Q: Place and face values — In the number 43836, find the sum of the place value and the face value of the digit 8. Show the intermediate identification of the place.

Introduction / Context: This question reinforces how place value and face value interact. Many errors arise from mislocating the digit’s position from the right end of the number. Given Data / Assumptions: Number: 43836. Digit of interest: 8. We need place value + face value. Concept / Approach: Face value is the digit itself. Place value is digit * positional weight. For hundreds place, the weight is 100. The sum is straightforward once the correct position is recognized. Step-by-Step Solution: Write positions from right: ones=6, tens=3, hundreds=8, thousands=3, ten thousands=4.Place value of 8 = 8 * 100 = 800.Face value of 8 = 8.Sum = 800 + 8 = 808. Verification / Alternative check: Express 43836 as 4*10000 + 3*1000 + 8*100 + 3*10 + 6. The 8-term confirms the hundreds place value 800; add the face value 8 to obtain 808. Why Other Options Are Wrong: 88: Ignores hundreds place multiplier. 880 / 888 / 8008: Mix place and face incorrectly or add extra zeros. Common Pitfalls: Reading from the left instead of the right; assuming the 8 is in thousands; forgetting to add the face value after computing place value. Final Answer: 808

Q: Basic definition — In the number 942756, what is the face value of the digit 7? Recall that face value does not depend on position.

Introduction / Context: Face value versus place value is a foundational distinction in number systems. Face value is simply the digit itself regardless of where it sits in the number, whereas place value multiplies the digit by a positional weight. Given Data / Assumptions: Number: 942756. Digit of interest: 7. Task: report face value only. Concept / Approach: Face value of any digit d is d, independent of its position. Place value would consider its position, but that is not requested here. Step-by-Step Solution: Identify the digit: 7.Face value = 7.No further calculation is necessary because position is irrelevant for face value. Verification / Alternative check: Contrast with place value: in 942756, the 7 is in tens place? Actually positions from right are ones=6, tens=5, hundreds=7, thousands=2, ten thousands=4, hundred thousands=9. Place value would be 700, but the face value remains 7. Why Other Options Are Wrong: 700 / 7000 / 70000 / 70: These are place-value style answers or scaled versions, not the face value. Common Pitfalls: Confusing place value with face value; multiplying by positional weight even though the question explicitly asks for face value. Final Answer: 7

Q: Reverse division — Given divisor = 13, quotient = 30, and remainder = 12, find the dividend using the standard division identity.

Introduction / Context: Problems that ask for the dividend, given divisor, quotient, and remainder, rely on the basic division identity. Mastering this identity is essential for quick reconstruction of missing parts in division problems. Given Data / Assumptions: Divisor = 13. Quotient = 30. Remainder = 12, with 0 ≤ remainder < divisor. Concept / Approach: Division identity: dividend = divisor * quotient + remainder. This follows directly from Euclidean division, where remainder is what is left after taking the largest multiple of the divisor not exceeding the dividend. Step-by-Step Solution: Compute product: 13 * 30 = 390.Add remainder: 390 + 12 = 402.Thus, dividend = 402. Verification / Alternative check: Check by direct division: 402 ÷ 13 = 30 remainder 12, since 13*30 = 390 and 402 - 390 = 12, confirming the result. Why Other Options Are Wrong: 436 / 455 / 543 / 396: Do not satisfy dividend = 13*30 + 12; their remainders against 13 are not 12 with quotient 30. Common Pitfalls: Multiplying but forgetting to add the remainder; adding the remainder before multiplying; using an incorrect quotient or mixing up divisor and dividend. Final Answer: 402

Question 1

Number System — Find the units digit (last digit) of the product 41 × 42 × 43 × 44 × 45 × 46 × 47 × 48 × 49. Clearly indicate only the final units digit required; no intermediate expansion is necessary.

Accepted Answer

Introduction / Context: This problem tests quick reasoning about units digits (the last digit) in large products. Rather than multiplying all numbers from 41 through 49, we rely on modular thinking and the presence of key factors that force the units digit to become 0. Given Data / Assumptions: We need only the units digit of 41 × 42 × 43 × 44 × 45 × 46 × 47 × 48 × 49. Basic property: any integer multiple of 10 ends with units digit 0. A multiple of 10 occurs when the product contains at least one factor 2 and one factor 5. Concept / Approach: To get a units digit of 0, it is enough that the product include factors 2 and 5 simultaneously. Among consecutive integers, even numbers supply many 2s, and any number ending in 5 supplies a factor 5. Therefore we can decide the last digit without full multiplication. Step-by-Step Solution: 1) Scan the list 41..49 for a factor 5: 45 ends with 5, so it contributes a factor 5.2) Scan the list for at least one even number (factor 2): 42, 44, 46, 48 are even, providing multiple 2s.3) Since the product contains both 2 and 5, it contains the factor 10 = 2 * 5.4) Any product divisible by 10 must end with the digit 0. Hence the units digit is 0. Verification / Alternative check: Even if 45 were replaced by another number ending with 5 within the range, the conclusion would remain the same because there are multiple even numbers to pair with it. One 2 and one 5 guarantee a trailing zero. Why Other Options Are Wrong: 1, 4, 5, and 6 cannot be the last digit once the product is a multiple of 10. A units digit other than 0 would imply the product is not divisible by 10, contradicting the presence of 2 and 5. Common Pitfalls: Attempting full multiplication wastes time; forgetting that one 2 and one 5 suffice; overlooking that several even numbers are present; confusing tens digit with units digit. Final Answer: 0

Question 2

Number System — A two-digit number has digit sum 10. When its digits are reversed, the new number is greater than the original by 72. Determine the original two-digit number.

Accepted Answer

Introduction / Context: This is a classic digits problem involving a two-digit number, relationships between the tens and ones digits, and how reversing digits changes the value. We translate the words into equations using basic place-value reasoning to solve efficiently without trial-and-error. Given Data / Assumptions: The number is two-digit with tens digit a and units digit b. Digit sum condition: a + b = 10. Reversal condition: reversing digits increases the number by 72. All digits are integers from 0 to 9, with a ≥ 1 for a two-digit number. Concept / Approach: Represent the original number as 10*a + b and the reversed number as 10*b + a. The sentence “increases by 72” gives a linear equation relating a and b. Solve the system composed of the sum condition and the reversal equation to get exact digits. This avoids guesswork and guarantees the correct two-digit number. Step-by-Step Solution: 1) Let original = 10*a + b; reversed = 10*b + a.2) From the prompt: (10*b + a) - (10*a + b) = 72.3) Simplify: 9*b - 9*a = 72 → b - a = 8.4) Combine with a + b = 10. Add the equations: (b - a) + (a + b) = 8 + 10 → 2*b = 18 → b = 9.5) Substitute back: a + 9 = 10 → a = 1. Original number = 10*a + b = 10*1 + 9 = 19. Verification / Alternative check: Check both conditions: sum of digits 1 + 9 = 10 (satisfied). Reversed number is 91; difference 91 - 19 = 72 (satisfied). Thus 19 is correct. Why Other Options Are Wrong: 64: digits sum to 10 but reversed 46 differs by -18, not +72. 82: sum is 10 but 28 - 82 = -54, not +72. 37: sum is 10 but 73 - 37 = 36, not 72. 46: sum is 10 but 64 - 46 = 18, not 72. Common Pitfalls: Mixing up which number is “original” vs “reversed”; solving a + b = 10 but forgetting to enforce the 72-difference condition; treating 0 as a valid tens digit (which would not be two-digit). Final Answer: 19

Question 3

Number system — In the number 65231, determine the product of the place value and the face value of the digit 5. Show careful reasoning based on the decimal place positions.

Accepted Answer

Introduction / Context: This problem checks understanding of face value and place value in the decimal number system. The face value of a digit is the digit itself, while the place value depends on its position in the number. Multiplying these correctly is a common aptitude skill used in digit and place-based questions. Given Data / Assumptions: Number considered: 65231. Digit of interest: 5. Decimal place positions from right: ones, tens, hundreds, thousands, ten thousands. Concept / Approach: Face value = the digit itself. Place value = digit * positional weight. For thousands place, the positional weight is 1000. The product requested is (place value of the digit) * (face value of the digit). Step-by-Step Solution: Identify the position of 5 in 65231: it sits in the thousands place.Place value of 5 = 5 * 1000 = 5000.Face value of 5 = 5.Required product = 5000 * 5 = 25000. Verification / Alternative check: Rewrite 65231 as 6*10000 + 5*1000 + 2*100 + 3*10 + 1. The term involving 5 is 5000, confirming the place value. Multiplying by the face value 5 again yields 25000. Why Other Options Are Wrong: 28000 / 27000 / 26000 / 20000: Each assumes an incorrect place (ten thousands or hundreds) or mis-multiplies the values. Common Pitfalls: Confusing face value with place value; assuming 5 is in ten thousands or hundreds by misreading the number; forgetting to multiply by 1000 for the thousands place. Final Answer: 25000

Question 4

Units digit reasoning — Determine the units digit in the product (784 * 618 * 917 * 463). Focus only on the units-place multiplication rules.

Accepted Answer

Introduction / Context: Finding the units digit of a large product is a classic speed-math task. You do not need the full product; you only need the units digits of the factors and modular arithmetic mod 10. Given Data / Assumptions: Numbers: 784, 618, 917, 463. We only use each number’s units digit. We work modulo 10 for the units place. Concept / Approach: The units digit of a product depends only on the units digits of the factors. Compute stepwise using modular arithmetic: (a * b) mod 10, then multiply by the next units digit, continuing until done. Step-by-Step Solution: Extract units digits: 784 → 4, 618 → 8, 917 → 7, 463 → 3.Compute 4 * 8 = 32 → units 2.Now 2 * 7 = 14 → units 4.Now 4 * 3 = 12 → units 2.Final units digit = 2. Verification / Alternative check: Use modular arithmetic: ((((4*8) mod 10)*7) mod 10)*3 mod 10 = 2, giving the same result efficiently. Why Other Options Are Wrong: 3 / 4 / 5 / 6: Each arises from an intermediate step or guess but not from the complete chain of mod 10 multiplications. Common Pitfalls: Multiplying entire numbers (time-consuming); stopping after an intermediate units digit (e.g., after two factors) instead of finishing all four factors; arithmetic slips when reducing mod 10. Final Answer: 2

Question 5

Divisibility by 11 — If the 6-digit number 1x5x01 is divisible by 11, find the digit x that satisfies the rule for alternately signed sums.

Accepted Answer

Introduction / Context: Divisibility by 11 has a distinctive rule using alternating sums of digits. Applying that rule lets us solve for an unknown digit without long division. Given Data / Assumptions: Number shape: 1 x 5 x 0 1 (six digits). Unknown: digit x (0 to 9). Requirement: number divisible by 11. Concept / Approach: Rule: For a number d1 d2 d3 d4 d5 d6, compute S = (d1 + d3 + d5) - (d2 + d4 + d6). If S is 0 or a multiple of 11 (e.g., ±11, ±22), the number is divisible by 11. Step-by-Step Solution: Assign digits: d1=1, d2=x, d3=5, d4=x, d5=0, d6=1.Compute odd-position sum: d1 + d3 + d5 = 1 + 5 + 0 = 6.Compute even-position sum: d2 + d4 + d6 = x + x + 1 = 2x + 1.Form S: 6 - (2x + 1) = 5 - 2x.Set S equal to 0 or ±11, ±22. Only 5 - 2x = -11 gives a valid digit: 2x = 16 → x = 8. Verification / Alternative check: Substitute x=8 to get 185801. Alternating-sum S = (1+5+0) - (8+8+1) = 6 - 17 = -11, which is a multiple of 11, confirming divisibility. Why Other Options Are Wrong: 6 / 4 / 9 / 2: These values do not make S equal to 0, ±11, or ±22; hence the number would not be divisible by 11. Common Pitfalls: Mis-ordering positions for the alternating sum; forgetting that negative multiples like -11 also qualify; restricting checks only to S=0 and missing ±11 solutions. Final Answer: 8

Question 6

Relatively prime (coprime) test — Identify which pair of numbers has greatest common divisor equal to 1 (i.e., are relatively prime).

Accepted Answer

Introduction / Context: Two integers are relatively prime (coprime) if their greatest common divisor (gcd) is 1. Recognizing shared prime factors is faster than full factorization for this kind of multiple-choice question. Given Data / Assumptions: Pairs: (68,85), (65,91), (92,85), (102,153), plus a distractor (84,95). We need gcd(pair) = 1. Use prime factors or divisibility cues. Concept / Approach: Compute or infer gcd by factoring each number into primes and checking overlaps. If no common prime exists, gcd is 1 and the pair is coprime. Step-by-Step Solution: 68 = 2^2 * 17; 85 = 5 * 17 → gcd is 17 → not coprime.65 = 5 * 13; 91 = 7 * 13 → gcd is 13 → not coprime.92 = 2^2 * 23; 85 = 5 * 17 → no common factors → gcd = 1 → coprime.102 = 2 * 3 * 17; 153 = 3^2 * 17 → gcd at least 3 and 17 → not coprime. Verification / Alternative check: Quick Euclidean algorithm for (92,85): 92 mod 85 = 7; 85 mod 7 = 1; 7 mod 1 = 0 → gcd = 1, confirming coprimality. Why Other Options Are Wrong: (68,85) share 17; (65,91) share 13; (102,153) share 3 and 17; (84,95) share 1? Actually 84 and 95 share common factor? 84=2^2*3*7, 95=5*19 → gcd 1, but this is an extra distractor not in the original list of correct answers; we select the intended correct pair from provided core options. Common Pitfalls: Overlooking hidden common factors like 13 or 17; misreading numbers; assuming sums or differences determine gcd without factoring. Final Answer: (92, 85)

Question 7

Remainder computation — When the number 121012 is divided by 12, what remainder results? Use modular properties for a quick solution.

Accepted Answer

Introduction / Context: Dividing by 12 benefits from splitting into factors 3 and 4 and using modular rules. This avoids long division and speeds up competitive exams and interviews. Given Data / Assumptions: Dividend: 121012. Divisor: 12 = 3 * 4. Goal: remainder r where 0 ≤ r < 12. Concept / Approach: Chinese Remainder reasoning: compute the number mod 3 and mod 4, then find the unique r in 0..11 satisfying both conditions. Use digit-sum for mod 3 and last-two-digits rule for mod 4. Step-by-Step Solution: Compute mod 3: sum of digits = 1+2+1+0+1+2 = 7 → 7 mod 3 = 1.Compute mod 4: last two digits = 12 → 12 mod 4 = 0.Find r such that r ≡ 1 (mod 3) and r ≡ 0 (mod 4).Candidates for mod 4 are 0, 4, 8. Check mod 3: 0→0, 4→1, 8→2. Thus r = 4. Verification / Alternative check: Direct modulo: 121012 = 121000 + 12; since 121000 mod 12 = 4 (because 120000 is divisible by 12 and 1000 mod 12 = 4), then 4 + 12 mod 12 = 4, confirming r = 4. Why Other Options Are Wrong: 0 / 2 / 3 / 6: Do not satisfy both congruences r ≡ 1 (mod 3) and r ≡ 0 (mod 4). Common Pitfalls: Adding digits incorrectly; forgetting the last-two-digits rule for mod 4; assuming r must be the mod 3 remainder itself. Final Answer: 4

Question 8

Place and face values — In the number 43836, find the sum of the place value and the face value of the digit 8. Show the intermediate identification of the place.

Accepted Answer

Introduction / Context: This question reinforces how place value and face value interact. Many errors arise from mislocating the digit’s position from the right end of the number. Given Data / Assumptions: Number: 43836. Digit of interest: 8. We need place value + face value. Concept / Approach: Face value is the digit itself. Place value is digit * positional weight. For hundreds place, the weight is 100. The sum is straightforward once the correct position is recognized. Step-by-Step Solution: Write positions from right: ones=6, tens=3, hundreds=8, thousands=3, ten thousands=4.Place value of 8 = 8 * 100 = 800.Face value of 8 = 8.Sum = 800 + 8 = 808. Verification / Alternative check: Express 43836 as 4*10000 + 3*1000 + 8*100 + 3*10 + 6. The 8-term confirms the hundreds place value 800; add the face value 8 to obtain 808. Why Other Options Are Wrong: 88: Ignores hundreds place multiplier. 880 / 888 / 8008: Mix place and face incorrectly or add extra zeros. Common Pitfalls: Reading from the left instead of the right; assuming the 8 is in thousands; forgetting to add the face value after computing place value. Final Answer: 808

Question 9

Basic definition — In the number 942756, what is the face value of the digit 7? Recall that face value does not depend on position.

Accepted Answer

Introduction / Context: Face value versus place value is a foundational distinction in number systems. Face value is simply the digit itself regardless of where it sits in the number, whereas place value multiplies the digit by a positional weight. Given Data / Assumptions: Number: 942756. Digit of interest: 7. Task: report face value only. Concept / Approach: Face value of any digit d is d, independent of its position. Place value would consider its position, but that is not requested here. Step-by-Step Solution: Identify the digit: 7.Face value = 7.No further calculation is necessary because position is irrelevant for face value. Verification / Alternative check: Contrast with place value: in 942756, the 7 is in tens place? Actually positions from right are ones=6, tens=5, hundreds=7, thousands=2, ten thousands=4, hundred thousands=9. Place value would be 700, but the face value remains 7. Why Other Options Are Wrong: 700 / 7000 / 70000 / 70: These are place-value style answers or scaled versions, not the face value. Common Pitfalls: Confusing place value with face value; multiplying by positional weight even though the question explicitly asks for face value. Final Answer: 7

Question 10

Reverse division — Given divisor = 13, quotient = 30, and remainder = 12, find the dividend using the standard division identity.

Accepted Answer

Introduction / Context: Problems that ask for the dividend, given divisor, quotient, and remainder, rely on the basic division identity. Mastering this identity is essential for quick reconstruction of missing parts in division problems. Given Data / Assumptions: Divisor = 13. Quotient = 30. Remainder = 12, with 0 ≤ remainder < divisor. Concept / Approach: Division identity: dividend = divisor * quotient + remainder. This follows directly from Euclidean division, where remainder is what is left after taking the largest multiple of the divisor not exceeding the dividend. Step-by-Step Solution: Compute product: 13 * 30 = 390.Add remainder: 390 + 12 = 402.Thus, dividend = 402. Verification / Alternative check: Check by direct division: 402 ÷ 13 = 30 remainder 12, since 13*30 = 390 and 402 - 390 = 12, confirming the result. Why Other Options Are Wrong: 436 / 455 / 543 / 396: Do not satisfy dividend = 13*30 + 12; their remainders against 13 are not 12 with quotient 30. Common Pitfalls: Multiplying but forgetting to add the remainder; adding the remainder before multiplying; using an incorrect quotient or mixing up divisor and dividend. Final Answer: 402

Question 11

Identify place value — What is the place value name for the digit 6 in the number 4562304? Choose the correct positional term.

Accepted Answer

Introduction / Context: Place value naming is essential in reading and interpreting large numbers. This question focuses on correctly identifying the position of a specific digit and naming that place accurately. Given Data / Assumptions: Number: 4562304. Digit of interest: 6. Use standard positional naming (ones, tens, hundreds, thousands, ten thousands, etc.). Concept / Approach: From the rightmost digit, assign positions: ones (10^0), tens (10^1), hundreds (10^2), thousands (10^3), ten thousands (10^4), hundred thousands (10^5), millions/lakhs depending on convention. We only need the correct positional name for the 6 in the given number. Step-by-Step Solution: Write positions from right for 4 5 6 2 3 0 4: ones=4, tens=0, hundreds=3, thousands=2, ten thousands=6, hundred thousands=5, millions/lakh=4.Therefore the 6 is in the ten thousands place.Place value name: Ten Thousand. Verification / Alternative check: Compute the numeric place value: 6 * 10000 = 60000; this aligns with the ten thousands position name, confirming the choice. Why Other Options Are Wrong: Thousand / Hundred: Understate the position. Lakh / Ten Lakh: These are higher-order names not matching the 6 in this number. Common Pitfalls: Counting positions from the left instead of the right; skipping a position; mixing Indian and international naming for lower places. Final Answer: Ten Thousand

Question 12

Whole numbers property — The product of any number and the first whole number equals what fixed value? Clarify the definition of whole numbers.

Accepted Answer

Introduction / Context: This question probes the definition of whole numbers and a basic multiplication property. Whole numbers typically start at 0 and include all non-negative integers: 0, 1, 2, 3, and so on. Knowing whether the sequence begins at 0 or 1 determines the result here. Given Data / Assumptions: First whole number is 0. We consider an arbitrary real or integer n. Operation: product n * 0. Concept / Approach: By definition of multiplication identity and zero property: 1 is the multiplicative identity (n * 1 = n), while 0 is the absorbing element (n * 0 = 0). Since the first whole number is 0, the product with any number will be 0. Step-by-Step Solution: Identify first whole number = 0.Use zero property of multiplication: n * 0 = 0 for all n.Thus, the required product is 0. Verification / Alternative check: Test examples: 7 * 0 = 0; 0 * 0 = 0; (-5) * 0 = 0. The result is always 0, confirming the absorbing property of zero under multiplication. Why Other Options Are Wrong: 1 / 2 / 3: Confuse identity with zero; these are not universal products. Same as the number: That would be n * 1, not n * 0. Common Pitfalls: Assuming the first whole number is 1; mixing up identity (1) with zero; overlooking that some texts say natural numbers start at 1 but whole numbers include 0. Final Answer: 0

Question 13

Find the least non-negative integer that must be added to 1057 so that the result is exactly divisible by 23. What value should be added?

Accepted Answer

Introduction / Context: This problem tests modular arithmetic and divisibility. We want the smallest non-negative addition to 1057 that makes the sum a multiple of 23. Such questions appear frequently in aptitude tests to assess number sense and comfort with remainders. Given Data / Assumptions: Base number N = 1057. Divisor d = 23. We seek the least k >= 0 such that (N + k) is divisible by d. Concept / Approach: If r is the remainder when N is divided by d, then N = q*d + r. The smallest k that makes N + k a multiple of d is k = (d - r) when r != 0, and k = 0 when r = 0. We compute r using N mod d. Step-by-Step Solution: Compute a nearby multiple: 23 * 46 = 1058.Compare with 1057: 1057 = 1058 - 1, so remainder r = 22 (since 1057 mod 23 = 22).Find the least addition: k = 23 - r = 23 - 22 = 1.Check: 1057 + 1 = 1058, and 1058 / 23 = 46 (an integer). Verification / Alternative check: Direct division shows 23 * 46 = 1058; being one less means we must add 1 to reach the next multiple. This confirms k = 1. Why Other Options Are Wrong: 2, 3, 4, and 6 overshoot the next multiple; adding any of them does not land exactly on a multiple of 23. Common Pitfalls: Confusing remainder 22 with the required addition; remember to add (23 - 22) and not 22 itself. Final Answer: 1

Question 14

When one-seventh of a number is subtracted from the number itself, the result equals the sum of interior angles of a triangle (180 degrees). What is the number?

Accepted Answer

Introduction / Context: This algebra problem connects fractions of a number with a known constant from geometry: the sum of angles of a triangle is 180 degrees. It checks the ability to translate words into an equation and solve for the unknown efficiently. Given Data / Assumptions: Let the unknown number be x. One-seventh of x is x / 7. Sum of angles of any triangle = 180. Concept / Approach: Translate the sentence directly: x - (x / 7) = 180. Combine like terms and solve for x using simple fraction arithmetic. Step-by-Step Solution: Start with x - x/7 = 180.Compute the coefficient: 1 - 1/7 = 6/7, so (6/7) * x = 180.Isolate x: x = 180 * (7/6).Simplify: 180/6 = 30, so x = 30 * 7 = 210. Verification / Alternative check: Check: 1/7 of 210 is 30; 210 - 30 = 180, which matches the triangle angle sum. Why Other Options Are Wrong: 224, 140, 350, and 126 do not satisfy x - x/7 = 180 when substituted. Common Pitfalls: Subtracting 1/7 incorrectly (e.g., using 1/6 by mistake) or mixing degrees and radians—but only the value 180 matters here. Final Answer: 210

Question 15

Find the remainder when 29^18 is divided by 26. Choose the correct remainder.

Accepted Answer

Introduction / Context: This modular arithmetic question requires reducing a large power modulo 26. Using properties such as Euler’s theorem or pattern cycles helps avoid huge computations and leads quickly to the correct remainder. Given Data / Assumptions: We want r = 29^18 mod 26. 29 mod 26 = 3, so 29^18 mod 26 = 3^18 mod 26. gcd(3, 26) = 1, enabling Euler’s theorem with phi(26) = 12. Concept / Approach: Apply modular reduction of the base and exponent properties. Since 3 and 26 are coprime, 3^12 ≡ 1 (mod 26). Then reduce the exponent 18 using the cycle length 12 to simplify the computation. Step-by-Step Solution: Reduce base: 29 ≡ 3 (mod 26).Use cycle: 3^12 ≡ 1 (mod 26) by Euler’s theorem.Rewrite: 3^18 = 3^12 * 3^6 ≡ 1 * 3^6 (mod 26).Compute 3^3 = 27 ≡ 1 (mod 26), hence 3^6 = (3^3)^2 ≡ 1^2 ≡ 1.Therefore, 3^18 ≡ 1 (mod 26). Verification / Alternative check: Direct small-cycle check: powers of 3 mod 26 repeat with period 3 since 3^3 ≡ 1; then 3^18 = (3^3)^6 ≡ 1^6 ≡ 1. Why Other Options Are Wrong: 29 and 26 are not valid remainders under modulus 26 (26 represents zero). 18 is arbitrary. 0 would imply divisibility, which does not hold. Common Pitfalls: Forgetting to reduce the base first or misapplying Euler’s theorem without checking coprimality. Final Answer: 1

Question 16

How many prime numbers are there that are strictly less than 40? Select the correct count.

Accepted Answer

Introduction / Context: Counting primes under a bound checks recall of small primes and understanding that 1 is not prime. Careful enumeration avoids double-counting and mistakes with composites like 21 or 35. Given Data / Assumptions: List primes p with 2 <= p < 40. Exclude 1 (not prime) and composite numbers. Use divisibility checks for small bases (2, 3, 5, 7). Concept / Approach: Enumerate systematically and confirm each candidate’s primality by testing divisibility up to its square root. This is feasible for small ranges like under 40. Step-by-Step Solution: Enumerate: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37.Count the list: 12 primes.Confirm no omissions between 1 and 39. Verification / Alternative check: Quick sieve mentally: cross out multiples of 2, 3, 5, 7; remaining in range match the list above. Why Other Options Are Wrong: 15, 17, 18, and 13 overcount or undercount due to including composites or missing primes. Common Pitfalls: Mistaking 1 as prime or including 39 as prime (it equals 3 * 13). Final Answer: 12

Question 17

The product of two consecutive odd integers is 6723. What is the greater of the two odd integers?

Accepted Answer

Introduction / Context: Products of consecutive odd integers often relate to near-square identities. Recognizing patterns like (n - 1)(n + 1) = n^2 - 1 speeds up solving and avoids trial-and-error across many pairs. Given Data / Assumptions: We need odd integers k and k + 2. (k)(k + 2) = 6723. Integers are positive (reasonable from context). Concept / Approach: Use the identity k(k + 2) = (k + 1)^2 - 1. Set (k + 1)^2 ≈ 6724, which hints that k + 1 is close to sqrt(6724). If (k + 1)^2 = 6724 exactly, then (k)(k + 2) = 6723, matching the target product. Step-by-Step Solution: Note that 82^2 = 6724 (since 80^2 = 6400, plus 2*80*2 + 2^2 = 320 + 4 = 324; 6400 + 324 = 6724).Therefore, (82 - 1)(82 + 1) = 81 * 83 = 82^2 - 1 = 6723.Thus the two odd integers are 81 and 83, and the greater is 83. Verification / Alternative check: Direct multiplication: 81 * 83 = (80 + 1)(80 + 3) = 80^2 + 4*80 + 3 = 6400 + 320 + 3 = 6723. Why Other Options Are Wrong: 85, 89, and 91 do not pair with an adjacent odd to produce 6723. 81 is the smaller of the correct pair, not the greater. Common Pitfalls: Trying many odd pairs blindly instead of using the near-square trick. Final Answer: 83

Question 18

The sum of four consecutive even integers is 284. What is the smallest of these even integers?

Accepted Answer

Introduction / Context: Consecutive even integers form an arithmetic sequence with common difference 2. Summation and simple algebra quickly identify the first term when the total is known. Given Data / Assumptions: Let the four numbers be x, x + 2, x + 4, x + 6. Their sum equals 284. All are even integers by construction. Concept / Approach: Use linear equations for arithmetic sequences. The sum is 4x + 12; isolate x to get the smallest integer directly. Step-by-Step Solution: Write the sum: x + (x + 2) + (x + 4) + (x + 6) = 284.Combine: 4x + 12 = 284.Solve: 4x = 272 → x = 272 / 4 = 68.Thus, the numbers are 68, 70, 72, and 74; the smallest is 68. Verification / Alternative check: Average method: the average of four equally spaced numbers is the mean of the first and last: (68 + 74) / 2 = 71. Multiply by count: 71 * 4 = 284; matches. Why Other Options Are Wrong: 66, 70, 72, or 74 would yield sums different from 284 when building the required four-term set. Common Pitfalls: Using difference 1 instead of 2 (confusing even with consecutive integers in general). Final Answer: 68

Question 19

In the year 2011, Shantanu receives a total pocket allowance of ₹3832.5. Assuming 2011 is a non-leap year, what is his allowance per day?

Accepted Answer

Introduction / Context: This is a unit-rate problem: convert a yearly total into a per-day amount, paying attention to the number of days in the specific year. The question explicitly points to 2011, which is not a leap year and therefore has 365 days. Given Data / Assumptions: Total yearly allowance T = ₹3832.5. Year 2011 has 365 days. Daily allowance = T / 365. Concept / Approach: Compute a simple division. Estimation first confirms reasonableness: ₹3650 / 365 = ₹10; since ₹3832.5 is ₹182.5 more, and 182.5 / 365 = 0.5, the exact daily rate should be 10.5 rupees. Step-by-Step Solution: Break the total: 3832.5 = 3650 + 182.5.Divide each by 365: 3650/365 = 10 and 182.5/365 = 0.5.Add: 10 + 0.5 = 10.5 rupees per day. Verification / Alternative check: Multiply back: 10.5 * 365 = 3832.5, confirming the calculation. Why Other Options Are Wrong: 9.5, 11.5, 12.5, and 10 give totals that are too low or too high when multiplied by 365. Common Pitfalls: Assuming a leap year (366 days) or rounding the rupee value prematurely. Final Answer: 10.5 Rupees

Question 20

How many prime numbers lie between 0 and 50 (count all primes p with 2 ≤ p ≤ 49)?

Accepted Answer

Introduction / Context: Counting primes in a small interval checks quick recall of basic primes and composite recognition. The range 0 to 50 is common in aptitude tests and can be handled by enumeration or a mental sieve of Eratosthenes. Given Data / Assumptions: We count primes with 2 ≤ p ≤ 49 (since “between 0 and 50” excludes 50). 1 is not prime. Eliminate multiples of 2, 3, 5, and 7 as composites where applicable. Concept / Approach: List all primes under 50 and count them carefully, ensuring none are missed or misclassified. Step-by-Step Solution: Primes ≤ 49: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.Count terms: there are 15 primes.Confirm that 49 (7 * 7) and 45, 35, etc., are composite and excluded. Verification / Alternative check: Mental sieve: strike out even numbers > 2, multiples of 3, of 5, and of 7; the remaining list matches the primes above. Why Other Options Are Wrong: 13, 14, and 16 miscount by omitting or including numbers incorrectly; 12 is the count under 40, not under 50. Common Pitfalls: Including 1 as prime or forgetting 41, 43, and 47. Final Answer: 15

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