Question 1

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 2

The difference between two numbers is 1365. When the larger number is divided by the smaller one, the quotient is 6 and the remainder is 15. What is the smaller number?

Accepted Answer

Concept & Formula: This problem requires translating two distinct conditions into a system of linear equations using the standard Division Algorithm. Dividend = (Divisor × Quotient) + Remainder Step-by-Step Solution: • Given: Let the larger number be L and the smaller number be S. • Condition 1: The difference is 1365. L - S = 1365 — (Equation 1) • Condition 2: Division algorithm variables. Dividend = L, Divisor = S, Quotient = 6, Remainder = 15. L = 6S + 15 — (Equation 2) • Calculation: • Substitute the value of L from Equation 2 into Equation 1: (6S + 15) - S = 1365 • Combine like terms: 5S + 15 = 1365 • Isolate the variable term: 5S = 1365 - 15 5S = 1350 • Solve for the smaller number: S = 1350 / 5 S = 270 Exam Strategy & Shortcut: Once you establish the equation 5S = 1350, look at the options before doing full division. The number 1350 ends in a 0. When divided by 5, the result must end in an even number or zero. 135 \div 5 = 27, so appending the zero gives 270. It quickly matches option (b). Common Pitfall: Students sometimes reverse the variables in the difference equation (writing S - L = 1365), leading to negative numbers and confusion. Always explicitly define which variable represents the larger value before building equations. Final Answer: Therefore, the correct answer is 270.

Question 3

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 4

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 5

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 6

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 7

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 8

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 9

If $17^{200}$ is divided by $18$, the remainder is

Accepted Answer

Concept & Formula: This problem is solved using the Remainder Theorem and the concept of Negative Remainders. When a number N is divided by a divisor D, we can express N in terms of D. If the number is slightly less than the divisor, using a negative remainder makes power calculations extremely easy. (ax - 1)^n \pmod x = (-1)^n Step-by-Step Solution: • Given: * The expression is 17^{200}. • The divisor is 18. • Calculation: • We can rewrite the base 17 in terms of the divisor 18: 17 = 18 - 1 • Substitute this into the expression: \frac{(18 - 1)^{200}}{18} • When expanding (18 - 1)^{200} using the binomial theorem, every term will contain a multiple of 18 except for the last term. • The remainder depends entirely on the last term: Remainder = (-1)^{200} • Since 200 is an even power, raising a negative number to an even power results in a positive number: (-1)^{200} = 1 Exam Strategy & Shortcut: Memorize this standard format: (x^n/x+1). If n is even, the remainder is always 1. If n is odd, the remainder is always x. Here, x = 17 and n = 200 (even). Therefore, you can instantly conclude the remainder is 1 without writing anything down. Common Pitfall: Students often try to find the cyclicity of 17, which is a massive waste of time for this specific structure. Cyclicity is useful when the base and divisor don't have a difference of 1, but whenever you see numbers like 17 and 18, immediately look for the negative remainder shortcut. Final Answer: Therefore, the correct answer is 1.

Question 10

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 11

$(7^{19} + 2)$ 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 12

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 13

$(3^{25} + 3^{26} + 3^{27} + 3^{28})$ is divisible by

Accepted Answer

Concept & Formula: The core strategy here is to factor out the term with the smallest exponent. By applying the distributive property of exponents, we can simplify the expression into a product of a prime power and an integer, which will clearly reveal its divisors. x^a + x^{a+1} + x^{a+2} = x^a(1 + x + x^2) Step-by-Step Solution: • Given: The expression (3^{25} + 3^{26} + 3^{27} + 3^{28}). • Factor out the smallest power: Extract 3^{25} as a common factor from all terms. 3^{25} (1 + 3^1 + 3^2 + 3^3) • Evaluate the bracketed sum: 3^{25} (1 + 3 + 9 + 27) 3^{25} (40) • Decompose to match options: The options are 11, 16, 25, 30. We need to see if 3^{25} × 40 contains one of these. Let us borrow a 3 from 3^{25}: 3^{24} × 3 × 40 3^{24} × 120 • Deduction: Since 120 is exactly divisible by 30 (120 \div 30 = 4), the entire expression is divisible by 30. Exam Strategy & Shortcut: Whenever you see a sum of consecutive powers, immediately factor out the lowest power. Calculate the numerical value of the remaining bracket. The answer will almost always be a factor of that bracketed sum, or a factor of the bracketed sum multiplied by the base (in this case, 40 × 3 = 120, which reveals 30). Common Pitfall: Students often incorrectly try to add the bases or exponents directly (e.g., thinking 3^{25} + 3^{26} = 6^{51}), which violates fundamental exponent rules. Always factor instead of adding. Final Answer: Therefore, the correct answer is 30.

Question 14

$(9^6 + 1)$ 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 15

If $n$ is even, $(6^n - 1)$ is divisible by

Accepted Answer

Concept & Formula: This relies on standard algebraic identities regarding differences of powers. Specifically, for any integer x and y, the expression x^n - y^n is perfectly divisible by both (x + y) and (x - y) strictly when n is an even integer. Step-by-Step Solution: • Given: The expression 6^n - 1 where n is an even number. This can be written as 6^n - 1^n. • Apply Algebraic Rule: Because n is even, x^n - y^n is divisible by (x + y) and (x - y). • Here, x = 6 and y = 1. • Find the factors: * Factor 1: (6 + 1) = 7 • Factor 2: (6 - 1) = 5 • Determine compound divisibility: Since the expression is divisible by both 7 and 5, and because 7 and 5 are co-prime (they share no common factors), the expression must be divisible by their product. 7 × 5 = 35 Exam Strategy & Shortcut: The fastest way to solve "If n is..." questions is to substitute the smallest valid value for n. The smallest even positive integer is n = 2. Plug it in: 6^2 - 1 = 36 - 1 = 35. Look at the options: 35 is clearly divisible by 35. Common Pitfall: Students might accidentally substitute an odd number like n=1, yielding 6^1 - 1 = 5. While 5 is a factor, it doesn't map to the provided options and fails to capture the full divisibility condition granted by an *even* exponent. Final Answer: Therefore, the correct answer is 35.

Question 16

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 17

If $(67^{67} + 67)$ is divided by $68$, the remainder is

Accepted Answer

Concept & Logic: This relies heavily on modular arithmetic, specifically evaluating each term in an additive sequence independently against the divisor. (A + B) \pmod C \equiv (A \pmod C + B \pmod C) \pmod C Step-by-Step Solution: • Given: Find the remainder when (67^{67} + 67) is divided by 68. • Split the expression: Apply the modulo 68 operation to both terms independently. Term 1: 67^{67} \pmod{68} Term 2: 67 \pmod{68} • Evaluate Term 1 (The Exponent): Using negative remainders, 67 is (68 - 1). 67 \equiv -1 \pmod{68} (-1)^{67} = -1 So, the remainder for the first term is -1. • Evaluate Term 2 (The Constant): The number 67 divided by 68 simply leaves a remainder of 67. Alternatively, we can use the negative remainder -1 to make addition easier. • Combine Remainders: Add the remainders of the two terms together. Total Remainder = -1 + 67 = 66 Exam Strategy & Shortcut: Use negative remainders for all terms close to the divisor. 67 \pmod{68} is -1. Substitute -1 into the entire expression: (-1)^{67} + (-1) = -1 - 1 = -2. Convert the negative remainder back to positive: 68 - 2 = 66. This avoids large numbers entirely. Common Pitfall: Students often correctly evaluate 67^{67} \pmod{68} as -1 but forget to carry down the second term (+ 67) from the original problem, leading them to incorrectly select 67 as the final answer. Final Answer: Therefore, the correct answer is 66.

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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