Question 1

If all the numbers from 501 to 700 are written, what is the total number of times the digit 6 appears?

Accepted Answer

Concept & Strategy: When asked to find how many times a specific digit appears in a given range of numbers, it is most efficient to count the occurrences of that digit in each positional place value independently (hundreds, tens, and units). Because we are counting the literal number of times the digit is written, a number like 666 will correctly be counted three times using this positional method. Step-by-Step Solution: • Given: The numerical range is from 501 to 700. We need to count the frequency of the digit 6. • Step 1: Count occurrences in the Hundreds place. The digit 6 appears in the hundreds place for all numbers from 600 to 699. Total occurrences = 100 times. • Step 2: Count occurrences in the Tens place. The digit 6 appears in the tens place for the 560s (560 to 569) and the 660s (660 to 669). There are 10 occurrences in the 500s and 10 occurrences in the 600s. Total occurrences = 20 times. • Step 3: Count occurrences in the Units place. The digit 6 appears in the units place whenever a number ends in 6 (e.g., 506, 516, 526, \dots, 596). In the 500s block, this happens 10 times. In the 600s block, this also happens 10 times (606, 616, \dots, 696). Total occurrences = 20 times. • Step 4: Sum the occurrences. Total = 100 (Hundreds) + 20 (Tens) + 20 (Units) = 140. Exam Strategy & Shortcut: Memorize the standard distribution of digits: In any full 100-number block (like 500 to 599), any given digit from 0-9 will appear exactly 20 times in the tens and units places combined. For 501 to 700, we have exactly two 100-number blocks (500s and 600s). Tens/Units count = 20 × 2 = 40. Then, just add the 100 occurrences where 6 is the leading digit in the 600s. 100 + 40 = 140. Solved in 5 seconds. Common Pitfall: A very common mistake is trying to list out numbers with 6 and mistakenly treating a number like 666 as a single occurrence, or conversely, subtracting overlaps. The question asks how many times the *digit* is written, not how many *numbers* contain the digit. Positional counting avoids this trap entirely. Final Answer: Therefore, the correct answer is 140.

Question 2

The total number of integers between 200 and 400, each of which either begins with 3 or ends with 3 or both is

Accepted Answer

Concept & Strategy: This problem relies on set theory, specifically the Principle of Inclusion-Exclusion, or simple logical division of a range to avoid double counting. We must count numbers that start with 3, numbers that end with 3, and ensure the overlap (numbers that start AND end with 3) is only counted once. Step-by-Step Solution: • Given: The target range is between 200 and 400. We are looking for integers that either begin with 3, end with 3, or both. • Step 1: Count numbers that begin with 3. Within the 200 to 400 range, all numbers in the 300s begin with 3. These span from 300 to 399. Total numbers = 100. *(Note: This block already includes all numbers that begin AND end with 3, like 303, 313, etc.)* • Step 2: Count numbers that end with 3, but do NOT begin with 3. Since we already completely covered the 300s, we only need to look at the remaining part of the range: the 200s (from 200 to 299). Numbers in this block ending in 3 are: 203, 213, 223, 233, 243, 253, 263, 273, 283, 293. Total numbers = 10. • Step 3: Combine the counts. Total valid integers = (Numbers starting with 3) + (Numbers ending with 3 in the remaining range) Total = 100 + 10 = 110. Exam Strategy & Shortcut: Break the problem into mutually exclusive sets to bypass inclusion-exclusion formulas entirely. Set A: The entire 300 series (100 numbers). Set B: The 200 series ending in 3. Since it's exactly one 100-number block, any specific unit digit appears exactly 10 times. 100 + 10 = 110. This mental math takes less than 10 seconds. Common Pitfall: Using the union formula n(A \cup B) = n(A) + n(B) - n(A \cap B) but miscalculating the sets. Students might incorrectly say there are 20 numbers ending in 3 between 200 and 400, add that to 100, and then subtract the 10 overlaps. While mathematically sound (100 + 20 - 10 = 110), it introduces unnecessary steps and room for arithmetic error. Final Answer: Therefore, the correct answer is 110.

Question 3

Which of the following prime numbers while dividing 2176 leaves 9 as remainder?

Accepted Answer

Concept & Strategy: By the division algorithm, if a number N divided by a divisor D leaves a remainder R, then the number (N - R) must be perfectly divisible by D. Once we find (N - R), we can test the options to see which prime perfectly divides it. Step-by-Step Solution: • Given: The original number is 2176 and the target remainder is 9. • Step 1: Subtract the remainder from the number. Perfectly divisible number = 2176 - 9 = 2167. The target prime number must be a factor of 2167. • Step 2: Check for prime factors using basic divisibility rules. Instead of testing large options immediately, quickly scan 2167 for small prime factors. Test for divisibility by 11: (Sum of digits in odd places) - (Sum of digits in even places). (2 + 6) - (1 + 7) = 8 - 8 = 0. Since the result is 0, 2167 is perfectly divisible by 11. • Step 3: Divide by the found factor to find the paired factor. 2167 \div 11 = 197. Thus, 2167 = 11 × 197. Both 11 and 197 are prime numbers that would leave a remainder of 9. • Step 4: Check options. 197 is present in the options. Exam Strategy & Shortcut: Subtract the remainder first to get 2167. Then, look at the unit digits of the options. • 17 × (something ending in 1) \rightarrow 7 (e.g. 17 × 100 = 1700, too small. 17 × 127?) • 29 × (something ending in 3) \rightarrow 7 A faster route is running the divisibility rule for 11 on 2167. The moment you spot it is divisible by 11, division yields 197 directly, matching the last option. Common Pitfall: Attempting to perform long division of 2176 by 167 and 197 directly without subtracting the remainder first. This makes the arithmetic significantly clunkier. Final Answer: Therefore, the correct answer is 197.

Question 4

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 5

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 6

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 7

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 8

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 9

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 10

$(7^{19} + 2)$ 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 11

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.

Question 12

$n$ being any odd number greater than 1, $n^{65} - n$ is always divisible by

Accepted Answer

Concept & Logic: This problem utilizes properties of consecutive numbers and factorization. The key insight is that for any odd number n, (n-1) and (n+1) are consecutive even numbers, guaranteeing divisibility by higher powers of 2. n^{65} - n = n(n^{64} - 1) Step-by-Step Solution: • Calculation: First, factor out n from the expression: n^{65} - n = n(n^{64} - 1) • Using the difference of squares, expand (n^{64} - 1) recursively: n(n^{32} + 1)(n^{16} + 1)(n^8 + 1)(n^4 + 1)(n^2 + 1)(n - 1)(n + 1) • Look closely at the terms n(n - 1)(n + 1). This can be rearranged as (n - 1)n(n + 1), which is the product of three consecutive integers. • The product of any three consecutive integers is always divisible by 3! = 6. • Since n is explicitly given as an odd number, (n - 1) and (n + 1) are consecutive even numbers. • Among any two consecutive even numbers, one is a multiple of 2 and the other is a multiple of 4. Therefore, their product (n - 1)(n + 1) is always divisible by 2 × 4 = 8. • Combining these facts: the term (n-1)n(n+1) is divisible by 3 (from being 3 consecutive integers) and divisible by 8 (from the even numbers property). • Since the lowest common multiple of 3 and 8 is 24, the entire expression is always divisible by 24. Exam Strategy & Shortcut: The fastest method for "always divisible by" questions with variables is Value Substitution. • Pick the smallest valid odd number greater than 1: let n = 3. • Evaluate the expression for the first few terms: n(n-1)(n+1) = 3(2)(4) = 24. • Since 24 must be a factor of the full expression 3^{65}-3, check the options. 24 is directly present. (Note: While 5 is also a factor due to Fermat's Little Theorem, 24 is the intended answer because it directly relies on the specific "odd number" constraint provided in the prompt). Common Pitfall: A major pitfall is picking n=1 to test, which gives 1^{65} - 1 = 0, leading to confusion since 0 is divisible by everything. Always read constraints carefully: the question specifies n > 1. Final Answer: Therefore, the correct answer is 24.

Question 13

These questions are based on the following information:

Given $N = 1! + 2! + 3! + ........ + 99! + 100!$.

Find the last two digits of $N$.

Accepted Answer

Concept & Logic: To find the last two digits of a number, we must find its remainder when divided by 100. In a factorial series, factorials quickly become multiples of 100 as they accumulate prime factors of 2 and 5. N \pmod{100} Step-by-Step Solution: • Given: N = 1! + 2! + 3! + ........ + 99! + 100!. (Note: The original text utilizes classical Indian L-bracket notation for factorials, universally translated to modern notation here). • Calculation: A factorial n! ends in at least two zeros (meaning it is perfectly divisible by 100) if it contains at least two 5s and two 2s in its prime factorization. • Let's look at 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1. This contains a 5 and a 10 (which provides another 5). Thus, it has two 5s. For any n \ge 10, n! \equiv 0 \pmod{100}. • Therefore, we only need to calculate the sum of the factorials from 1! to 9! and find their remainder modulo 100. • Let's evaluate them modulo 100: • 1! = 1 • 2! = 2 • 3! = 6 • 4! = 24 • 5! = 120 \implies 20 • 6! = 720 \implies 20 • 7! = 5040 \implies 40 • 8! = 40320 \implies 20 • 9! = 362880 \implies 80 • Sum these last two digits together: Sum = 1 + 2 + 6 + 24 + 20 + 20 + 40 + 20 + 80 = 213 • The last two digits of 213 are 13. Exam Strategy & Shortcut: Memorize the last two digits of factorials up to 9!. Actually, you only need to calculate them iteratively on the fly: 5! = 20. 6! = 20 × 6 = 120 \rightarrow 20. 7! = 20 × 7 = 140 \rightarrow 40. 8! = 40 × 8 = 320 \rightarrow 20. 9! = 20 × 9 = 180 \rightarrow 80. Add them up. Once you hit 10!, every single term contributes 00, so you can completely ignore the rest of the infinite-looking series. Common Pitfall: A common mistake is stopping at 5!, thinking that since it has a zero, all subsequent terms will have the same zero impact. However, 5! to 9! only have ONE trailing zero. You need TWO trailing zeros to drop the term entirely when calculating the last two digits. Final Answer: Therefore, the correct answer is 13.

Question 14

If $a^2 + b^2 + c^2 = 1$, what is the maximum value of $abc$?

Accepted Answer

Concept & Formula: This problem is solved using the AM-GM Inequality (Arithmetic Mean - Geometric Mean Inequality). For any set of non-negative real numbers, the arithmetic mean is always greater than or equal to the geometric mean. (x + y + z/3) \ge \sqrt[3]{xyz} Step-by-Step Solution: • Given: a^2 + b^2 + c^2 = 1 • Calculation: Let's apply the AM-GM inequality to the three non-negative terms: a^2, b^2, and c^2. (a^2 + b^2 + c^2/3) \ge \sqrt[3]{a^2 × b^2 × c^2} • Substitute the given value (a^2 + b^2 + c^2 = 1) into the left side of the inequality: (1/3) \ge \sqrt[3]{(abc)^2} • To isolate abc, first cube both sides to remove the cube root: ((1/3))^3 \ge (abc)^2 (1/27) \ge (abc)^2 • Now, take the square root of both sides. Since we are looking for the maximum positive value, we take the positive root: \sqrt{(1/27)} \ge abc (1/\sqrt{9 × 3)} \ge abc (1/3\sqrt{3)} \ge abc • Therefore, the maximum possible value of abc is (1/3\sqrt{3)}. Exam Strategy & Shortcut: In optimization problems involving squares or symmetric conditions (like a^2 + b^2 + c^2 = 1), the maximum or minimum product almost always occurs when all variables are equal. Assume Symmetry: a = b = c. Substitute this into the given equation: 3a^2 = 1 \implies a^2 = 1/3 \implies a = 1/\sqrt{3}. The maximum product abc is simply a × a × a = a^3. Calculate: (1/\sqrt{3})^3 = 1 / (3\sqrt{3}). You arrive at the answer in seconds without formal inequalities. Common Pitfall: Students often try to use calculus (Lagrange multipliers) or trigonometry (spherical coordinates) to solve this. While technically correct, those methods are far too time-consuming for an aptitude exam. Always check for symmetry or AM-GM first. Final Answer: Therefore, the correct answer is (1/3\sqrt{3)}.

Question 15

Find the unit's digit in the sum of the fifth powers of the first 100 natural numbers.

Accepted Answer

Concept & Logic: This question relies on the Cyclicity of Unit Digits. For any integer n, raising it to a power creates a repeating pattern in its unit digit. The fundamental rule of cyclicity is that every number repeats its unit digit cycle every 4th power. n^{4k+1} \equiv n^1 \pmod{10} This means the unit digit of n^5 is identical to the unit digit of n^1. Step-by-Step Solution: • Given: The sum 1^5 + 2^5 + 3^5 + ... + 100^5. • Calculation: According to the cyclicity rule (n^5 has the same unit digit as n^1), we can simplify the expression if we only care about the final unit digit. • The unit digit of the entire sum will be exactly the same as the unit digit of the simplified sum: 1^1 + 2^1 + 3^1 + ... + 100^1 • This is simply the sum of the first 100 natural numbers. Use the standard formula for the sum of an arithmetic progression: Sum = (n(n + 1)/2) • Substitute n = 100: Sum = (100(101)/2) Sum = 50 × 101 = 5050 • The total sum is 5050. The unit digit of 5050 is 0. Exam Strategy & Shortcut: Memorize the cyclicity of 5: any number to the 5th power ends in the same digit as the base number itself. The moment you see "fifth powers", mentally delete the exponent. The question instantly becomes "Find the unit digit of the sum of the first 100 numbers." 50 × 101 obviously ends in a zero. The whole process takes less than 15 seconds mentally. Common Pitfall: A common trap is trying to evaluate the first few fifth powers (1, 32, 243, 1024...) and looking for a new pattern within those sums, which burns time and increases the chance of a simple arithmetic mistake. Final Answer: Therefore, the correct answer is 0.

Question 16

If $p, q, r$ are all real numbers then $(p - q)^3 + (q - r)^3 + (r - p)^3$ is equal to

Accepted Answer

Concept & Formula: This problem relies on a very specific and highly tested algebraic identity involving the sum of three cubes. The conditional algebraic identity states that if a + b + c = 0, then: a^3 + b^3 + c^3 = 3abc Step-by-Step Solution: • Given: The expression to evaluate is (p - q)^3 + (q - r)^3 + (r - p)^3. • Substitution: Let's define three new variables to map this to our standard identity: a = p - q b = q - r c = r - p • Calculation: Check the sum of these new variables (a + b + c). a + b + c = (p - q) + (q - r) + (r - p) a + b + c = p - q + q - r + r - p = 0 • Deduction: Since the condition a + b + c = 0 is perfectly satisfied, we can directly apply the rule a^3 + b^3 + c^3 = 3abc. • Substitute a, b, and c back into the formula: (p - q)^3 + (q - r)^3 + (r - p)^3 = 3(p - q)(q - r)(r - p) Exam Strategy & Shortcut: Whenever you see a cyclic sum of cubes like (x - y)^3 + (y - z)^3 + (z - x)^3 in a competitive exam, you should instantly recognize it as a zero-sum condition. You don't need to write out the proof; memorize that the result will always be 3 × (product of the terms). Common Pitfall: A frequent mistake is choosing 0 (Option d) as the answer. Students correctly identify that the inner terms sum to 0, but mistakenly assume that the sum of their *cubes* will also be 0. Always remember the 3abc outcome. Final Answer: Therefore, the correct answer is 3(p - q)(q - r)(r - p).

Question 17

The least number of five digit is exactly divisible by 88 is

Accepted Answer

Concept & Logic: This problem is based on the concept of finding a number within a specific range that is completely divisible by a given divisor. To find the least n-digit number divisible by a divisor d: 1. Identify the smallest n-digit number. 2. Divide it by d to find the remainder. 3. Add the difference of the divisor and the remainder (d - r) to the smallest n-digit number. Step-by-Step Solution: • Given: We need the least 5-digit number divisible by 88. • The smallest 5-digit number is 10000. • Calculation: Divide 10000 by 88. 10000 \div 88 88 × 113 = 9944 • The remainder (r) is: 10000 - 9944 = 56 • Deduction: If we subtract 56 from 10000, the number becomes a 4-digit number (9944), which is divisible by 88. • To get the next multiple of 88 (which will be the smallest 5-digit multiple), we must add the divisor 88 to 9944: 9944 + 88 = 10032 • Alternatively, use the standard formula: Least Number + (Divisor - Remainder) 10000 + (88 - 56) = 10000 + 32 = 10032 Exam Strategy & Shortcut: Use the Divisibility Rule Option Elimination method. A number is divisible by 88 if it is divisible by both 8 and 11. • Rule of 8: The last 3 digits must be divisible by 8. • (a) 032 \div 8 = 4 (Divisible) • (b) 132 \div 8 = 16.5 (Not divisible) • (c) 088 \div 8 = 11 (Divisible) • (d) 023 (Not even, so not divisible) • We are left with (a) 10032 and (c) 10088. • The question asks for the *least* number. 10032 is smaller than 10088. • Verify 10032 with the Rule of 11 (Difference of alternating sums is 0 or 11): (1+0+2) - (0+3) = 3 - 3 = 0. It works! Common Pitfall: A very common mistake is simply subtracting the remainder (56) from 10000 and arriving at 9944, then forgetting that 9944 is a 4-digit number. Always remember to add the divisor back or use the (d - r) addition rule. Final Answer: Therefore, the correct answer is 10032.

Question 18

The number of three digit numbers which are multiples of 9 are

Accepted Answer

Concept & Formula: This problem can be solved by modeling the multiples of 9 as an Arithmetic Progression (A.P.). The formula for finding the n-th term of an A.P. is: T_n = a + (n - 1)d Where T_n is the last term, a is the first term, and d is the common difference. Step-by-Step Solution: • Given: We need to find all 3-digit multiples of 9. • The smallest 3-digit number is 100. Dividing 100 by 9 gives a remainder of 1. The first multiple is 100 + (9 - 1) = 108. • The largest 3-digit number is 999. Since 9 + 9 + 9 = 27 (divisible by 9), 999 itself is the last multiple. • Calculation: Set up the variables for the Arithmetic Progression. First term (a) = 108 Last term (T_n) = 999 Common difference (d) = 9 • Apply the A.P. formula: 999 = 108 + (n - 1)9 891 = (n - 1)9 n - 1 = (891/9) n - 1 = 99 n = 100 Exam Strategy & Shortcut: A much faster way to find the total count of multiples in a continuous range starting from 1 is simple division. • Total multiples of 9 from 1 to 999: (999/9) = 111 • Total multiples of 9 from 1 to 99 (the 1 and 2 digit numbers): (99/9) = 11 • To find strictly the 3-digit multiples, subtract the two values: 111 - 11 = 100 This requires almost zero calculation time. Common Pitfall: When using the A.P. formula, a common error is forgetting to add 1 at the final step, leading students to select 99 as the answer. When finding the number of terms between boundaries, it is an inclusive count. Final Answer: Therefore, the correct answer is 100.

Question 19

Two consecutive even positive integers, sum of the squares of which is 1060, are

Accepted Answer

Concept & Logic: This problem maps to forming a quadratic equation based on consecutive even numbers. However, in an exam setting, relying on standard quadratic factoring is slow; understanding unit digits and perfect squares is the primary logical key. Consecutive even integers can be algebraically represented as x and x + 2. Step-by-Step Solution: • Given: The sum of the squares of two consecutive even integers is 1060. • Let the numbers be x and x + 2. • Set up the mathematical equation: x^2 + (x + 2)^2 = 1060 • Expand the binomial: x^2 + x^2 + 4x + 4 = 1060 2x^2 + 4x - 1056 = 0 • Simplify by dividing by 2: x^2 + 2x - 528 = 0 • Calculation: Factor the quadratic equation. We need two numbers that multiply to -528 and add to 2. (x + 24)(x - 22) = 0 • Since the question specifies *positive* integers, x = -24 is invalid. Therefore, x = 22. • The consecutive even number is x + 2 = 24. Exam Strategy & Shortcut: Never solve the quadratic equation in the exam. Use Option Elimination via Unit Digits. We need the sum of the squares to end in a 0 (1060). Let's check the unit digits of the squares of the options: • (a) 12 and 14: 2^2 + 4^2 = 4 + 16 = 20. Unit digit is 0. Sum is 144 + 196 = 340. Too small. • (b) 20 and 22: 0^2 + 2^2 = 0 + 4 = 4. Unit digit is 4, not 0. Reject immediately. • (c) 22 and 24: 2^2 + 4^2 = 4 + 16 = 20. Unit digit is 0. Let's estimate: 20^2 + 20^2 = 800. 25^2 + 25^2 = 1250. The answer must be closely bounded between these. Let's calculate: 484 + 576 = 1060. This matches perfectly. • (d) 15 and 18 are not consecutive even integers. Reject immediately. Common Pitfall: The main pitfall is wasting 2-3 minutes trying to manually factorize large numbers like 528 during the quadratic step. Another mistake is forgetting the definition of "consecutive even" and representing the numbers as x and x+1. Final Answer: Therefore, the correct answer is 22 and 24.

Question 20

Directions: Read the table carefully and answer the following question.  The table shows the data about distance (km) covered from one point to another point by the vehicle used between each point and speed of each vehicle (km/h). Note: (i) If one p…

Accepted Answer

Concept & Factorization: This problem requires reverse-engineering a multiplication equation by testing single-digit factors of a given product. PQR × X = 2634 Step-by-Step Solution: 1. Find single-digit factors of 2634: We need X to be a single digit (1 \le X \le 9) such that dividing 2634 by X results in a three-digit integer (PQR). - If X = 2: 2634 / 2 = 1317 (This is a 4-digit number, so X 
eq 2) - If X = 3: 2634 / 3 = 878. Check condition: Is the tens digit (7) equal to half of X (1.5)? No. - If X = 6: 2634 / 6 = 439. Check condition: Is the tens digit (3) equal to half of X (6 / 2 = 3)? Yes! 2. We have identified the numbers: PQR = 439 and X = 6. 3. Evaluate the given statements about 439: - (i) Unit digit is 9: The unit digit of 439 is indeed 9. (True) - (ii) It is a prime number: The square root of 439 is approximately 20.9. Testing prime factors up to 19 (2, 3, 5, 7, 11, 13, 17, 19): 439 is not divisible by any of them. Thus, it is prime. (True) - (iii) Sum of the digits is 18: 4 + 3 + 9 = 16 
eq 18. (False) 4. Therefore, only statements (i) and (ii) are true. Exam Strategy & Shortcut: To quickly find single-digit factors of 2634, use divisibility rules. The number ends in 4 (even, so divisible by 2). Sum of digits 2+6+3+4 = 15 (divisible by 3). Since it is divisible by both 2 and 3, it is definitely divisible by 6. Testing 6 first directly gives you the correct framework. Common Pitfall: Failing to verify prime numbers correctly. Remember you only need to test divisibility by prime numbers up to the square root of the target number to definitively prove it is prime. Final Answer: Therefore, the correct answer is Only (i) and (ii).

Number System Questions

Questions

Practice smarter