Q: In a division sum, the quotient, dividend and remainder are $15$, $940$ and $25$ respectively. The divisor is

Concept & Formula: The core principle relies on the standard Division Algorithm, which establishes the fundamental relationship between the four components of any division operation. Dividend = (Divisor × Quotient) + Remainder Step-by-Step Solution: • Given: * Quotient = 15 • Dividend = 940 • Remainder = 25 • Calculation: • Substitute the known values into the division formula: • 940 = (Divisor × 15) + 25 • Isolate the term with the Divisor by subtracting the Remainder from both sides: • 940 - 25 = Divisor × 15 • 915 = Divisor × 15 • Divide by the Quotient to find the Divisor: • Divisor = 915 / 15 • Divisor = 61 Exam Strategy & Shortcut: Instead of fully dividing 915 by 15, use the unit digit method to quickly verify the options. We need a number (Divisor) that, when multiplied by 15, gives a product ending in 5 (from 915). Looking at the options, only 61 ends in 1. 1 × 5 = 5. Thus, Option (d) is the only logical choice. Common Pitfall: Students often confuse the placement of the dividend and divisor in the formula, mistakenly trying to divide the dividend by the remainder or setting up the equation backwards. Always remember that the Dividend is the largest total block being broken down. Final Answer: Therefore, the correct answer is 61.

Q: In doing a question of division with zero remainder, a candidate took $12$ as divisor instead of $21$. The quotient obtained by him was $35$. The correct quotient is

Concept & Formula: When a remainder is zero, the Dividend is simply the product of the Divisor and the Quotient. If the divisor is misread, the underlying Dividend itself does not change; it remains a constant anchor we can use to find the correct quotient. Dividend = Divisor × Quotient Step-by-Step Solution: • Given: • Remainder = 0 • Incorrect Divisor = 12 • Incorrect Quotient = 35 • Correct Divisor = 21 • Calculation: • First, deduce the constant Dividend based on the candidate's incorrect reading: • Dividend = 12 × 35 • Dividend = 420 • Now, recalculate using the correct Divisor to find the true Quotient: • Correct Quotient = Dividend / Correct Divisor • Correct Quotient = 420 / 21 • Correct Quotient = 20 Exam Strategy & Shortcut: Since the Dividend is constant, you can set up an inverse proportion equation directly: D_1 × Q_1 = D_2 × Q_2 12 × 35 = 21 × Q_2 Simplify the ratio before multiplying: Q_2 = (12 × 35) / 21. Divide 35 and 21 by their common factor 7 to get 5 and 3. Then divide 12 by 3 to get 4. Finally, 4 × 5 = 20. This avoids large multiplications entirely. Common Pitfall: Calculating 12 × 35 manually can lead to arithmetic errors under time pressure. Delay multiplication until the final step. Setting it up as a fraction (12 × 35) / 21 allows for rapid cross-cancellation. Final Answer: Therefore, the correct answer is 20.

Q: In a division problem, the divisor is $7$ times of quotient and $5$ times of remainder. If the dividend is $6$ times of remainder, then the quotient is equal to

Concept & Logic: This problem is solved by translating word statements into algebraic equations and substituting them into the standard Division Algorithm. Dividend = (Divisor × Quotient) + Remainder Step-by-Step Solution: • Given Relationships: • Divisor (D) = 5 × Remainder (R) • Divisor (D) = 7 × Quotient (Q) • Dividend = 6 × Remainder (R) • Calculation: • Substitute the variables expressed in terms of R into the division formula: • 6R = (5R × Q) + R • Subtract R from both sides of the equation: • 6R - R = 5RQ • 5R = 5RQ • Assuming R is not zero (as that would mean the divisor is zero, which is undefined), we can divide both sides by 5R: • Q = 1 Exam Strategy & Shortcut: Algebraic substitution is the fastest path here. By recognizing that both the Dividend and Divisor can be expressed in terms of the Remainder (R), you can quickly set up the equation 6R = 5RQ + R. The R terms cancel out almost instantly in your head, leaving 5 = 5Q, which immediately gives Q = 1. Common Pitfall: A common trap is attempting to equate 7Q = 5R and solving for a numerical value prematurely without using the Dividend information. Always write down all given conditions before attempting to solve the system of equations. Final Answer: Therefore, the correct answer is 1.

Q: A number when divided by $195$ leaves a remainder $47$. If the same number is divided by $15$, the remainder will be

Concept & Rule: This relies on the transitive property of divisibility. If a large block (the original divisor) can be perfectly broken down into smaller blocks (the new divisor), you only need to evaluate the "leftover" pieces (the original remainder) to find the final remainder. New Remainder = Old Remainder \pmod{New Divisor} Step-by-Step Solution: • Given: • First Divisor = 195 • First Remainder = 47 • Second Divisor = 15 • Calculation: • Step 1: Confirm that 195 is a multiple of 15. • 15 × 10 = 150, and 15 × 3 = 45. 150 + 45 = 195. Yes, it divides perfectly. • Step 2: Since the base divisor is compatible, divide the original remainder by the new divisor. • Divide 47 by 15. • 15 × 3 = 45. • 47 - 45 = 2. • The new remainder is 2. Exam Strategy & Shortcut: Bypass verifying the first divisor if you are severely short on time. In 99% of competitive exam questions formatted exactly like this, the first divisor is deliberately designed to be a multiple of the second. Jump straight to dividing the old remainder by the new divisor: 47 / 15 \rightarrow remainder is 2. Common Pitfall: A fatal error occurs if a student divides the first divisor (195) by the new divisor (15), gets a quotient of 13, and mistakenly marks a derivation of 13 as the answer. Always focus operations strictly on the *remainder* once divisibility is confirmed. Final Answer: Therefore, the correct answer is 2.

Q: A number when divided by 5 leaves the remainder 3. What is the remainder when the square of the same number is divided by 5?

Concept & Logic: According to the properties of remainders in division, whatever operation is performed on the original number (addition, multiplication, squaring) can be directly applied to its remainder to find the new remainder. New Remainder = (R_1)^2 \pmod{D} Step-by-Step Solution: • Let the unknown number be N. • When N is divided by 5, the remainder is 3. We can write this algebraically as: N = 5k + 3 (where k is an integer quotient) • The question asks for the remainder when the square of the number (N^2) is divided by 5. Let's square our equation: N^2 = (5k + 3)^2 N^2 = 25k^2 + 30k + 9 • Now, divide this entire expression by 5. • The terms 25k^2 and 30k are completely divisible by 5, leaving no remainder. • We only need to find the remainder when 9 is divided by 5. • 9 \div 5 gives a quotient of 1 and a remainder of 4. Exam Strategy & Shortcut: Bypass the algebra entirely. When a problem asks for the remainder of a squared number, simply square the original remainder and divide that by the divisor. Original remainder = 3. Square it = 3^2 = 9. Divide 9 by 5. The remainder is 4. This takes less than 3 seconds. Common Pitfall: A frequent mistake is squaring the remainder (getting 9) and selecting it as the answer, forgetting that a remainder can never be larger than the divisor (which is 5). You must perform that final division step. Final Answer: Therefore, the correct answer is 4.

Q: When $n$ is divided by 4, the remainder is 3. What is the remainder when $2n$ is divided by 4?

Concept & Logic: Similar to squaring, if a number is multiplied by a constant, its remainder when divided by the same divisor is also multiplied by that constant. New Remainder = (2 × R_1) \pmod{D} Step-by-Step Solution: • Algebraic Method: • Let the number be n. • According to the division rule, we can express n as: n = 4k + 3 (where k is some integer) • The question asks about 2n. Multiply the entire equation by 2: 2n = 2(4k + 3) 2n = 8k + 6 • Now, we divide 2n by 4 to find the remainder. • The term 8k is perfectly divisible by 4. • We only evaluate the constant term, 6, divided by 4. • 6 \div 4 yields a quotient of 1 with a remainder of 2. Exam Strategy & Shortcut: The absolute fastest method for variable remainder problems is to assume the smallest possible valid integer for the variable. If n divided by 4 leaves a remainder of 3, assume the quotient is 0. Thus, n = 3. The problem asks for 2n. Double it: 2 × 3 = 6. Now divide 6 by 4. The remainder is 2. Common Pitfall: Multiplying the remainder (3 × 2 = 6) and marking 6 as the answer (Option d) is a very common trap. A remainder can never equal or exceed its divisor (4). Always complete the final modular reduction step. Final Answer: Therefore, the correct answer is 2.

Q: When a number is divided by 13, the remainder is 11. When the same number is divided by 17, the remainder is 9. What is the number?

Concept & Strategy: This problem represents a system of linear congruences. While there are formal algebraic methods to solve this (like the Chinese Remainder Theorem), they are far too time-consuming for competitive exams. The optimal strategy is direct option elimination. Step-by-Step Solution: • Let the unknown number be N. We have two conditions: 1) N \div 13 \rightarrow Remainder 11 2) N \div 17 \rightarrow Remainder 9 • Test Option (a) 339: • Divide by 13: 339 = 13 × 26 + 1. Remainder is 1 (Fails condition 1). • Test Option (b) 349: • Divide by 13: 349 = 13 × 26 + 11. Remainder is 11 (Passes condition 1). • Divide by 17: 349 = 17 × 20 + 9. Remainder is 9 (Passes condition 2). • Since 349 perfectly satisfies both conditions, we can stop testing here. Exam Strategy & Shortcut: Use mental estimation to speed up the division checks. For 13: You know 13 × 10 = 130, so 13 × 20 = 260. 349 - 260 = 89. The closest multiple of 13 to 89 is 13 × 6 = 78. 89 - 78 = 11. Condition 1 confirmed. For 17: You know 17 × 20 = 340. 349 - 340 = 9. Condition 2 confirmed instantly. Common Pitfall: Students often fall into the trap of trying to build and solve a general algebraic formula like N = 13x + 11 = 17y + 9. This turns into a tedious Diophantine equation and drains precious minutes. When the options contain specific numbers, always use them to back-solve. Final Answer: Therefore, the correct answer is 349.

Q: After the division of a number successively by $3$, $4$ and $7$, the remainders obtained are $2$, $1$ and $4$ respectively. What will be the remainder if $84$ divides the same number?

Concept & Logic: This problem is based on the concept of Successive Division. In successive division, the quotient of the first division is taken as the dividend for the second division, and so on. The general relation is: Dividend = (Divisor × Quotient) + Remainder Step-by-Step Solution: • Given: • Divisors in sequence: 3, 4, 7 • Respective remainders: 2, 1, 4 • Deduction (Working Backwards): • Let's assume the final quotient for the last division (by 7) is 0 to find the smallest possible valid number. • Step 1: The number before being divided by 7 (let's call it quotient q_2). q_2 = (7 × 0) + 4 = 4 • Step 2: The number before being divided by 4 (let's call it quotient q_1). q_1 = (4 × 4) + 1 = 16 + 1 = 17 • Step 3: The original number N before being divided by 3. N = (3 × 17) + 2 = 51 + 2 = 53 • Calculation: • The smallest possible original number N is 53. • The question asks for the remainder when this number is divided by 84. • Since 53 is less than 84, dividing 53 by 84 simply leaves a remainder of 53. Exam Strategy & Shortcut: Whenever you are asked to divide the number by the product of the successive divisors (here, 3 × 4 × 7 = 84), the smallest number you derive by assuming the final quotient as 0 will always be the remainder itself. Just work backwards once: 7(0)+4=4, 4(4)+1=17, 3(17)+2=53. Mark 53 and move on. Common Pitfall: A major mistake is treating "successive division" like normal division (trying to find an LCM or solving simultaneous equations). Always remember the chain reaction: the quotient of one step is the dividend of the next. Final Answer: Therefore, the correct answer is 53.

Q: A number when divided by $3$ leaves a remainder $1$. When the quotient is divided by $2$, it leaves a remainder $1$. What will be the remainder when the number is divided by $6$?

Concept & Logic: This is a classic Successive Division problem disguised with slightly different phrasing ("When the quotient is divided by..."). We find the smallest possible original number by working backward and then apply the final division condition. Step-by-Step Solution: • Given: • First divisor = 3, Remainder = 1. • Second divisor = 2, Remainder = 1. • Deduction: • We work backward starting from the second division. To find the smallest valid number, assume the final quotient is 0. • Let the intermediate quotient be q. q = (2 × 0) + 1 = 1 • Now, reconstruct the original number N using the first divisor and remainder, along with our intermediate quotient q. N = (3 × q) + 1 N = (3 × 1) + 1 = 4 • Calculation: • We have determined the original number N is 4. • The question asks for the remainder when N is divided by 6. • 4 \div 6 = 0 with a remainder of 4. Exam Strategy & Shortcut: You can verify this quickly by assuming the final quotient is 1 instead of 0. q = (2 × 1) + 1 = 3. N = (3 × 3) + 1 = 10. Now divide 10 by 6. 10 \div 6 \rightarrow Remainder 4. The remainder remains consistent regardless of the final quotient you choose. Always pick 0 for the fastest mental math. Common Pitfall: A frequent trap is reading this as simultaneous independent conditions (e.g., N \pmod 3 = 1 and N \pmod 2 = 1). That would give N=7, which leaves a remainder of 1 when divided by 6 (an option not present, luckily). Pay close attention to the phrase "When the quotient is divided...", indicating successive division. Final Answer: Therefore, the correct answer is 4.

Q: When the square of any odd number, greater than 1, is divided by $8$, it always leaves remainder

Concept & Formula: This problem relies on the Algebraic Properties of Odd Numbers. Any odd integer can be universally represented by the algebraic expression (2k + 1), where k is any integer. Step-by-Step Solution: • Deduction via Algebra: • Let an odd number be represented as n = 2k + 1. • The square of this odd number is: n^2 = (2k + 1)^2 • Expand the square using the identity (a+b)^2 = a^2 + 2ab + b^2: n^2 = 4k^2 + 4k + 1 • Factor out 4k from the first two terms: n^2 = 4k(k + 1) + 1 • Calculation: • Look at the term k(k + 1). This represents the product of two consecutive integers. • The product of any two consecutive integers is always even (i.e., divisible by 2). • Therefore, we can write k(k + 1) = 2m, for some integer m. • Substitute this back into our equation: n^2 = 4(2m) + 1 n^2 = 8m + 1 • This final form shows that n^2 is exactly 1 more than a multiple of 8. Thus, dividing by 8 will always leave a remainder of 1. Exam Strategy & Shortcut: Do not use algebra in a timed exam for universal property questions. Use Substitution. Pick the smallest odd numbers greater than 1: Take 3: 3^2 = 9. 9 \div 8 leaves remainder 1. Take 5: 5^2 = 25. 25 \div 8 leaves remainder 1. Take 7: 7^2 = 49. 49 \div 8 leaves remainder 1. The pattern is instantly obvious. Common Pitfall: Overthinking the algebraic proof can eat up valuable time. While the proof is mathematically elegant and important to understand, experimental verification with two simple test cases is vastly superior for competitive exams. Final Answer: Therefore, the correct answer is 1.

Question 1

Match List I with List II and select the correct answer:  List I ($a, b$ as given in Euclidean algorithm $a = bq + r$) A. $a = -112, b = -7$ B. $a = 118, b = -9$ C. $a = -109, b = 6$ D. $a = 115, b = 8$  List II (Values of $q$ and $r$) 1. $q = -13, …

Accepted Answer

Concept & Formula: The Euclidean division algorithm states that for any two integers a and b (where b 
eq 0), there exist unique integers q (quotient) and r (remainder) such that: a = bq + r Crucial Constraint: The remainder r must always be non-negative and strictly less than the absolute value of the divisor. Mathematically: 0 \le r < |b|. Step-by-Step Solution: Let's evaluate each pair to find valid q and r values that satisfy 0 \le r < |b|. • Evaluate A: a = -112, b = -7 -112 \div -7 = 16 exactly. Equation: -112 = -7(16) + 0. Here, q = 16, r = 0. This maps A to 4. • Evaluate D (Easiest positive numbers next): a = 115, b = 8 115 \div 8 = 14 with a remainder. Equation: 115 = 8(14) + r \implies 115 = 112 + 3. Here, q = 14, r = 3. This maps D to 2. • Evaluate B: a = 118, b = -9 We need a quotient q such that -9 × q is just below 118. Try q = -13: -9(-13) = 117. Equation: 118 = 117 + r \implies r = 1. (Since 0 \le 1 < |-9|, this is valid). Here, q = -13, r = 1. This maps B to 1. • Evaluate C: a = -109, b = 6 We need a quotient q such that 6 × q \le -109. Try q = -18: 6(-18) = -108 \implies r = -1 (Invalid, remainder must be positive). Try q = -19: 6(-19) = -114. Equation: -109 = -114 + r \implies r = 5. (Since 0 \le 5 < |6|, this is valid). Here, q = -19, r = 5. This maps C to 3. The final matching sequence is A-4, B-1, C-3, D-2. Exam Strategy & Shortcut: Use Option Elimination. Start with the easiest calculations: 1. D (115, 8) is basic division. 115 = 8 × 14 + 3. D matches 2. Looking at the options, only (a) and (c) end in 2. 2. Next easiest is A (-112, -7). -112 = -7 × 16 + 0. A matches 4. Looking at the remaining options (a) and (c), only option (c) starts with 4. You can select (c) immediately without calculating B or C. Common Pitfall: When dealing with negative dividends like C (-109 divided by 6), students often mistakenly choose quotient -18 and remainder -1. While -108 - 1 = -109, it violates the strict rule that remainders in Euclidean algorithms cannot be negative. You must push the quotient one integer further down (to -19) to force a positive remainder. Final Answer: Therefore, the correct answer is 4 1 3 2.

Question 2

In a division sum, the quotient, dividend and remainder are $15$, $940$ and $25$ respectively. The divisor is

Accepted Answer

Concept & Formula: The core principle relies on the standard Division Algorithm, which establishes the fundamental relationship between the four components of any division operation. Dividend = (Divisor × Quotient) + Remainder Step-by-Step Solution: • Given: * Quotient = 15 • Dividend = 940 • Remainder = 25 • Calculation: • Substitute the known values into the division formula: • 940 = (Divisor × 15) + 25 • Isolate the term with the Divisor by subtracting the Remainder from both sides: • 940 - 25 = Divisor × 15 • 915 = Divisor × 15 • Divide by the Quotient to find the Divisor: • Divisor = 915 / 15 • Divisor = 61 Exam Strategy & Shortcut: Instead of fully dividing 915 by 15, use the unit digit method to quickly verify the options. We need a number (Divisor) that, when multiplied by 15, gives a product ending in 5 (from 915). Looking at the options, only 61 ends in 1. 1 × 5 = 5. Thus, Option (d) is the only logical choice. Common Pitfall: Students often confuse the placement of the dividend and divisor in the formula, mistakenly trying to divide the dividend by the remainder or setting up the equation backwards. Always remember that the Dividend is the largest total block being broken down. Final Answer: Therefore, the correct answer is 61.

Question 3

In doing a question of division with zero remainder, a candidate took $12$ as divisor instead of $21$. The quotient obtained by him was $35$. The correct quotient is

Accepted Answer

Concept & Formula: When a remainder is zero, the Dividend is simply the product of the Divisor and the Quotient. If the divisor is misread, the underlying Dividend itself does not change; it remains a constant anchor we can use to find the correct quotient. Dividend = Divisor × Quotient Step-by-Step Solution: • Given: • Remainder = 0 • Incorrect Divisor = 12 • Incorrect Quotient = 35 • Correct Divisor = 21 • Calculation: • First, deduce the constant Dividend based on the candidate's incorrect reading: • Dividend = 12 × 35 • Dividend = 420 • Now, recalculate using the correct Divisor to find the true Quotient: • Correct Quotient = Dividend / Correct Divisor • Correct Quotient = 420 / 21 • Correct Quotient = 20 Exam Strategy & Shortcut: Since the Dividend is constant, you can set up an inverse proportion equation directly: D_1 × Q_1 = D_2 × Q_2 12 × 35 = 21 × Q_2 Simplify the ratio before multiplying: Q_2 = (12 × 35) / 21. Divide 35 and 21 by their common factor 7 to get 5 and 3. Then divide 12 by 3 to get 4. Finally, 4 × 5 = 20. This avoids large multiplications entirely. Common Pitfall: Calculating 12 × 35 manually can lead to arithmetic errors under time pressure. Delay multiplication until the final step. Setting it up as a fraction (12 × 35) / 21 allows for rapid cross-cancellation. Final Answer: Therefore, the correct answer is 20.

Question 4

In a division problem, the divisor is $7$ times of quotient and $5$ times of remainder. If the dividend is $6$ times of remainder, then the quotient is equal to

Accepted Answer

Concept & Logic: This problem is solved by translating word statements into algebraic equations and substituting them into the standard Division Algorithm. Dividend = (Divisor × Quotient) + Remainder Step-by-Step Solution: • Given Relationships: • Divisor (D) = 5 × Remainder (R) • Divisor (D) = 7 × Quotient (Q) • Dividend = 6 × Remainder (R) • Calculation: • Substitute the variables expressed in terms of R into the division formula: • 6R = (5R × Q) + R • Subtract R from both sides of the equation: • 6R - R = 5RQ • 5R = 5RQ • Assuming R is not zero (as that would mean the divisor is zero, which is undefined), we can divide both sides by 5R: • Q = 1 Exam Strategy & Shortcut: Algebraic substitution is the fastest path here. By recognizing that both the Dividend and Divisor can be expressed in terms of the Remainder (R), you can quickly set up the equation 6R = 5RQ + R. The R terms cancel out almost instantly in your head, leaving 5 = 5Q, which immediately gives Q = 1. Common Pitfall: A common trap is attempting to equate 7Q = 5R and solving for a numerical value prematurely without using the Dividend information. Always write down all given conditions before attempting to solve the system of equations. Final Answer: Therefore, the correct answer is 1.

Question 5

A number when divided by $195$ leaves a remainder $47$. If the same number is divided by $15$, the remainder will be

Accepted Answer

Concept & Rule: This relies on the transitive property of divisibility. If a large block (the original divisor) can be perfectly broken down into smaller blocks (the new divisor), you only need to evaluate the "leftover" pieces (the original remainder) to find the final remainder. New Remainder = Old Remainder \pmod{New Divisor} Step-by-Step Solution: • Given: • First Divisor = 195 • First Remainder = 47 • Second Divisor = 15 • Calculation: • Step 1: Confirm that 195 is a multiple of 15. • 15 × 10 = 150, and 15 × 3 = 45. 150 + 45 = 195. Yes, it divides perfectly. • Step 2: Since the base divisor is compatible, divide the original remainder by the new divisor. • Divide 47 by 15. • 15 × 3 = 45. • 47 - 45 = 2. • The new remainder is 2. Exam Strategy & Shortcut: Bypass verifying the first divisor if you are severely short on time. In 99% of competitive exam questions formatted exactly like this, the first divisor is deliberately designed to be a multiple of the second. Jump straight to dividing the old remainder by the new divisor: 47 / 15 \rightarrow remainder is 2. Common Pitfall: A fatal error occurs if a student divides the first divisor (195) by the new divisor (15), gets a quotient of 13, and mistakenly marks a derivation of 13 as the answer. Always focus operations strictly on the *remainder* once divisibility is confirmed. Final Answer: Therefore, the correct answer is 2.

Question 6

A number when divided by 5 leaves the remainder 3. What is the remainder when the square of the same number is divided by 5?

Accepted Answer

Concept & Logic: According to the properties of remainders in division, whatever operation is performed on the original number (addition, multiplication, squaring) can be directly applied to its remainder to find the new remainder. New Remainder = (R_1)^2 \pmod{D} Step-by-Step Solution: • Let the unknown number be N. • When N is divided by 5, the remainder is 3. We can write this algebraically as: N = 5k + 3 (where k is an integer quotient) • The question asks for the remainder when the square of the number (N^2) is divided by 5. Let's square our equation: N^2 = (5k + 3)^2 N^2 = 25k^2 + 30k + 9 • Now, divide this entire expression by 5. • The terms 25k^2 and 30k are completely divisible by 5, leaving no remainder. • We only need to find the remainder when 9 is divided by 5. • 9 \div 5 gives a quotient of 1 and a remainder of 4. Exam Strategy & Shortcut: Bypass the algebra entirely. When a problem asks for the remainder of a squared number, simply square the original remainder and divide that by the divisor. Original remainder = 3. Square it = 3^2 = 9. Divide 9 by 5. The remainder is 4. This takes less than 3 seconds. Common Pitfall: A frequent mistake is squaring the remainder (getting 9) and selecting it as the answer, forgetting that a remainder can never be larger than the divisor (which is 5). You must perform that final division step. Final Answer: Therefore, the correct answer is 4.

Question 7

When $n$ is divided by 4, the remainder is 3. What is the remainder when $2n$ is divided by 4?

Accepted Answer

Concept & Logic: Similar to squaring, if a number is multiplied by a constant, its remainder when divided by the same divisor is also multiplied by that constant. New Remainder = (2 × R_1) \pmod{D} Step-by-Step Solution: • Algebraic Method: • Let the number be n. • According to the division rule, we can express n as: n = 4k + 3 (where k is some integer) • The question asks about 2n. Multiply the entire equation by 2: 2n = 2(4k + 3) 2n = 8k + 6 • Now, we divide 2n by 4 to find the remainder. • The term 8k is perfectly divisible by 4. • We only evaluate the constant term, 6, divided by 4. • 6 \div 4 yields a quotient of 1 with a remainder of 2. Exam Strategy & Shortcut: The absolute fastest method for variable remainder problems is to assume the smallest possible valid integer for the variable. If n divided by 4 leaves a remainder of 3, assume the quotient is 0. Thus, n = 3. The problem asks for 2n. Double it: 2 × 3 = 6. Now divide 6 by 4. The remainder is 2. Common Pitfall: Multiplying the remainder (3 × 2 = 6) and marking 6 as the answer (Option d) is a very common trap. A remainder can never equal or exceed its divisor (4). Always complete the final modular reduction step. Final Answer: Therefore, the correct answer is 2.

Question 8

When a number is divided by 13, the remainder is 11. When the same number is divided by 17, the remainder is 9. What is the number?

Accepted Answer

Concept & Strategy: This problem represents a system of linear congruences. While there are formal algebraic methods to solve this (like the Chinese Remainder Theorem), they are far too time-consuming for competitive exams. The optimal strategy is direct option elimination. Step-by-Step Solution: • Let the unknown number be N. We have two conditions: 1) N \div 13 \rightarrow Remainder 11 2) N \div 17 \rightarrow Remainder 9 • Test Option (a) 339: • Divide by 13: 339 = 13 × 26 + 1. Remainder is 1 (Fails condition 1). • Test Option (b) 349: • Divide by 13: 349 = 13 × 26 + 11. Remainder is 11 (Passes condition 1). • Divide by 17: 349 = 17 × 20 + 9. Remainder is 9 (Passes condition 2). • Since 349 perfectly satisfies both conditions, we can stop testing here. Exam Strategy & Shortcut: Use mental estimation to speed up the division checks. For 13: You know 13 × 10 = 130, so 13 × 20 = 260. 349 - 260 = 89. The closest multiple of 13 to 89 is 13 × 6 = 78. 89 - 78 = 11. Condition 1 confirmed. For 17: You know 17 × 20 = 340. 349 - 340 = 9. Condition 2 confirmed instantly. Common Pitfall: Students often fall into the trap of trying to build and solve a general algebraic formula like N = 13x + 11 = 17y + 9. This turns into a tedious Diophantine equation and drains precious minutes. When the options contain specific numbers, always use them to back-solve. Final Answer: Therefore, the correct answer is 349.

Question 9

After the division of a number successively by $3$, $4$ and $7$, the remainders obtained are $2$, $1$ and $4$ respectively. What will be the remainder if $84$ divides the same number?

Accepted Answer

Concept & Logic: This problem is based on the concept of Successive Division. In successive division, the quotient of the first division is taken as the dividend for the second division, and so on. The general relation is: Dividend = (Divisor × Quotient) + Remainder Step-by-Step Solution: • Given: • Divisors in sequence: 3, 4, 7 • Respective remainders: 2, 1, 4 • Deduction (Working Backwards): • Let's assume the final quotient for the last division (by 7) is 0 to find the smallest possible valid number. • Step 1: The number before being divided by 7 (let's call it quotient q_2). q_2 = (7 × 0) + 4 = 4 • Step 2: The number before being divided by 4 (let's call it quotient q_1). q_1 = (4 × 4) + 1 = 16 + 1 = 17 • Step 3: The original number N before being divided by 3. N = (3 × 17) + 2 = 51 + 2 = 53 • Calculation: • The smallest possible original number N is 53. • The question asks for the remainder when this number is divided by 84. • Since 53 is less than 84, dividing 53 by 84 simply leaves a remainder of 53. Exam Strategy & Shortcut: Whenever you are asked to divide the number by the product of the successive divisors (here, 3 × 4 × 7 = 84), the smallest number you derive by assuming the final quotient as 0 will always be the remainder itself. Just work backwards once: 7(0)+4=4, 4(4)+1=17, 3(17)+2=53. Mark 53 and move on. Common Pitfall: A major mistake is treating "successive division" like normal division (trying to find an LCM or solving simultaneous equations). Always remember the chain reaction: the quotient of one step is the dividend of the next. Final Answer: Therefore, the correct answer is 53.

Question 10

A number when divided by $3$ leaves a remainder $1$. When the quotient is divided by $2$, it leaves a remainder $1$. What will be the remainder when the number is divided by $6$?

Accepted Answer

Concept & Logic: This is a classic Successive Division problem disguised with slightly different phrasing ("When the quotient is divided by..."). We find the smallest possible original number by working backward and then apply the final division condition. Step-by-Step Solution: • Given: • First divisor = 3, Remainder = 1. • Second divisor = 2, Remainder = 1. • Deduction: • We work backward starting from the second division. To find the smallest valid number, assume the final quotient is 0. • Let the intermediate quotient be q. q = (2 × 0) + 1 = 1 • Now, reconstruct the original number N using the first divisor and remainder, along with our intermediate quotient q. N = (3 × q) + 1 N = (3 × 1) + 1 = 4 • Calculation: • We have determined the original number N is 4. • The question asks for the remainder when N is divided by 6. • 4 \div 6 = 0 with a remainder of 4. Exam Strategy & Shortcut: You can verify this quickly by assuming the final quotient is 1 instead of 0. q = (2 × 1) + 1 = 3. N = (3 × 3) + 1 = 10. Now divide 10 by 6. 10 \div 6 \rightarrow Remainder 4. The remainder remains consistent regardless of the final quotient you choose. Always pick 0 for the fastest mental math. Common Pitfall: A frequent trap is reading this as simultaneous independent conditions (e.g., N \pmod 3 = 1 and N \pmod 2 = 1). That would give N=7, which leaves a remainder of 1 when divided by 6 (an option not present, luckily). Pay close attention to the phrase "When the quotient is divided...", indicating successive division. Final Answer: Therefore, the correct answer is 4.

Question 11

When the square of any odd number, greater than 1, is divided by $8$, it always leaves remainder

Accepted Answer

Concept & Formula: This problem relies on the Algebraic Properties of Odd Numbers. Any odd integer can be universally represented by the algebraic expression (2k + 1), where k is any integer. Step-by-Step Solution: • Deduction via Algebra: • Let an odd number be represented as n = 2k + 1. • The square of this odd number is: n^2 = (2k + 1)^2 • Expand the square using the identity (a+b)^2 = a^2 + 2ab + b^2: n^2 = 4k^2 + 4k + 1 • Factor out 4k from the first two terms: n^2 = 4k(k + 1) + 1 • Calculation: • Look at the term k(k + 1). This represents the product of two consecutive integers. • The product of any two consecutive integers is always even (i.e., divisible by 2). • Therefore, we can write k(k + 1) = 2m, for some integer m. • Substitute this back into our equation: n^2 = 4(2m) + 1 n^2 = 8m + 1 • This final form shows that n^2 is exactly 1 more than a multiple of 8. Thus, dividing by 8 will always leave a remainder of 1. Exam Strategy & Shortcut: Do not use algebra in a timed exam for universal property questions. Use Substitution. Pick the smallest odd numbers greater than 1: Take 3: 3^2 = 9. 9 \div 8 leaves remainder 1. Take 5: 5^2 = 25. 25 \div 8 leaves remainder 1. Take 7: 7^2 = 49. 49 \div 8 leaves remainder 1. The pattern is instantly obvious. Common Pitfall: Overthinking the algebraic proof can eat up valuable time. While the proof is mathematically elegant and important to understand, experimental verification with two simple test cases is vastly superior for competitive exams. Final Answer: Therefore, the correct answer is 1.

Question 12

The numbers from $1$ to $29$ are written side by side as follows:  $1234567891011121314..........2829$  If this number is divided by $9$, then what is the remainder?

Accepted Answer

Concept & Rule: This problem utilizes the Divisibility Rule of 9. A number is perfectly divisible by 9 if the sum of its digits is divisible by 9. Furthermore, the remainder obtained when dividing a number by 9 is strictly equal to the remainder obtained when dividing the *sum of its digits* by 9. Step-by-Step Solution: • Given: • A large number formed by writing 1 through 29 sequentially. • We need to find the remainder when this massive number is divided by 9. • Calculation: • We must calculate the total sum of all the individual digits from 1 to 29. • Group 1 (Single digits 1 to 9): Sum = 1 + 2 + 3 + ... + 9 = 45 • Group 2 (Two-digit numbers 10 to 19): Each number has a ten's digit of 1. There are ten such numbers. Sum of ten's digits = 10 × 1 = 10 Sum of unit's digits (0 to 9) = 45 Total sum for this group = 10 + 45 = 55 • Group 3 (Two-digit numbers 20 to 29): Each number has a ten's digit of 2. There are ten such numbers. Sum of ten's digits = 10 × 2 = 20 Sum of unit's digits (0 to 9) = 45 Total sum for this group = 20 + 45 = 65 • Total Sum of all digits: Total Sum = 45 + 55 + 65 = 165 • Now, divide this total sum by 9 to find the remainder: 165 \div 9 gives a quotient of 18 (18 × 9 = 162). 165 - 162 = 3 • The remainder is 3. Exam Strategy & Shortcut: You can apply the divisibility rule recursively to speed things up. Once you get the sum 165, you don't even need to divide it by 9. Just sum the digits of 165 again: 1 + 6 + 5 = 12. Sum the digits of 12 again: 1 + 2 = 3. The final single digit is your remainder. Common Pitfall: The most common mistake is attempting to sum the numbers themselves (e.g., using the arithmetic progression sum formula N(N+1)/2 to find the sum of 1+2+...+29) rather than summing the *individual digits* that make up the number string. The rule of 9 applies strictly to the individual digits. Final Answer: Therefore, the correct answer is 3.

Question 13

What is the remainder when $2^{31}$ is divided by $5$?

Accepted Answer

Concept & Logic: This problem requires understanding the Cyclicity of Remainders (or power cycles). When powers of a number are divided by a divisor, the remainders repeat in a fixed cycle. Finding the length of this cycle allows us to reduce huge exponents to manageable numbers. Step-by-Step Solution: • Calculation: • Let's find the pattern of remainders when powers of 2 are divided by 5: • 2^1 = 2 \implies 2 \pmod 5 = 2 • 2^2 = 4 \implies 4 \pmod 5 = 4 • 2^3 = 8 \implies 8 \pmod 5 = 3 • 2^4 = 16 \implies 16 \pmod 5 = 1 • Once we hit a remainder of 1, the cycle repeats. The cyclicity of 2 when divided by 5 is 4. • Now, divide the given exponent (31) by the cyclicity (4) to find the remaining cycle steps. 31 \div 4 = 7 with a remainder of 3 • This means the pattern repeats exactly 7 times, and we need the 3^{rd} value in the cycle. • The 3^{rd} value corresponds to 2^3. 2^{31} \pmod 5 \equiv 2^3 \pmod 5 = 8 \pmod 5 = 3 Exam Strategy & Shortcut: An alternative faster method is using Fermat's Little Theorem, which states that if p is a prime number, then a^{p-1} \pmod p = 1. Here, p = 5, so 2^4 \pmod 5 = 1. Break the exponent 31 into multiples of 4: 2^{31} = 2^{28} \cdot 2^3 = (2^4)^7 \cdot 2^3 Since 2^4 \pmod 5 = 1, the expression becomes 1^7 \cdot 8 = 8. Dividing 8 by 5 leaves 3. Common Pitfall: A common error is confusing the unit digit cycle with the remainder cycle. While they often align (because finding the unit digit is essentially finding the remainder when divided by 10), you must explicitly check the cycle for the specific divisor given in the question (in this case, 5). Final Answer: Therefore, the correct answer is 3.

Question 14

$(7^{19} + 2)$ is divided by $6$. The remainder is

Accepted Answer

Concept & Logic: This problem uses the Additive Property of Remainders. If you are dividing a sum of two terms by a divisor, you can find the remainder of each term individually and then add those remainders together. (A + B) \pmod D = [A \pmod D + B \pmod D] \pmod D Step-by-Step Solution: • Given: • Expression: 7^{19} + 2 • Divisor: 6 • Calculation: • Break the expression into two parts and divide each by 6. • Part 1: Find the remainder of 7^{19} \div 6. • Since 7 is exactly 1 more than the divisor 6, we can write 7 = 6 + 1. • 7^{19} \pmod 6 = (6 + 1)^{19} \pmod 6. • As per the remainder theorem, this simply becomes 1^{19} = 1. • Part 2: Find the remainder of 2 \div 6. • Since 2 is smaller than 6, the remainder is simply 2. • Combine: Add the individual remainders together. Total Remainder = 1 + 2 = 3 Exam Strategy & Shortcut: Whenever the base is exactly 1 more than the divisor (like 7 and 6), the remainder of that term to *any* power is always exactly 1. You can mentally skip all the math: 1 + 2 = 3. This should take no more than 3 seconds to solve on test day. Common Pitfall: A careless mistake is trying to evaluate the power or add the 2 to the base before dealing with the exponent (e.g., falsely thinking it acts like 9^{19}). Always isolate terms separated by a plus or minus sign when applying the remainder theorem. Final Answer: Therefore, the correct answer is 3.

Question 15

If $(10^{12} + 25)^2 - (10^{12} - 25)^2 = 10^n$, then the value of $n$ is

Accepted Answer

Concept & Formula: This problem is a direct application of standard Algebraic Identities. Instead of trying to calculate massively huge exponents, recognize the structure of the equation. The expression is in the form of: (a + b)^2 - (a - b)^2 The algebraic shortcut for this expansion is: (a + b)^2 - (a - b)^2 = 4ab Step-by-Step Solution: • Given: • The equation is (10^{12} + 25)^2 - (10^{12} - 25)^2 = 10^n • By matching the pattern, let a = 10^{12} and b = 25. • Calculation: • Apply the identity (a + b)^2 - (a - b)^2 = 4ab: = 4 \cdot (10^{12}) \cdot (25) • Rearrange the multiplication to group the constants: = (4 \cdot 25) \cdot 10^{12} • Simplify the constants: = 100 \cdot 10^{12} • Convert 100 into a power of 10: = 10^2 \cdot 10^{12} • Apply the law of exponents (x^m \cdot x^n = x^{m+n}): = 10^{2 + 12} = 10^{14} • The problem states this equals 10^n. Therefore: 10^{14} = 10^n \implies n = 14 Exam Strategy & Shortcut: Memorize the twin identities: 1. (a+b)^2 + (a-b)^2 = 2(a^2 + b^2) 2. (a+b)^2 - (a-b)^2 = 4ab When you see the minus sign between the squares, immediately multiply the two inner terms by 4. 4 × 25 is 100 (which is 10^2). Adding this power of 2 to the existing power of 12 instantly gives you 14. Common Pitfall: Students who forget the identity often attempt to partially expand the squares and get lost in the zero-counting, leading to arithmetic mistakes. Another error is writing 100 \cdot 10^{12} as 100^{12} or misapplying exponent laws when multiplying bases. Final Answer: Therefore, the correct answer is 14.

Question 16

$(9^6 + 1)$ when divided by $8$, would leave a remainder of

Accepted Answer

Concept & Formula: This question utilizes the Remainder Theorem and Modular Arithmetic. When a number (x + a)^n is divided by x, the remainder is strictly dependent on a^n. (x + a)^n \pmod x \equiv a^n \pmod x Step-by-Step Solution: • Given: We need to find the remainder of (9^6 + 1) divided by 8. • Rewrite the base: Express the base 9 in terms of the divisor 8. 9 = 8 + 1 So, the expression becomes ((8 + 1)^6 + 1) \div 8. • Apply the Remainder Theorem: The remainder of (8 + 1)^6 when divided by 8 is simply the remainder of 1^6. 1^6 = 1 • Calculate total remainder: We must not forget the "+ 1" present in the original expression. Total Remainder = 1 + 1 = 2 Exam Strategy & Shortcut: Use modular arithmetic directly. Since 9 \equiv 1 \pmod 8, any power of 9 will also be congruent to 1 \pmod 8. So, (9^6 + 1) \pmod 8 \equiv (1^6 + 1) \pmod 8 \equiv 2. This can be solved in under 5 seconds without writing anything down. Common Pitfall: The most common error is finding the remainder of 9^6 (which is 1) and forgetting to add the trailing +1 from the original question, leading a student to incorrectly choose option (b). Final Answer: Therefore, the correct answer is 2.

Question 17

If $(12^n + 1)$ is divisible by $13$, then $n$ is

Accepted Answer

Concept & Formula: This problem tests the algebraic identity for the sum of n-th powers. The expression x^n + y^n is divisible by (x + y) if and only if n is an odd integer. Step-by-Step Solution: • Given: The expression (12^n + 1) is perfectly divisible by 13. • Format to match identity: We can rewrite 13 as (12 + 1) and the expression as (12^n + 1^n). • Apply the rule: We know that x^n + a^n is divisible by (x + a) only when n is odd. • Here, x = 12 and a = 1. • So, (12^n + 1^n) is divisible by (12 + 1) = 13 only when n is an odd integer. Exam Strategy & Shortcut: Use modular arithmetic to prove it instantly. We want 12^n + 1 \equiv 0 \pmod{13}. Since 12 \equiv -1 \pmod{13}, substitute this into the equation: (-1)^n + 1 \equiv 0 \pmod{13} (-1)^n \equiv -1 \pmod{13} For (-1)^n to equal -1, n must absolutely be an odd number (like 1, 3, 5). Common Pitfall: A student might test n = 1, see that 12^1 + 1 = 13 works, and hastily select "1 only" (option a). Always remember that algebraic divisibility rules apply universally to sets of numbers (like all odds or all evens), not just a single instance. Final Answer: Therefore, the correct answer is any odd integer.

Question 18

One less than $(49)^{15}$ is exactly divisible by

Accepted Answer

Concept & Formula: This question tests the divisibility rules of algebraic polynomials. For any positive integer n, the difference of two n-th powers, x^n - y^n, is always perfectly divisible by (x - y). (x^n - y^n/x - y) = Integer Step-by-Step Solution: • Given: The mathematical expression for "One less than (49)^{15}" is (49^{15} - 1). • Format to match identity: We can rewrite the constant 1 as 1^{15} without changing its value. 49^{15} - 1^{15} • Apply the divisibility rule: Since x^n - y^n is divisible by (x - y), substitute x = 49 and y = 1. The expression is divisible by (49 - 1) = 48. • Find factors of 48: If a massive number is exactly divisible by 48, it must also be exactly divisible by all the fundamental factors of 48. Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. • Deduction: Check the given options (8, 14, 50, 51). The only option that is a factor of 48 is 8. Exam Strategy & Shortcut: When asked for a static divisor of an algebraic expression, substituting n = 1 often reveals the core factor. Let the exponent be 1. 49^1 - 1 = 48. Scan the options for a number that perfectly divides 48. Only 8 works. Common Pitfall: Seeing (49)^{15} - 1, a student might associate 49 with 7^2 and mistakenly guess that 14 is a divisor, or falsely apply the x^n + y^n rule to assume it is divisible by (49 + 1) = 50. Strictly follow the minus sign rule. Final Answer: Therefore, the correct answer is 8.

Question 19

The remainder when $7^{84}$ is divided by $342$ is

Accepted Answer

Concept & Strategy: When dividing a huge exponent by a specific number, the goal is to rewrite the base as a power that is exactly 1 more or 1 less than the divisor (or a multiple of the divisor). This allows us to use the Remainder Theorem: (a \cdot x \pm 1)^n \pmod x \equiv (\pm 1)^n. Step-by-Step Solution: • Given: Find the remainder for 7^{84} \div 342. • Find a strategic base: We need a power of 7 that is close to 342. Let us test the first few powers of 7: 7^1 = 7 7^2 = 49 7^3 = 343 • Rewrite the expression: Notice that 343 is exactly 1 more than the divisor 342. We must restructure 7^{84} to use 7^3. Divide the exponent 84 by 3. 84 \div 3 = 28 7^{84} = (7^3)^{28} = 343^{28} • Apply modular arithmetic: Now divide 343^{28} by 342. 343 \equiv 1 \pmod{342} 1^{28} = 1 Exam Strategy & Shortcut: Memorizing the powers of single-digit numbers up to their cubes is crucial. If you know 7^3 = 343, you immediately see the +1 relationship with 342. Because 343 leaves a remainder of 1, and 1 raised to *any* power is 1, you don't even need to calculate the new exponent (28). The answer must be 1. Common Pitfall: Students who don't know 7^3 = 343 will try to find a repeating cycle for the units digit of 7, which does not help when dividing by a large three-digit number like 342. Final Answer: Therefore, the correct answer is 1.

Question 20

By how many of the following numbers is $2^{12} - 1$ divisible?  $2, 3, 5, 7, 10, 11, 13, 14$

Accepted Answer

Concept & Formula: This problem tests algebraic factorization, specifically using the difference of two squares to break down large exponents into prime factors. x^2 - y^2 = (x - y)(x + y) Step-by-Step Solution: • Given: The expression 2^{12} - 1 and a list of potential divisors: 2, 3, 5, 7, 10, 11, 13, 14. • Rewrite as difference of squares: We can express 2^{12} as (2^6)^2. (2^6)^2 - 1^2 • Factorize: Apply the formula (x - y)(x + y). (2^6 - 1)(2^6 + 1) • Evaluate the brackets: We know 2^6 = 64. (64 - 1)(64 + 1) 63 × 65 • Find prime factors: Break down 63 and 65 into their prime factorization. • 63 = 9 × 7 = 3^2 × 7 • 65 = 5 × 13 So, the complete factorization of 2^{12} - 1 is 3^2 × 5 × 7 × 13. • Compare with the given list: * The individual prime factors present are 3, 5, 7, and 13. • Let's check the composites in the list: 10 requires a factor of 2 (none present), 11 is not present, 14 requires a factor of 2 (none present). • Therefore, the numbers from the list that divide the expression are exactly four: 3, 5, 7, 13. Exam Strategy & Shortcut: Memorize powers of 2 up to 2^{12}. 2^{10} = 1024, so 2^{12} = 4096. The question asks for factors of 4096 - 1 = 4095. Any number ending in 5 is odd, eliminating all even divisors immediately (2, 10, 14). 4095 ends in 5, so 5 is a factor. Sum of digits: 4+0+9+5=18 (divisible by 9, thus by 3). Divide 4095 \div 45 = 91. 91 is famously 7 × 13. Thus, the factors from the list are 3, 5, 7, 13 (a total of 4 numbers). Common Pitfall: A common mistake is forgetting to factor further after reaching 63 × 65 and trying to manually divide 4095 by every single number on the list, which wastes valuable time and risks arithmetic errors. Final Answer: Therefore, the correct answer is 4.

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