If $A=7$, you can't fit three unique odd digits in the front (you'd need to use $5, 7$, and what else? $1$ is taken by $E$). Thus, $A$ must be $9$. ### Common Pitfall Ignoring the "no repetition" rule. If repetition were allowed, you could have multiple $9$s, but since all digits must be unique, the limited pool of odd digits forces the largest one to take the front position. ### Final Answer Therefore, the correct answer is 9.

Question 2

Directions: For a 5-digit number, without repetition of digits, the following information is available.  (i) The first digit is more than 5 times the last digit. (ii) The two-digit number formed by the last two digits is the product of two prime num…

Accepted Answer

### Concept & Logic To determine the factors of the number, we first need to deduce all five digits. Once the number is known, we can apply standard divisibility rules (like the sum of digits for $3$ and $9$) to check the options. ### Step-by-Step Solution Let the 5-digit number be $A B C D E$. From previous deductions: 1. $E = 1$ (because $A > 5E$ and $0$ is not allowed). 2. $\{A, B, C\} = \{5, 7, 9\}$ (because they are odd, distinct, and $1$ is used). 3. $A = 9$ (because it's the largest). So, $B$ and $C$ are $5$ and $7$ in some order. Now we must find $D$. The remaining available even digits are $2, 4, 6, 8$. Statement (ii) says the two-digit number $D E$ is the product of two prime numbers. Since $E = 1$, the number is $D1$. Let's test the possibilities: * If $D = 2 \implies 21 = 3 	imes 7$ (Both $3$ and $7$ are primes. Valid.) * If $D = 4 \implies 41$ (Prime number, not a product. Invalid.) * If $D = 6 \implies 61$ (Prime number. Invalid.) * If $D = 8 \implies 81 = 3 	imes 3 	imes 3 	imes 3$ (Product of four primes. Invalid.) Thus, $D$ must be $2$. The 5-digit number is either $95721$ or $97521$. Now, check the given options for factors: * **Check 2 and 4:** The number ends in $1$ (an odd number), so it is not divisible by $2$ or $4$. * **Check 3:** The divisibility rule for $3$ states the sum of digits must be a multiple of $3$. Sum = $9 + 5 + 7 + 2 + 1 = 24$. Since $24$ is divisible by $3$, the number is divisible by $3$. * **Check 9:** The sum is $24$, which is not divisible by $9$. ### Exam Strategy & Shortcut Once you deduce that the sum of the five digits is always $9 + 5 + 7 + 2 + 1 = 24$, regardless of the internal order of $5$ and $7$, you can immediately apply the divisibility rule for $3$. A sum of $24$ guarantees divisibility by $3$ but eliminates $9$. ### Common Pitfall A common error is misinterpreting "product of two prime numbers" in clue (ii). Some students might test $81$ thinking $9 	imes 9$, but $9$ is not prime. Being strict with the definition of prime numbers ($2, 3, 5, 7, 11...$) is crucial for deducing $D=2$. ### Final Answer Therefore, the correct answer is 3.

Question 3

The least prime number is

Accepted Answer

### Concept & Logic The core concept is the fundamental definition of a prime number. A prime number is defined as a natural number greater than $1$ that has exactly two distinct positive divisors: $1$ and the number itself. ### Step-by-Step Solution Let's evaluate the given options based on the definition of prime numbers: * **Option (a) 0:** $0$ is not a natural number (in most standard definitions used in these exams) and it is divisible by every number, so it is not prime. * **Option (b) 1:** $1$ has only one positive divisor (itself). Since it does not have exactly two distinct divisors, it is strictly classified as neither prime nor composite. * **Option (c) 2:** $2$ is a natural number greater than $1$. Its only divisors are $1$ and $2$. Thus, it is a prime number. * **Option (d) 3:** $3$ is also a prime number, but it is larger than $2$. Therefore, the smallest (least) prime number is $2$. ### Exam Strategy & Shortcut This is a pure knowledge-based question that should take less than two seconds. Memorize the basic classification of numbers: $2$ is the first prime number and the *only* even prime number. ### Common Pitfall The most common mistake is selecting $1$ because it is the first positive integer or the smallest odd number. Remember that $1$ is a special case in number theory and is excluded from being prime to maintain the uniqueness of prime factorization. ### Final Answer Therefore, the correct answer is 2.

Question 4

Consider the following statements:  1. If $x$ and $y$ are composite numbers, then $x + y$ is always composite. 2. There does not exist a natural number which is neither prime nor composite.  Which of the above statements is/are correct?

Accepted Answer

### Concept & Logic This problem tests your theoretical understanding of number classifications. To disprove an "always" statement in mathematics, you only need to find a single counter-example. To evaluate existence statements, you must recall boundary cases like the number $1$. ### Step-by-Step Solution Let's evaluate each statement individually. **Statement 1 Evaluation:** * Claim: If $x$ and $y$ are composite, $x + y$ is always composite. * Let's find a counter-example. * Let $x = 4$ (a composite number). * Let $y = 9$ (a composite number). * Calculate the sum: $4 + 9 = 13$. * The sum $13$ is a prime number, not a composite number. * Because we found a case where the sum is prime, Statement 1 is false. **Statement 2 Evaluation:** * Claim: There does not exist a natural number which is neither prime nor composite. * The natural numbers begin at $1, 2, 3, 4, ...$ * By definition, prime numbers have exactly two divisors. Composite numbers have more than two divisors. * The number $1$ has exactly one divisor. Therefore, $1$ is classified as neither prime nor composite. * Because $1$ exists, Statement 2 is false. ### Exam Strategy & Shortcut For Statement 1, quickly mentally add small composite numbers ($4, 6, 8, 9, 10$) together until you hit a prime. $4 + 9 = 13$ is the fastest trigger. For Statement 2, always remember the unique status of the number $1$ when dealing with primes. ### Common Pitfall Students often evaluate Statement 1 by only adding even composite numbers (e.g., $4 + 6 = 10$, $8 + 4 = 12$). Since the sum of two even numbers is always even (and therefore composite if $>2$), they incorrectly assume the statement is true. Always test a mix of even and odd composites. ### Final Answer Therefore, the correct answer is Neither 1 nor 2.

Question 5

The number of prime numbers between 0 and 50 is

Accepted Answer

### Concept & Logic This relies on direct counting of prime numbers. A prime number is a number greater than $1$ with exactly two divisors. For exam efficiency, the count of primes in standard ranges (like $1-50$ and $50-100$) should be memorized. ### Step-by-Step Solution Let's list all the prime numbers starting from $2$ up to $50$: * Single-digit primes: $2, 3, 5, 7$ (Total: 4) * Teens: $11, 13, 17, 19$ (Total: 4) * Twenties: $23, 29$ (Total: 2) * Thirties: $31, 37$ (Total: 2) * Forties: $41, 43, 47$ (Total: 3) Now, sum the counts: $$4 + 4 + 2 + 2 + 3 = 15$$ There are exactly $15$ prime numbers between $0$ and $50$. ### Exam Strategy & Shortcut Do not calculate this during the exam. Memorize these standard milestones: * Primes between $1$ and $50$: **15** * Primes between $50$ and $100$: **10** * Primes between $1$ and $100$: **25** Knowing this fact instantly saves you a minute of manual calculation and eliminates the risk of missing a number. ### Common Pitfall When manually listing primes, students frequently mistakenly include numbers divisible by $3$ or $7$ that look prime at a glance. Common culprits are $39$ ($3 	imes 13$) and $49$ ($7 	imes 7$). ### Final Answer Therefore, the correct answer is 15.

Question 6

The prime numbers dividing 143 and leaving a remainder of 3 in each case are

Accepted Answer

### Concept & Strategy This problem combines divisibility with remainder logic. If a number $N$, when divided by a divisor $d$, leaves a remainder $R$, then the number $(N - R)$ must be perfectly divisible by $d$. $$N \equiv R \pmod d \implies (N - R) 	ext{ is a multiple of } d$$ ### Step-by-Step Solution Given $N = 143$ and the remainder $R = 3$. Subtract the remainder from the main number to find the perfectly divisible value: $$143 - 3 = 140$$ The required prime numbers must be factors of $140$. Let's find the prime factorization of $140$: $$140 = 14 	imes 10$$ $$140 = (2 	imes 7) 	imes (2 	imes 5)$$ $$140 = 2^2 	imes 5 	imes 7$$ The unique prime factors of $140$ are $2, 5$, and $7$. This means dividing $143$ by $2, 5$, or $7$ will leave a remainder of $3$ (provided the divisor is greater than the remainder, so $2$ is technically invalid as a divisor leaving a remainder of $3$, but $5$ and $7$ are valid). Now, let's look at the given options to find a pair of these valid prime factors: * (a) 2 and 11: 11 is not a factor of 140. * (b) 11 and 13: Neither are factors of 140. * (c) 3 and 7: 3 is not a factor of 140. * (d) 5 and 7: Both are prime factors of 140. ### Exam Strategy & Shortcut Instead of factoring, use the Options Elimination technique by testing the divisors directly on $143$. Test option (d) 5 and 7: * $143 \div 5 = 28$ with remainder $3$. (Works!) * $143 \div 7 ightarrow 140$ is a multiple of $7$, so $143$ leaves remainder $3$. (Works!) This avoids algebraic setup and lands you the answer in seconds. ### Common Pitfall A common mistake is trying to divide $143$ directly by all options without mentally subtracting the $3$ first. Mentally checking if $140$ is a multiple of the options is vastly quicker than calculating remainders for $143$. ### Final Answer Therefore, the correct answer is 5 and 7.

Question 7

The sum of the first four primes is

Accepted Answer

### Concept & Logic This problem tests your basic knowledge of the sequence of prime numbers. You must accurately identify the first four primes and compute their sum. ### Step-by-Step Solution Identify the first four prime numbers in sequential order. A prime number is a natural number $>1$ with exactly two divisors. * 1st prime: $2$ (The only even prime) * 2nd prime: $3$ * 3rd prime: $5$ * 4th prime: $7$ Now, add these numbers together: $$Sum = 2 + 3 + 5 + 7$$ $$Sum = 17$$ ### Exam Strategy & Shortcut This is a pure arithmetic calculation. The fastest way is to group the numbers making ten: $(3 + 7) = 10$, then add the remaining $(10 + 5 + 2) = 17$. ### Common Pitfall The single most common error is including $1$ in the list of primes. If you mistakenly count $1$ as a prime, your list would be $1, 2, 3, 5$, and your sum would be $11$ (Option B). Remember, $1$ is neither prime nor composite. ### Final Answer Therefore, the correct answer is 17.

Question 8

The sum of all the prime numbers from 1 to 20 is

Accepted Answer

### Concept & Logic This requires listing all prime numbers within a defined boundary ($1$ to $20$) and calculating their sum. Accuracy in identifying primes is critical here. ### Step-by-Step Solution First, list all the prime numbers between $1$ and $20$. The primes are: $2, 3, 5, 7, 11, 13, 17$, and $19$. Next, calculate the total sum. To do this efficiently, group the numbers into pairs that end in round numbers: * Pair 1: $11 + 19 = 30$ * Pair 2: $13 + 17 = 30$ * Remaining primes: $2 + 3 + 5 + 7 = 17$ Add the grouped sums together: $$Sum = 30 + 30 + 17$$ $$Sum = 77$$ ### Exam Strategy & Shortcut Do not add the numbers sequentially ($2+3=5, 5+5=10...$) as this increases cognitive load and the chance of a minor arithmetic error under time pressure. Always look for pairs that sum to multiples of $10$ (like the $13+17$ trick shown above) to optimize your calculation speed. ### Common Pitfall Accidentally including $9$ or $15$ in the sum. Because they are odd numbers, students rushing through the exam often mistakenly categorize them as primes. $9$ ($3 	imes 3$) and $15$ ($3 	imes 5$) are composite. Including $9$ would incorrectly lead to a sum of $86$. ### Final Answer Therefore, the correct answer is 77.

Question 9

A prime number $N$, in the range 10 to 50, remains unchanged when its digits are reversed. The square of such a number is

Accepted Answer

### Concept & Logic This problem combines number properties (palindromes) with prime identification. If a two-digit number remains unchanged when reversed, its tens digit must be equal to its units digit (e.g., $11, 22$). ### Step-by-Step Solution The condition states the number $N$ is between $10$ and $50$. It also states that reversing the digits yields the exact same number. Let the two-digit number be $xy$. If reversed, it becomes $yx$. For $xy = yx$, it must be true that $x = y$. Let's list all two-digit numbers with identical digits between $10$ and $50$: * $11$ * $22$ * $33$ * $44$ The problem specifies that $N$ is a **prime number**. Let's evaluate our list: * $22$ is divisible by $2$ and $11$. * $33$ is divisible by $3$ and $11$. * $44$ is divisible by $2, 4$, and $11$. The only prime number in this list is $11$. Therefore, $N = 11$. The question asks for the square of such a number: $$N^2 = 11^2 = 121$$ ### Exam Strategy & Shortcut Any two-digit number with repeating digits (like $22, 33, 44...$) is inherently a multiple of $11$. Therefore, none of them can be prime, *except* for $11$ itself. You can bypass listing the numbers and immediately realize $N$ must be $11$. Since $11^2 = 121$, the answer is found instantly. ### Common Pitfall A student might read the question too quickly, find $N = 11$, and look for $11$ in the options. Realizing it's missing, they panic. Always double-check what the question is asking for—in this case, it asks for the *square* of the number, not the number itself. ### Final Answer Therefore, the correct answer is 121.

Question 10

The remainder obtained when any prime number greater than 6 is divided by 6 must be

Accepted Answer

### Concept & Strategy This question tests the general algebraic form of prime numbers. A fundamental theorem in number theory states that every prime number greater than $3$ can be expressed in the form $6k \pm 1$, where $k$ is a natural number. ### Step-by-Step Solution Let's understand why the $6k \pm 1$ rule works. Any integer can be categorized by its remainder when divided by $6$. The possible forms are: 1. $6k$ (Divisible by $6$) 2. $6k + 1$ 3. $6k + 2$ (Factors out to $2(3k+1)$, divisible by $2$) 4. $6k + 3$ (Factors out to $3(2k+1)$, divisible by $3$) 5. $6k + 4$ (Factors out to $2(3k+2)$, divisible by $2$) 6. $6k + 5$ (Can also be written as $6k - 1$) Because primes greater than $3$ cannot be divisible by $2$ or $3$, we must immediately eliminate the forms $6k, 6k+2, 6k+3$, and $6k+4$. The only remaining possible forms for a prime number greater than $3$ (and therefore greater than $6$) are $6k + 1$ and $6k + 5$. * If a prime is of the form $6k + 1$, dividing by $6$ leaves a remainder of $1$. * If a prime is of the form $6k + 5$, dividing by $6$ leaves a remainder of $5$. Therefore, the remainder must be either $1$ or $5$. ### Exam Strategy & Shortcut You don't need to prove the theorem during the exam. Simply pick two or three prime numbers greater than $6$ and divide them by $6$ to find the pattern experimentally. * Prime $7$: $7 \div 6 = 1$ with Remainder **1** * Prime $11$: $11 \div 6 = 1$ with Remainder **5** * Prime $13$: $13 \div 6 = 2$ with Remainder **1** The pattern $1$ and $5$ emerges immediately, pointing directly to option (c). ### Common Pitfall Some students try to memorize complex divisibility rules instead of just testing real numbers. If you forget the $6k \pm 1$ formula under pressure, simply testing $7$ and $11$ safely guarantees the correct answer without theoretical knowledge. ### Final Answer Therefore, the correct answer is either 1 or 5.

Question 11

Which of the following is not a prime number?

Accepted Answer

### Concept & Logic A prime number is a positive integer greater than 1 that has no positive divisors other than 1 and itself. A composite number has at least one positive divisor other than 1 or itself. ### Step-by-Step Solution * **Check Option (a):** $21$ The sum of the digits is $2 + 1 = 3$. According to the divisibility rule of 3, 21 is divisible by 3. $$21 = 3 	imes 7$$ Since 21 has factors other than 1 and itself, it is not a prime number. * **Check Remaining Options:** $23$, $29$, and $43$ are only divisible by 1 and themselves. They are all prime numbers. ### Exam Strategy & Shortcut When asked to find the non-prime (composite) number among small values, always check for basic divisibility rules first: particularly for 2 (even numbers), 3 (sum of digits), and 5 (ends in 0 or 5). 21 immediately triggers the rule for 3. ### Common Pitfall Failing to check basic divisibility rules and instead trying to mentally divide by larger, random numbers, which wastes time. Always start with the smallest primes. ### Final Answer **Therefore, the correct answer is 21.**

Question 12

Which of the following is a prime number?

Accepted Answer

### Concept & Logic To identify a prime number, confirm that it has exactly two distinct factors: 1 and itself. Any even number greater than 2 is composite, and basic divisibility rules can quickly eliminate other composite numbers. ### Step-by-Step Solution * **Option (b) 20 and Option (d) 22:** Both are even numbers greater than 2. Thus, they are divisible by 2 and are composite. * **Option (c) 21:** The sum of the digits ($2 + 1 = 3$) is a multiple of 3. Thus, it is divisible by 3 ($21 = 3 	imes 7$) and is composite. * **Option (a) 19:** 19 has no positive divisors other than 1 and itself. It is a prime number. ### Exam Strategy & Shortcut Use the process of elimination. Immediately cross out all even numbers (20, 22). Next, cross out numbers matching basic multiplication tables you know ($21 = 3 	imes 7$). The remaining number (19) must be the answer. ### Common Pitfall Second-guessing if a smaller odd number like 19 has hidden factors. For numbers under 25, simply knowing your multiplication tables up to $5 	imes 5$ is enough to be confident. ### Final Answer **Therefore, the correct answer is 19.**

Question 13

Which of the following is a prime number?

Accepted Answer

### Concept & Strategy To verify if a number $N$ is prime, you only need to check for divisibility by prime numbers up to the approximate square root of $N$, denoted as $\sqrt{N}$. ### Step-by-Step Solution * **Option (a) 115:** The number ends in 5, making it divisible by 5. Thus, it is composite. $$115 = 5 	imes 23$$ * **Option (b) 119:** The approximate square root is slightly less than 11 (since $11^2 = 121$). Test primes: 2, 3, 5, 7. It is not divisible by 2, 3, or 5. Check 7: $119 = 7 	imes 17$. It is composite. * **Option (c) 127:** The approximate square root is slightly more than 11. Test primes: 2, 3, 5, 7, 11. It is not even, sum of digits (10) isn't divisible by 3, doesn't end in 5, $127 \div 7$ leaves remainder 1, and $127 \div 11$ leaves remainder 6. Since it isn't divisible by any prime up to its square root, 127 is prime. ### Exam Strategy & Shortcut Quickly eliminate obvious composites using the rules for 2, 3, and 5. For 119 and 127, memorizing multiples of 7 (like $7 	imes 10 = 70$, $70 + 49 = 119$) helps you instantly eliminate 119. ### Common Pitfall Assuming 119 is prime because it doesn't immediately look divisible by small numbers like 2, 3, or 5. Forgetting to test the number 7 is the most common mistake here. ### Final Answer **Therefore, the correct answer is 127.**

Question 14

Which of the following is a prime number?

Accepted Answer

### Concept & Formula To determine if a larger number $N$ is prime, find the integer $k$ such that $k > \sqrt{N}$. Then, test $N$ for divisibility by all prime numbers less than or equal to $k$. ### Step-by-Step Solution * **Option (a) 143:** Applying the alternating digit sum rule for 11: $(1 + 3) - 4 = 0$. So, 143 is divisible by 11. $$143 = 11 	imes 13$$ * **Option (b) 289:** You should recognize this as a perfect square. $$289 = 17^2$$ * **Option (c) 117:** Sum of the digits is $1 + 1 + 7 = 9$. It is divisible by 9 (and 3). $$117 = 9 	imes 13$$ * **Option (d) 359:** Approximate square root: $\sqrt{361} = 19$. Check primes up to 19: 2, 3, 5, 7, 11, 13, 17. 359 is not divisible by any of these primes. Thus, 359 is a prime number. ### Exam Strategy & Shortcut Rely on memorized perfect squares (up to 30) and basic divisibility tricks. 289 is instantly eliminated as $17^2$. 117 is eliminated by the rule of 9. 143 is instantly recognized as $11 	imes 13$. Process of elimination leaves 359 as the only possible answer without requiring heavy calculation. ### Common Pitfall Getting bogged down trying to prove 359 is prime manually. It is much faster to prove that 143, 289, and 117 are composite through quick elimination tests. ### Final Answer **Therefore, the correct answer is 359.**

Question 15

The smallest value of natural number $n$, for which $2n + 1$ is not a prime number, is

Accepted Answer

### Concept & Strategy Evaluate the given algebraic expression by substituting consecutive natural numbers starting from the smallest ($n = 1$) until the result evaluates to a composite (non-prime) number. ### Step-by-Step Solution * **Given Expression:** $$E = 2n + 1$$ where $n$ is a natural number ($n = 1, 2, 3, \dots$). * **Test Values:** Let's substitute natural numbers sequentially: For $n = 1$: $E = 2(1) + 1 = 3$ (Prime) For $n = 2$: $E = 2(2) + 1 = 5$ (Prime) For $n = 3$: $E = 2(3) + 1 = 7$ (Prime) For $n = 4$: $E = 2(4) + 1 = 9$ (Composite, divisible by 3) * The condition is met at $n = 4$, which results in 9 (not a prime number). ### Exam Strategy & Shortcut Rather than testing all numbers from $n = 1$, you can directly use the provided options. Test $n = 3$ first, which yields 7 (prime). Next test $n = 4$, which yields 9 (not prime). You have found the answer instantly. ### Common Pitfall Confusing "natural numbers" (which start from 1) with "whole numbers" (which start from 0). If you plug in $n = 0$, you get 1, which is not prime, but 0 is not a natural number. Another mistake is forgetting the prompt asks for when the result is *not* prime. ### Final Answer **Therefore, the correct answer is 4.**

Question 16

The smallest three-digit prime number is

Accepted Answer

### Concept & Logic To find the smallest three-digit prime number, start with the first three-digit number ($100$) and check subsequent numbers for primality. A number is prime if its only divisors are 1 and itself. ### Step-by-Step Solution * **Check 100:** 100 is an even number, thus divisible by 2. It is composite. * **Check 101:** We only need to test for divisibility by prime numbers up to the approximate square root of 101. $$\sqrt{101} \approx 10$$ The primes to test are 2, 3, 5, and 7. - Not divisible by 2 (it is odd). - Not divisible by 3 (sum of digits $1 + 0 + 1 = 2$). - Not divisible by 5 (does not end in 0 or 5). - Not divisible by 7 ($101 \div 7$ leaves a remainder of 3). Since no prime up to its square root divides it evenly, 101 is a prime number. ### Exam Strategy & Shortcut There is no complex calculation needed here. Memorizing the first few prime numbers immediately following 100 (101, 103, 107, 109) is highly recommended for quantitative aptitude exams to save precious seconds. 101 is the smallest. ### Common Pitfall Overthinking or confusing 101 with a number that looks prime but isn't. Some students guess "None of these" assuming a trap, but basic sequential testing proves 101 is the answer. ### Final Answer **Therefore, the correct answer is 101.**

Question 17

How many of the integers between 110 and 120 are prime numbers?

Accepted Answer

### Concept & Strategy To find prime numbers within a small range, write out the odd integers and systematically eliminate those divisible by 3, 5, and 7. Even integers greater than 2 are never prime. ### Step-by-Step Solution * **List potential primes:** The odd integers between 110 and 120 are: 111, 113, 115, 117, 119. * **Apply Divisibility Rules:** - **111:** Sum of digits is $1 + 1 + 1 = 3$. Divisible by 3. - **115:** Ends in 5. Divisible by 5. - **117:** Sum of digits is $1 + 1 + 7 = 9$. Divisible by 3 (and 9). - **119:** Test divisibility by 7. $119 = 7 	imes 17$. It is composite. * **Remaining Number:** - **113:** Its approximate square root is slightly less than 11. It is not divisible by 2, 3, 5, or 7. Therefore, 113 is a prime number. * Only one integer in this range is prime. ### Exam Strategy & Shortcut Instantly filter out the evens and the number ending in 5. You are left checking only 111, 113, 117, and 119. 111 and 117 obviously fall to the rule of 3. You only have to actively remember that 119 is $7 	imes 17$. This leaves 113 as the sole prime. ### Common Pitfall Failing to recognize that 119 is a composite number. Because 119 does not fall to the simple divisibility rules of 2, 3, or 5, students frequently count it as a prime, mistakenly choosing option (c) 2. ### Final Answer **Therefore, the correct answer is 1.**

Question 18

Four prime numbers are arranged in ascending order. The product of first three is 385 and that of last three is 1001. The largest prime number is

Accepted Answer

### Concept & Formula When given the products of overlapping subsets of a consecutive sequence, setting up a ratio will cancel out the common terms, revealing the relationship between the non-overlapping terms. Let the primes be $a, b, c, d$. $$\frac{a 	imes b 	imes c}{b 	imes c 	imes d} = \frac{a}{d}$$ ### Step-by-Step Solution * **Set up the variables:** Let the four prime numbers in ascending order be $a, b, c,$ and $d$ ($a < b < c < d$). * **Write the equations based on given data:** Product of first three: $a 	imes b 	imes c = 385$ Product of last three: $b 	imes c 	imes d = 1001$ * **Divide the two equations:** $$\frac{a 	imes b 	imes c}{b 	imes c 	imes d} = \frac{385}{1001}$$ $$\frac{a}{d} = \frac{385}{1001}$$ * **Simplify the fraction:** Notice that 385 ends in 5, so it is divisible by 5. $385 = 5 	imes 77$. Since 385 is the product of three primes and 5 is one of them, the primes are 5, 7, and 11 (as $7 	imes 11 = 77$). Thus, $a = 5$, $b = 7$, $c = 11$. Now simplify the fraction by dividing numerator and denominator by 77: $$\frac{385 \div 77}{1001 \div 77} = \frac{5}{13}$$ So, $\frac{a}{d} = \frac{5}{13}$. Since $a$ and $d$ must be prime numbers, $a = 5$ and $d = 13$. * The largest prime number $d$ is 13. ### Exam Strategy & Shortcut Look at the first product: 385. It ends in 5. Since these are ascending primes, one of them must be 5, and it has to be the smallest ($a=5$). If $5 	imes b 	imes c = 385$, then $b 	imes c = 77$. The next consecutive primes that multiply to 77 are 7 and 11. Now use the second product: $7 	imes 11 	imes d = 1001$. So $77 	imes d = 1001 \Rightarrow d = 13$. ### Common Pitfall Wasting time attempting to prime factorize 1001 entirely from scratch. You should leverage the common factors ($b$ and $c$) derived from the much easier number 385 to solve for $d$ quickly. ### Final Answer **Therefore, the correct answer is 13.**

Question 19

The sum of three prime numbers is 100. If one of them exceeds another by 36, then one of the numbers is

Accepted Answer

### Concept & Logic The sum of three odd numbers is always an odd number. Since the given sum of three prime numbers is 100 (an even number), one of the prime numbers *must* be even. The only even prime number is 2. ### Step-by-Step Solution * **Set up the variables:** Let the three prime numbers be $x, y,$ and $z$. Given: $x + y + z = 100$ * **Identify the even prime:** As established, for the sum to be even, one prime must be 2. Let $x = 2$. Substitute $x$ into the equation: $$2 + y + z = 100$$ $$y + z = 98$$ * **Use the second condition:** We are given that one prime exceeds another by 36. Let this relationship be between $z$ and $y$. $$z - y = 36$$ * **Solve the system of linear equations:** Equation 1: $z + y = 98$ Equation 2: $z - y = 36$ Add the two equations together to eliminate $y$: $$2z = 134$$ $$z = 67$$ * **Find the remaining prime:** Substitute $z = 67$ back into Equation 1: $$67 + y = 98$$ $$y = 31$$ The three prime numbers are 2, 31, and 67. Option (d) contains 67. ### Exam Strategy & Shortcut The realization that one prime *must* be 2 is the key to unlocking this problem in seconds. Once you deduce $y + z = 98$ and $z - y = 36$, you can simply scan the options. You need an option that, when you subtract 36 from it, gives another prime number. Only $67 - 36 = 31$ (which is prime) fits perfectly. ### Common Pitfall Assuming all prime numbers must be odd. Students who forget that 2 is a prime number will find it impossible to solve $x + y + z = 100$ because they will assume $x, y,$ and $z$ are all odd. ### Final Answer **Therefore, the correct answer is 67.**

Question 20

Which one of the following is a prime number?

Accepted Answer

### Concept & Strategy To verify if a larger number $N$ is prime, find its approximate square root $\sqrt{N}$. Check if $N$ is divisible by any prime number up to that integer limit. If it isn't divisible by any, it is a prime number. ### Step-by-Step Solution * **Option (a) 161:** The approximate square root is slightly less than 13 ($\sqrt{169} = 13$). Test primes up to 11. Testing 7: $161 = 7 	imes 23$. Thus, 161 is composite. * **Option (b) 221:** The approximate square root is slightly less than 15. Test primes up to 13. Testing 13: $221 = 13 	imes 17$. Thus, 221 is composite. * **Option (d) 437:** The approximate square root is roughly 21 ($\sqrt{400} = 20, \sqrt{441} = 21$). Test primes up to 19. Testing 19: $437 = 19 	imes 23$. Thus, 437 is composite. * **Option (c) 373:** The approximate square root is roughly 19.3. Test primes up to 19 (2, 3, 5, 7, 11, 13, 17, 19). 373 is odd, doesn't end in 5, sum of digits is 13 (not div by 3). $373 \div 7$ leaves remainder 2. $373 \div 11$ leaves remainder 10. $373 \div 13$ leaves remainder 9. $373 \div 17$ leaves remainder 16. $373 \div 19$ leaves remainder 12. It does not divide evenly by any of these primes. Therefore, 373 is prime. ### Exam Strategy & Shortcut Aptitude test setters intentionally use products of mid-range primes (7, 13, 17, 19, 23) as options to trap students. Memorize common combinations like $13 	imes 17 = 221$. When checking 161, recognize that $140 + 21 = 161$, making it instantly divisible by 7. By eliminating the easier composites, you can confidently select 373 without having to manually divide it by every single prime up to 19. ### Common Pitfall Giving up too early on division tests. Many students check for 2, 3, 5, 7, and 11, and if none work, they assume the number is prime. You must test primes all the way up to the approximate square root (like 17 and 19 for these larger options) to be accurate. ### Final Answer **Therefore, the correct answer is 373.**

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