Question 1

Which of the following numbers is divisible by 3, 7, 9 and 11?

Accepted Answer

Concept & Strategy: To find a number divisible by multiple prime and composite factors, we do not need to test every single number against every single rule. Instead, apply the easiest, most restrictive divisibility rules first (like 9 and 11) to rapidly eliminate incorrect options. Any number divisible by 9 is inherently divisible by 3. Step-by-Step Solution: • Test 1: Divisibility by 9 (Sum of digits must be a multiple of 9) • (a) 639 \rightarrow 6+3+9 = 18 (Divisible) • (b) 2079 \rightarrow 2+0+7+9 = 18 (Divisible) • (c) 3791 \rightarrow 3+7+9+1 = 20 (Not divisible - Eliminate) • (d) 37911 \rightarrow 3+7+9+1+1 = 21 (Not divisible - Eliminate) • Test 2: Divisibility by 11 (Alternating sum difference) We only need to test the remaining options (a) and (b). • (a) 639 \rightarrow Odd sum: 9+6=15. Even sum: 3. Difference: 15-3 = 12. (Not divisible - Eliminate) • (b) 2079 \rightarrow Odd sum: 9+0=9. Even sum: 7+2=9. Difference: 9-9 = 0. (Divisible) By elimination, 2079 is the only valid number. Exam Strategy & Shortcut: When a question lists 9 and 3, ignore 3 completely. It is redundant information designed to waste your time. Always start your elimination with the rule of 9 because calculating the digit sum is incredibly fast. Once reduced to two options, the rule of 11 acts as a quick tie-breaker. You almost never actually need to use the rule of 7, which is complex and time-consuming. Common Pitfall: The biggest trap is trying to find the Least Common Multiple (LCM) of 3, 7, 9, and 11 and then seeing which option matches it. Calculating 9 × 7 × 11 = 693 and then looking for multiples of 693 takes much longer than simply using the elimination method with basic digit rules. Final Answer: Therefore, the correct answer is 2079.

Question 2

How many of the following numbers are divisible by 3 but not by 9?  2133, 2343, 3474, 4131, 5286, 5340, 6336, 7347, 8115, 9276

Accepted Answer

Concept & Logic: This problem tests the relationship between the divisibility rules of 3 and 9. • Divisibility by 3: The sum of the digits must be a multiple of 3. • Divisibility by 9: The sum of the digits must be a multiple of 9. Every number divisible by 9 is also divisible by 3, but the reverse is not true. We need to find numbers whose digit sum is a multiple of 3, but strictly not a multiple of 9. Step-by-Step Solution: Calculation / Deduction: Let's calculate the sum of the digits for each number and evaluate both conditions: • 2133: 2 + 1 + 3 + 3 = 9. Divisible by 9. (Reject) • 2343: 2 + 3 + 4 + 3 = 12. Divisible by 3, NOT 9. (Accept) • 3474: 3 + 4 + 7 + 4 = 18. Divisible by 9. (Reject) • 4131: 4 + 1 + 3 + 1 = 9. Divisible by 9. (Reject) • 5286: 5 + 2 + 8 + 6 = 21. Divisible by 3, NOT 9. (Accept) • 5340: 5 + 3 + 4 + 0 = 12. Divisible by 3, NOT 9. (Accept) • 6336: 6 + 3 + 3 + 6 = 18. Divisible by 9. (Reject) • 7347: 7 + 3 + 4 + 7 = 21. Divisible by 3, NOT 9. (Accept) • 8115: 8 + 1 + 1 + 5 = 15. Divisible by 3, NOT 9. (Accept) • 9276: 9 + 2 + 7 + 6 = 24. Divisible by 3, NOT 9. (Accept) Counting the accepted numbers, we have exactly 6 numbers that meet the criteria. Exam Strategy & Shortcut: Instead of writing out the full sum, use the Digit Cancellation Method (Casting out 9s). Cross out any digits (or combinations of digits) that sum to 9. If the remaining digits sum to 3 or 6, the number is divisible by 3 but not 9. If the remaining sum is 0 (or 9), it is divisible by 9. • *Example 6336:* Cross out 6 and 3 (sums to 9), cross out 3 and 6 (sums to 9). Remainder is 0. Divisible by 9. Skip it. • *Example 7347:* 7+2=9 (borrow 2 from 3, leaves 1). 1+4+4(from 7)=9. Remaining is 3. Divisible by 3, not 9. Keep it. Common Pitfall: A major trap is calculating the sum of digits, confirming it's divisible by 3, and immediately counting it without checking if it's also a multiple of 9. The question specifically excludes multiples of 9, so filtering them out is the core task. Final Answer: Therefore, the correct answer is 6.

Question 3

If the number $357*25*$ is divisible by both 3 and 5, then the missing digits in the unit's place and the thousandth's place respectively are

Accepted Answer

Concept & Logic: To find multiple missing digits, tackle the rule that restricts the end of the number first (like 2, 5, or 10), then apply summation rules (like 3 or 9). • Divisibility by 5: The unit digit must be strictly `0` or `5`. • Divisibility by 3: The sum of all digits must be a multiple of `3`. Step-by-Step Solution: Given: The number is 357x25y, where x is the digit in the thousand's place (noted as thousandth's in the question) and y is the digit in the unit's place. The question asks for the values of (y, x) respectively. Calculation / Deduction: • Step 1: Check the unit's place (y) for divisibility by 5. For the number to be divisible by 5, the unit digit y must be either 0 or 5. • Step 2: Formulate the rule for divisibility by 3. Sum of the digits: 3 + 5 + 7 + x + 2 + 5 + y = 22 + x + y. This sum (22 + x + y) must be a multiple of 3. • Step 3: Evaluate the given options for (y, x) respectively. • *(a) 0, 6*: y=0, x=6. Condition for 5 is met. Sum = 22 + 6 + 0 = 28. 28 is not divisible by 3. (Reject) • *(b) 5, 1*: y=5, x=1. Condition for 5 is met. Sum = 22 + 1 + 5 = 28. 28 is not divisible by 3. (Reject) • *(c) 5, 4*: y=5, x=4. Condition for 5 is met. Sum = 22 + 4 + 5 = 31. 31 is not divisible by 3. (Reject) Since none of the provided pairs yield a digit sum that is a multiple of 3, none of the options are correct. Exam Strategy & Shortcut: Always scan the required order of the options. The prompt asks for unit's place and then thousand's place, meaning the options are ordered (y, x). Check the first digit of every option immediately to see if it satisfies the rule of 5 (it must be 0 or 5). Since all options start with 0 or 5, you quickly transition to adding the two digits to the base sum of 22. 22 + 0 + 6 = 28 (No) 22 + 5 + 1 = 28 (No) 22 + 5 + 4 = 31 (No) You can solve this in under 15 seconds without writing anything down. Common Pitfall: The most common mistake is ignoring the word "respectively" and swapping the variables, assuming the options are in left-to-right order (x, y). However, even if a student swapped them here, they would calculate the exact same sums (x+y = y+x) and still correctly arrive at "None of these". The real pitfall is math anxiety causing a student to second-guess themselves when none of the primary options work. Trust your math! Final Answer: Therefore, the correct answer is None of these.

Question 4

Which of the following numbers is exactly divisible by 24?

Accepted Answer

Concept & Logic: To check for divisibility by a composite number like 24, we must break it down into its co-prime factors. The co-prime factors of 24 are 8 and 3 (since 8 × 3 = 24 and they share no common factors other than 1). Therefore, a number is divisible by 24 if and only if it satisfies both the rule for 8 and the rule for 3. • Divisibility by 8: The last three digits of the number must be divisible by 8. • Divisibility by 3: The sum of all digits must be divisible by 3. Step-by-Step Solution: Calculation / Deduction: Always test the stricter rule first to eliminate options quickly. Testing divisibility by 8 on the last three digits is much faster than summing up a 7-digit number. • Step 1: Check rule of 8 (Last three digits) • *(a) 35718:* Last three digits are 718. 718 \div 8 = 89 remainder 6. (Reject) • *(b) 63810:* Last three digits are 810. 810 \div 8 = 101 remainder 2. (Reject) • *(c) 537804:* Last three digits are 804. 804 \div 8 = 100 remainder 4. (Reject) • *(d) 3125736:* Last three digits are 736. 736 \div 8 = 92 exactly. (Accept) • Step 2: Check rule of 3 for the remaining candidate (d) • Number: 3125736 • Sum of digits: 3 + 1 + 2 + 5 + 7 + 3 + 6 = 27 • Since 27 is a multiple of 3, the number is divisible by 3. Since 3125736 is divisible by both 8 and 3, it is exactly divisible by 24. Exam Strategy & Shortcut: Rule Stacking: Whenever you have a composite divisibility question, never test the addition rule (like 3 or 9) first. Addition rules require processing the entire number. Always test the "tail" rules first (like 2, 4, 8, 5, 25). By isolating the last three digits (718, 810, 804, 736) and testing for 8, we eliminated three options in seconds without doing a single addition. Common Pitfall: A fatal mistake is breaking 24 down into factors that are NOT co-prime, such as 12 and 2, or 6 and 4. If you check for divisibility by 6 and 4, you might get false positives. For example, the number 12 is divisible by 6 and 4, but it is certainly not divisible by 24! Always use co-prime factors (8 and 3). Final Answer: Therefore, the correct answer is 3125736.

Question 5

If the seven-figure number $30X0103$ is a multiple of 13, then $X$ is

Accepted Answer

Concept & Strategy: Use modulo arithmetic based on place values, or apply the block-of-three alternating sum rule for 13 to isolate the unknown digit X. Step-by-Step Solution: Given: The number 30X0103 is perfectly divisible by 13. Calculation / Deduction: • Step 1: Break the number into manageable parts based on place value. 30X0103 = 3000103 + (X × 10000) • Step 2: Find the remainder when 3000103 is divided by 13. 3000103 \div 13 = 230777 with a remainder of 2. • Step 3: Find the remainder when 10000 is divided by 13. 10000 \div 13 = 769 with a remainder of 3. • Step 4: Formulate the modulo equation. For the whole number to be divisible by 13, the sum of their remainders must be a multiple of 13. Therefore, (2 + 3X) must be a multiple of 13. • Step 5: Test the given options for X. • If X = 1: 2 + 3(1) = 5 (Not divisible) • If X = 6: 2 + 3(6) = 20 (Not divisible) • If X = 7: 2 + 3(7) = 23 (Not divisible) • If X = 8: 2 + 3(8) = 26 (Yes, 26 is a multiple of 13) Exam Strategy & Shortcut: An alternative faster trick for divisibility by 7, 11, and 13 is taking the alternating sum of blocks of 3 digits from right to left. Blocks: `103`, `0X0`, `3`. Alternating sum: 103 - (0X0) + 3 = 106 - 10X. Test the options directly into (106 - 10X) to see which yields a multiple of 13. For X=8 \implies 106 - 80 = 26. Divides perfectly! Common Pitfall: When using the block method, a frequent mistake is misinterpreting the place value of the unknown digit within its block. In `0X0`, the X represents the tens place, so its numerical value is 10X, not just X. Writing 103 - X + 3 will lead to an incorrect answer. Final Answer: Therefore, the correct answer is 8.

Question 6

If a number is divisible by both 11 and 13, then it must be necessarily

Accepted Answer

Concept & Logic: If a number is completely divisible by two co-prime integers, it must also be perfectly divisible by their product. Step-by-Step Solution: Calculation / Deduction: • Step 1: Identify the properties of the given divisors. Both 11 and 13 are prime numbers. • Step 2: Determine their relationship. Because they are both prime, they share no common factors other than 1. This classifies them as co-prime to each other. • Step 3: Apply number theory rules. If a number N is divisible by x and y, and the greatest common divisor gcd(x, y) = 1, then N is strictly divisible by (x × y). • Step 4: Conclusion. Therefore, any number divisible by both 11 and 13 must necessarily be divisible by their product, (11 × 13). Exam Strategy & Shortcut: This is a foundational theoretical rule of number systems. You should not need to perform any calculations. Recognize the prompt mentions "both primes" and instantly select the option containing their "product". Common Pitfall: A common trap is selecting option (a) `429` because 429 = 11 × 13 × 3. While 429 *is* divisible by both, a number divisible by 11 and 13 doesn't *have* to be 429 (for example, it could be 143 or 286). The question asks what condition must *necessarily* be true for all such numbers. Final Answer: Therefore, the correct answer is divisible by (11 × 13).

Question 7

Which of the following numbers are completely divisible by 7?  I. $195195$ II. $181181$ III. $120120$ IV. $891891$

Accepted Answer

Concept & Formula: Any six-digit number formed by repeating a three-digit block (format abcabc) is inherently a multiple of 1001. Since 1001 = 7 × 11 × 13, any such number is always divisible by 7. Step-by-Step Solution: Given: Four six-digit numbers: 195195, 181181, 120120, and 891891. Calculation / Deduction: • Step 1: Observe the structural pattern. All four numbers follow the exact format of repeating triplets: abcabc. • Step 2: Apply the expansion rule. Recall that abcabc = (abc × 1000) + abc = abc × 1001. • Step 3: Evaluate the prime factors. Factor 1001 into its primes: 7 × 11 × 13. • Step 4: Conclude. Because 7 is a prime factor of the base multiplier 1001, every single number formatted as abcabc is mathematically guaranteed to be completely divisible by 7. Thus, I, II, III, and IV all qualify. Exam Strategy & Shortcut: Never perform manual division for 7 on these types of numbers. The divisibility rule for 7 is notoriously tedious to execute under time pressure. The moment you spot the repeating 3-digit triplet, know instantly that it passes the divisibility tests for 7, 11, and 13. You can answer this question visually in 2 seconds. Common Pitfall: The biggest mistake is wasting precious exam time manually applying the standard divisibility rule for 7 to all four numbers sequentially instead of recognizing the overarching property of the number format. Final Answer: Therefore, the correct answer is All are divisible.

Question 8

The six-digit number $5ABB7A$ is a multiple of 33 for non-zero digits $A$ and $B$. Which of the following could be possible value of $A + B$?

Accepted Answer

Concept & Logic: For a number to be a multiple of 33, it must be completely divisible by its co-prime factors, which are 11 and 3. • Divisibility by 11: The difference between the sum of the digits in odd positions and the sum of the digits in even positions must be `0` or a multiple of `11`. • Divisibility by 3: The sum of all digits must be a multiple of `3`. Step-by-Step Solution: Given: The number is 5ABB7A. A and B are non-zero single digits. Calculation / Deduction: • Step 1: Test for divisibility by 11. Let's group the digits by alternating positions (from right to left): • Sum of odd positions (1st, 3rd, 5th): A + B + A = 2A + B • Sum of even positions (2nd, 4th, 6th): 7 + B + 5 = 12 + B • Calculate the difference: (12 + B) - (2A + B) = 12 - 2A For the number to be divisible by 11, this difference (12 - 2A) must be 0, 11, -11, etc. Since A is a non-zero digit, the only logical way this equals a multiple of 11 is if it equals 0: 12 - 2A = 0 \implies 2A = 12 \implies A = 6. • Step 2: Test for divisibility by 3. Now substitute A = 6 back into the number: 56BB76. Sum of all digits: 5 + 6 + B + B + 7 + 6 = 24 + 2B. This sum must be a multiple of 3. Let's test possible non-zero values for B: • If B = 3: Sum = 24 + 6 = 30 (Divisible by 3) • If B = 6: Sum = 24 + 12 = 36 (Divisible by 3) • If B = 9: Sum = 24 + 18 = 42 (Divisible by 3) • Step 3: Find the matching sum (A + B). We have three valid combinations for (A, B): • 6 + 3 = 9 • 6 + 6 = 12 • 6 + 9 = 15 Looking at the given options (8, 9, 10, 14), only 9 is a listed possible value. Exam Strategy & Shortcut: When formatting the alternating sum for the rule of 11, strategically align the equations so that variables cancel out. Notice how B appears exactly once in both the odd and even position sets. When you subtract them, B immediately cancels out completely (+B - B = 0), leaving an equation with only A. This lets you solve for A = 6 in a matter of seconds without any guesswork. Common Pitfall: A frequent mistake is finding A=6 and assuming B can be any number, thereby randomly picking an option. Because the question specifies "multiple of 33", ignoring the rule of 3 will lead to incorrect pairs like A=6, B=4 (which would make A+B=10, a trap option). You must verify both conditions. Final Answer: Therefore, the correct answer is 9.

Question 9

Which of the following numbers is divisible by 99?

Accepted Answer

Concept & Strategy: To determine if a large number is divisible by 99, check if it is divisible by its co-prime factors: 9 and 11. • Rule of 9: The sum of all digits must be a multiple of 9. • Rule of 11: The difference between the sum of digits in alternating positions must be 0 or a multiple of 11. Always test the rule of 9 first, as it is computationally faster and often eliminates options instantly. Step-by-Step Solution: Calculation / Deduction: • Step 1: Test all options using the Rule of 9. • *(a) 114345:* Sum = 1 + 1 + 4 + 3 + 4 + 5 = 18. (Divisible by 9, keep) • *(b) 913464:* Sum = 9 + 1 + 3 + 4 + 6 + 4 = 27. (Divisible by 9, keep) • *(c) 135792:* Sum = 1 + 3 + 5 + 7 + 9 + 2 = 27. (Divisible by 9, keep) • *(d) 3572404:* Sum = 3 + 5 + 7 + 2 + 4 + 0 + 4 = 25. (Reject) • Step 2: Test the remaining options using the Rule of 11. • *(a) 114345:* Odd positions (from right): 5 + 3 + 1 = 9 Even positions (from right): 4 + 4 + 1 = 9 Difference: 9 - 9 = 0. (Divisible by 11. Accept!) • *(b) 913464:* Odd positions (from right): 4 + 4 + 1 = 9 Even positions (from right): 6 + 3 + 9 = 18 Difference: 9 - 18 = -9. (Reject) • *(c) 135792:* Odd positions (from right): 2 + 7 + 3 = 12 Even positions (from right): 9 + 5 + 1 = 15 Difference: 12 - 15 = -3. (Reject) Exam Strategy & Shortcut: For the rule of 9, use the Casting Out Nines technique. Instead of adding every single number, cross out any individual 9s or any groups of numbers that sum to 9. • For (a) 114345: 4+5=9 (cross out). Remaining: 1+1+4+3 = 9 (cross out). Remainder is 0, so it passes. This saves mental energy for the rule of 11, which requires more careful tracking of alternating digits. Common Pitfall: The biggest time-waster here is starting with the rule of 11. While robust, writing out the odd/even sums for four different six/seven-digit numbers takes considerably longer than a rapid mental digit sum. Always sequence your divisibility tests from the simplest (2, 5, 10), to moderate (3, 9, 4), to complex (7, 11, 13). Final Answer: Therefore, the correct answer is 114345.

Question 10

If $37X3$ is a four-digit natural number divisible by 7, then the place marked as $X$ must have the value

Accepted Answer

Concept & Logic: While there is a formal divisibility rule for 7 (double the last digit and subtract it from the remaining number), for a small four-digit number with one unknown, it is often faster to use Place Value Expansion and standard remainder algebra. Step-by-Step Solution: Given: The number 37X3 is completely divisible by 7. Calculation / Deduction: • Step 1: Expand the number using place values. 37X3 can be written mathematically as 3700 + 10X + 3, which simplifies to 3703 + 10X. • Step 2: Find the remainder of the known portion when divided by 7. Let's divide 3703 by 7. 3500 \div 7 = 500 Remaining: 3703 - 3500 = 203. 203 \div 7 = 29 exactly (7 × 20 = 140, 7 × 9 = 63, 140+63=203). Since 3703 is perfectly divisible by 7 (it equals 7 × 529), its remainder is 0. • Step 3: Solve for X. Since 3703 + 10X must be divisible by 7, and 3703 is already divisible by 7, the remaining part, 10X, must also be a multiple of 7. For 10 × X to be a multiple of 7, X itself must be a multiple of 7 (since 10 and 7 are co-prime). Because X is a single digit (0-9), the only valid multiples of 7 are 0 and 7. • Step 4: Check the options. Looking at the choices (0, 3, 5, 9), only 0 is available. (If 7 were an option, both 3703 and 3773 would be valid answers, but only 0 is provided). Exam Strategy & Shortcut: Instead of expanding algebraically, you can test the options rapidly using the Osculator Method (Rule of 7: drop unit digit, multiply by 2, subtract). Let's test Option (a) X=0: Number is 3703. Drop 3, double it (6). Subtract from 370: 370 - 6 = 364. Drop 4, double it (8). Subtract from 36: 36 - 8 = 28. 28 is a known multiple of 7. The first option works, allowing you to move on immediately! Common Pitfall: A common trap is assuming X cannot be 0 because it "hides" the tens place, making students lean towards non-zero options. Remember that 0 is a perfectly valid digit in the middle of a natural number, and 0 × 10 = 0 is a perfectly valid multiple of any number (including 7). Final Answer: Therefore, the correct answer is 0.

Question 11

If a number $774958A96B$ is divisible by 8 and 9, the respective values of $A$ and $B$ will be

Accepted Answer

Concept & Logic: For a number to be divisible by both 8 and 9, it must independently satisfy the divisibility rules for each. Rule for 8: The last three digits must be divisible by 8. Rule for 9: The sum of all digits must be divisible by 9. Step-by-Step Solution: • Given: The number is 774958A96B. • Step 1: Apply the rule for 8. The last three digits are 96B. We need (96B/8) to be an integer. • Calculation: 960 is exactly divisible by 8 (960 \div 8 = 120). The next multiple is 968. Therefore, the single digit B can be either 0 or 8. • Step 2: Apply the rule for 9 (Case 1: B = 0). The number is 774958A960. • Sum of digits = 7+7+4+9+5+8+A+9+6+0 = 55 + A. • For 55 + A to be divisible by 9, the next multiple of 9 is 63. • 55 + A = 63 \implies A = 8. • This gives the pair (A, B) = (8, 0). This matches option (c). • Step 3: Check Case 2 (B = 8) just to be sure. The number is 774958A968. • Sum of digits = 55 + 8 + A = 63 + A. • For 63 + A to be divisible by 9, A could be 0 or 9. • The pairs would be (0, 8) or (9, 8). Neither of these are in the options. Exam Strategy & Shortcut: Use option elimination to save time. Since B is the last digit, check the given options for B. Options provide B=8, 8, 0. Since both 0 and 8 work for divisibility by 8, you must check the sum for 9. By quickly substituting B=0 and A=8 (from option c), sum = 55 + 8 = 63, which is divisible by 9. You can verify the correct answer in seconds without solving for every possibility. Common Pitfall: The most common mistake is miscalculating the large sum of digits. A great tip to avoid this is to "cast out nines". When adding 7+7+4+9+5+8+A+9+6+B, simply cross out any 9s or combinations of digits that add to 9 (like 4+5). This keeps your sum small and error-free! Final Answer: Therefore, the correct answer is 8 and 0.

Question 12

The difference between the squares of any two consecutive integers is equal to

Accepted Answer

Concept & Formula: This problem tests your knowledge of the algebraic expansion of consecutive integers and the difference of squares formula. (x + y)(x - y) = x^2 - y^2 Step-by-Step Solution: • Given: We need to find the difference between the squares of two consecutive integers. • Step 1: Define the variables. Let the smaller integer be n. The consecutive integer directly following it will be (n + 1). • Step 2: Set up the equation. We need to find (n + 1)^2 - n^2. • Step 3: Expand the expression. • (n + 1)^2 - n^2 = (n^2 + 2n + 1) - n^2 • = 2n + 1 • Step 4: Interpret the result. We can rewrite 2n + 1 by breaking it apart into our original numbers: • 2n + 1 = n + (n + 1) • This perfectly matches the sum of our original two consecutive integers. Exam Strategy & Shortcut: Use real numbers: Pick any two small, easy consecutive integers, like 3 and 4. Calculate the difference of their squares: 4^2 - 3^2 = 16 - 9 = 7. Now, evaluate the options using the numbers 3 and 4: (a) Is 7 an even number? No. (b) Is 7 the difference of the given numbers (4 - 3 = 1)? No. (c) Is 7 the sum of the given numbers (4 + 3 = 7)? Yes! Common Pitfall: Students often try to memorize this as a standalone rule rather than recognizing it as a simple application of a^2 - b^2. Because (a-b) for consecutive integers is always 1, a^2 - b^2 simplifies immediately to 1 × (a+b), which is just the sum! Final Answer: Therefore, the correct answer is sum of given numbers.

Question 13

The number $6n^2 + 6n$ for natural number $n$ is always divisible by

Accepted Answer

Concept & Logic: To find the universal divisors of an algebraic expression, factor it completely to reveal its core components. The product of any two consecutive integers is always an even number (a multiple of 2). Step-by-Step Solution: • Given: The algebraic expression is 6n^2 + 6n, where n is a natural number (n \ge 1). • Step 1: Factor the expression. Pull out the greatest common factor, which is 6n. • 6n^2 + 6n = 6n(n + 1) • Step 2: Analyze the components. We have a constant 6 multiplied by n(n + 1). • Step 3: Apply number properties. For any natural number n, the numbers n and (n + 1) are consecutive integers. Exactly one of them must be even. Therefore, their product n(n + 1) is always a multiple of 2. • Let n(n + 1) = 2k (where k is an integer). • Step 4: Substitute back. * 6n(n + 1) = 6(2k) = 12k. • Conclusion: The expression is always a multiple of 12. If a number is divisible by 12, it is fundamentally also divisible by its factors, including 6. Exam Strategy & Shortcut: The Plugin Method: Test the first two natural numbers to find the common divisors rapidly. Test n = 1: 6(1)^2 + 6(1) = 6 + 6 = 12. • 12 is divisible by 6 and 12, but NOT 18. (Eliminates option d). Test n = 2: 6(2)^2 + 6(2) = 24 + 12 = 36. • 36 is divisible by both 6 and 12. Since it works for both 6 and 12 across different test cases, option (b) is the robust answer. Common Pitfall: The word "only" in the options is a trap. A student might factor out the 6, see 6 × n(n+1), and immediately select "6 only" without realizing that the variable portion n(n+1) guarantees an additional factor of 2, bumping the true universal divisor up to 12. Final Answer: Therefore, the correct answer is 6 and 12.

Question 14

The sum of a number consisting of two digits and the number formed by interchanging the digits is always divisible by

Accepted Answer

Concept & Formula: To solve digit-manipulation problems, represent the two-digit number algebraically using its base-10 place values. A two-digit number = 10x + y Step-by-Step Solution: • Step 1: Define the primary number. Let the original two-digit number have a tens digit x and units digit y. Its value is 10x + y. • Step 2: Define the interchanged number. When the digits swap places, y becomes the tens digit and x becomes the units digit. The new value is 10y + x. • Step 3: Calculate their sum. * Sum = (10x + y) + (10y + x) • Combine like terms: = 11x + 11y • Step 4: Factor the expression. * = 11(x + y) • Conclusion: The algebraic sum is 11 multiplied by the sum of the digits. Therefore, this total will always be a multiple of 11. Exam Strategy & Shortcut: Pick a real number and test it: Let's choose a simple two-digit number, like 23. Reverse the digits to get 32. Add them together: 23 + 32 = 55. Check the options: 55 is clearly divisible by 11. It is not divisible by 7, 9, or 10. You can verify the theoretical rule with practical numbers instantly. Common Pitfall: The main danger here is mixing up this rule with the subtraction counterpart. • Sum of reversed numbers = divisible by 11. • Difference of reversed numbers = divisible by 9. Keeping these two properties distinctly memorized saves massive amounts of time on standard aptitude tests. Final Answer: Therefore, the correct answer is 11.

Question 15

If $n$ is a whole number greater than 1, then $n^2(n^2 - 1)$ is always divisible by

Accepted Answer

Concept & Logic: This problem requires algebraic factorization to reveal the consecutive integer components hidden within the expression. n^2(n^2 - 1) = n \cdot n \cdot (n - 1) \cdot (n + 1) = (n - 1)n(n + 1) \cdot n Step-by-Step Solution: • Given: The expression is n^2(n^2 - 1) where n is a whole number > 1. • Step 1: Factor the expression completely. As shown in the concept, the expression contains (n - 1)n(n + 1), which represents the product of three consecutive integers. • Step 2: Apply divisibility rules for consecutive integers. The product of any three consecutive integers is always divisible by 3! = 6. So, the expression has a guaranteed factor of 6. • Step 3: Analyze the remaining n. We have an extra n multiplying our sequence. • If n is even, n^2 contains a factor of 4. • If n is odd, then (n-1) and (n+1) are both even. Their product contains a factor of 2 × 2 = 4. • In all cases, the entire expression is guaranteed to be divisible by 4. • Step 4: Combine the factors. The expression is divisible by 3 (from the consecutive trio) and by 4. Therefore, it is always divisible by 3 × 4 = 12. Exam Strategy & Shortcut: The Plugin Method: The quickest way to solve "always divisible by" questions is to plug in the smallest allowed value for the variable. The condition is "n is a whole number greater than 1". The smallest valid number is n = 2. Substitute n = 2: 2^2(2^2 - 1) = 4(4 - 1) = 4 × 3 = 12. Look at the options: 12 is divisible by 12, but not by 8, 10, or 16. You have your answer instantly. Common Pitfall: A fatal mistake is ignoring the domain constraint "n is greater than 1" and plugging in n = 1. This results in 1(0) = 0. Since 0 is technically divisible by all numbers, it eliminates zero options and wastes your time. Final Answer: Therefore, the correct answer is 12.

Question 16

If $n$ is any odd number greater than 1, then $n(n^2 - 1)$ is

Accepted Answer

Concept & Logic: This problem relies on the specific behavior of consecutive even numbers. First, we must factor the algebraic expression. n(n^2 - 1) = (n - 1)n(n + 1) This represents the product of three consecutive integers. Step-by-Step Solution: • Given: n is an *odd* number greater than 1. • Step 1: Analyze the factors. We have three consecutive numbers: (n - 1), n, and (n + 1). • Step 2: Apply the odd constraint. Since n is odd, the numbers immediately before and after it—(n - 1) and (n + 1)—must both be *even* numbers. • Step 3: Use the consecutive even numbers rule. The product of any two consecutive even numbers is always a multiple of 8. (For example, 2 × 4 = 8, or 4 × 6 = 24). So, (n - 1)(n + 1) is a multiple of 8. • Step 4: Use the consecutive integers rule. In any sequence of three consecutive integers, exactly one of them must be a multiple of 3. • Conclusion: The entire product must be divisible by both 8 and 3. Since 8 and 3 are co-prime, the product is always divisible by 8 × 3 = 24. Exam Strategy & Shortcut: Test the Boundary Condition: Avoid abstract algebra during a timed exam by picking the smallest number that fits the prompt's rules. The condition says "n is an odd number greater than 1". The smallest valid number is n = 3. Substitute n = 3 into the expression: 3(3^2 - 1) = 3(9 - 1) = 3 × 8 = 24. The result is 24. The only option that 24 is always divisible by is 24 itself. Common Pitfall: Students often remember that the product of two even numbers is divisible by 4 (since 2 × 2 = 4). However, *consecutive* even numbers guarantee that one is a multiple of 2 and the other is a multiple of 4, meaning their product is actually divisible by 8. Underestimating this factor leads to incorrect minimum bounds. Final Answer: Therefore, the correct answer is divisible by 24 always.

Question 17

The sum of the digits of a 3-digit number is subtracted from the number. The resulting number is always

Accepted Answer

Concept & Formula: Any generic number can be expanded algebraically into its base-10 place values. This is the key to proving universal divisibility rules for digits. A 3-digit number = 100x + 10y + z Where x, y, and z are the hundreds, tens, and units digits respectively. Step-by-Step Solution: • Step 1: Set up the original number. Let our 3-digit number be represented as 100x + 10y + z. • Step 2: Set up the sum of the digits. The literal sum of its digits is simply (x + y + z). • Step 3: Perform the subtraction. The problem states to subtract the digit sum from the original number: • Result = (100x + 10y + z) - (x + y + z) • Combine like terms: Result = 100x - x + 10y - y + z - z • Result = 99x + 9y • Step 4: Factor the result. * Result = 9(11x + y) • Conclusion: Because the final algebraic expression has a global factor of 9, the resulting number will always be a perfect multiple of 9, regardless of what the original digits were. Exam Strategy & Shortcut: Pick a real number: Don't waste time setting up algebra if you can do quick mental math. Pick any random 3-digit number, for example, 123. Sum of its digits = 1 + 2 + 3 = 6. Subtract the sum from the number: 123 - 6 = 117. Check the options: Is 117 divisible by 9? Check the digit sum of 117: 1 + 1 + 7 = 9. Yes, it is perfectly divisible by 9. Common Pitfall: Students sometimes randomly guess divisibility by 6 (option d) because they pick a test number like 111 (Sum = 3. 111 - 3 = 108. 108 is divisible by both 9 and 6). You must pick a number that breaks the false patterns. 123 yields 117, which is NOT divisible by 6, proving option (d) false and leaving option (b) as the universal truth. Final Answer: Therefore, the correct answer is divisible by 9.

Question 18

A number is multiplied by 11 and 11 is added to the product. If the resulting number is divisible by 13, the smallest original number is

Accepted Answer

Concept & Strategy: This problem combines basic linear algebra with modular arithmetic (divisibility). Factoring out common multipliers is the fastest way to simplify the divisibility check. 11x + 11 \equiv 0 \pmod{13} Step-by-Step Solution: • Step 1: Translate the word problem into algebra. Let the original unknown number be x. • Multiply by 11: 11x • Add 11 to the product: 11x + 11 • Step 2: Factor the expression. * 11x + 11 = 11(x + 1) • Step 3: Apply the divisibility condition. We are given that 11(x + 1) is perfectly divisible by 13. • Step 4: Use co-prime logic. 11 and 13 are prime numbers; they share no common factors. Therefore, for the entire product 11(x + 1) to be divisible by 13, the (x + 1) portion *must* be the part that is divisible by 13. • Step 5: Find the minimum value. For (x + 1) to be a multiple of 13, the smallest positive integer multiple is 13 itself. • x + 1 = 13 • x = 12 Exam Strategy & Shortcut: Backsolving from Options: When asked for the "smallest" number, immediately start plugging the provided options into the described process, starting from the smallest option (a). Test Option (a) 12: • Multiply by 11: 12 × 11 = 132 • Add 11: 132 + 11 = 143 • Check divisibility by 13: 143 \div 13 = 11. It works perfectly! Since it's the smallest option provided, you don't even need to test the others. Common Pitfall: A common mistake is forgetting to factor the 11x + 11 and trying to solve 11x \equiv -11 \pmod{13} using complex modular inverses. While mathematically valid, it takes significantly more time than simple factoring or backsolving. Final Answer: Therefore, the correct answer is 12.

Question 19

The greatest number by which the product of three consecutive multiples of 3 is always divisible is

Accepted Answer

Concept & Logic: When dealing with consecutive multiples of a specific number, factor out the multiplier from each term first, then apply the standard rule for consecutive integers (n!). Step-by-Step Solution: • Let the three consecutive multiples of 3 be 3n, 3(n+1), and 3(n+2), where n is a natural number. • Multiply them together: 3n × 3(n+1) × 3(n+2) = 27[n(n+1)(n+2)] • The term n(n+1)(n+2) is the product of three consecutive natural numbers. As established, this is always divisible by 3!, which is 6. • Therefore, the entire expression is always divisible by 27 × 6 = 162. Exam Strategy & Shortcut: Bypass the algebra completely by finding the product of the smallest possible sequence. The smallest consecutive multiples of 3 are 3, 6, and 9. Their product is 3 × 6 × 9 = 162. Since the question asks for the greatest number that *always* divides the product, it cannot be larger than the smallest possible product itself. Common Pitfall: A common error is factoring out 3 only once instead of three times, mistakenly believing the product is 3 × [n(n+1)(n+2)] = 18. Remember that each of the three terms contributes a factor of 3. Final Answer: Therefore, the correct answer is 162.

Question 20

If $p$ is a prime number greater than 3, then $(p^2 - 1)$ is always divisible by

Accepted Answer

Concept & Logic: Any prime number greater than 3 can be expressed in the form 6k \pm 1. Alternatively, you can analyze the algebraic expansion of the difference of squares: a^2 - b^2 = (a-b)(a+b). Step-by-Step Solution: • Expand the given expression: p^2 - 1 = (p - 1)(p + 1) • Since p is a prime number greater than 3, p must be an odd number. • Therefore, both (p - 1) and (p + 1) are consecutive even numbers. • In any two consecutive even numbers, one is a multiple of 2 and the other is a multiple of 4. Their product is therefore divisible by 2 × 4 = 8. • Additionally, consider the three consecutive integers: (p - 1), p, and (p + 1). Exactly one of them must be a multiple of 3. • Since p is a prime > 3, p cannot be the multiple of 3. Thus, either (p - 1) or (p + 1) must be divisible by 3. • Combining these factors, the product (p - 1)(p + 1) is perfectly divisible by both 8 and 3, meaning it is divisible by 8 × 3 = 24. Exam Strategy & Shortcut: Use value substitution for instant results. Take the first prime number greater than 3, which is p = 5. Calculate 5^2 - 1 = 25 - 1 = 24. Check the next prime, p = 7. Calculate 7^2 - 1 = 49 - 1 = 48. Both results are divisible by 24, firmly pointing to option (c). Common Pitfall: Students often pick p = 2 or p = 3 without reading the constraint "> 3". If p = 3, 3^2 - 1 = 8, which is not divisible by 24, leading test-takers to incorrectly choose "None of these". Always honor the constraints in the prompt. Final Answer: Therefore, the correct answer is 24.

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