Question 1

Find the least positive multiple of 7 that leaves a remainder of 4 when divided by each of 6, 9, 15, and 18.

Accepted Answer

Introduction / Context: This is a simultaneous congruences problem. Because the required remainder is the same (4) for all divisors, we can combine the conditions using the least common multiple (LCM). We also require the result to be a multiple of 7. Given Data / Assumptions: N ≡ 4 (mod 6), N ≡ 4 (mod 9), N ≡ 4 (mod 15), N ≡ 4 (mod 18) N is a multiple of 7 Concept / Approach: First, condense the congruences to N ≡ 4 (mod LCM(6, 9, 15, 18)). Then among numbers of the form 4 + k*LCM, find the least one divisible by 7. Step-by-Step Solution: LCM(6, 9, 15, 18) = 2 * 3^2 * 5 = 90.General form: N = 4 + 90t.Require 7 | N ⇒ 4 + 90t ≡ 0 (mod 7). Since 90 ≡ 6 (mod 7), we need 4 + 6t ≡ 0 ⇒ 6t ≡ 3 ⇒ t ≡ 4 (mod 7).Smallest t = 4 ⇒ N = 4 + 90*4 = 364. Verification / Alternative check: 364 mod 6, 9, 15, 18 equals 4; and 364/7 = 52 is integral. Conditions satisfied. Why Other Options Are Wrong: 74, 94, 184, 454 do not satisfy both constraints simultaneously (either not ≡ 4 modulo the LCM or not a multiple of 7). Common Pitfalls: Forgetting to combine the congruences using the LCM; testing only one modulus; or missing the multiple-of-7 requirement. Final Answer: 364

Question 2

Consider the product of any four consecutive natural numbers, n(n + 1)(n + 2)(n + 3). What is the largest natural number that always divides this product?

Accepted Answer

Introduction / Context: This question asks for a universal divisor of every product of four consecutive integers. We use parity and divisibility arguments to identify guaranteed factors (powers of 2 and 3) present in any block of four consecutive numbers. Given Data / Assumptions: General product: n(n + 1)(n + 2)(n + 3) for integer n We seek a constant divisor valid for all n Concept / Approach: Among any four consecutive integers: (i) there are two even numbers, one of which is a multiple of 4, ensuring a factor of 8; (ii) there is always a multiple of 3, ensuring a factor of 3. Combine these guaranteed factors to find the fixed divisor. Step-by-Step Solution: Among n, n+1, n+2, n+3: exactly two are even; one of the even numbers is divisible by 4 ⇒ contributes 4.The other even number contributes an extra factor 2 ⇒ total factor from parity = 8.At least one of the four is divisible by 3 ⇒ factor 3 present.Fixed divisor = 8 * 3 = 24. Verification / Alternative check: Try n = 1: product = 1*2*3*4 = 24, divisible by 24. Try n = 2: 2*3*4*5 = 120, also divisible by 24. Why Other Options Are Wrong: 6 and 12 are too small (not maximal). 48 and 120 do not always divide the product (48 would require a guaranteed factor 16, which is not always present). Common Pitfalls: Overestimating the fixed power of 2 (thinking 16 is always present) or forgetting to include a guaranteed multiple of 3. Final Answer: 24

Question 3

Compute the total number of prime factors, counting multiplicity, of the product 8^20 × 15^24 × 7^15.

Accepted Answer

Introduction / Context: “Total number of prime factors” here means counting prime factors with multiplicity (the sum of exponents in the prime factorization). We factor each base, multiply exponents appropriately, and then add the exponents. Given Data / Assumptions: Expression: 8^20 × 15^24 × 7^15 We count prime factors with multiplicity Concept / Approach: Prime-factorize each base: 8 = 2^3, 15 = 3*5, and 7 is prime. Then raise to the given powers and add exponents across all primes involved. Step-by-Step Solution: 8^20 = (2^3)^20 = 2^60 ⇒ contributes 60 twos.15^24 = (3*5)^24 = 3^24 * 5^24 ⇒ contributes 24 threes and 24 fives.7^15 ⇒ contributes 15 sevens.Total prime factors with multiplicity = 60 + 24 + 24 + 15 = 123. Verification / Alternative check: No overlap among these primes beyond what is counted in exponents. Summation is straightforward. Why Other Options Are Wrong: 59 and 98 underestimate by dropping exponents; 138 overestimates; 117 misses some contribution. Common Pitfalls: Confusing “total prime factors” with “number of distinct primes” (which would be 4 here: 2, 3, 5, 7). Final Answer: 123

Question 4

Three traffic lights change at regular intervals of 48 s, 72 s, and 108 s. If they all change simultaneously at 08:20:00, at what time will they next change together?

Accepted Answer

Introduction / Context: Simultaneous events occurring at fixed intervals repeat together every least common multiple (LCM) of those intervals. We compute the LCM in seconds and add it to the given start time to get the next simultaneous change. Given Data / Assumptions: Intervals: 48 s, 72 s, 108 s Initial simultaneous time: 08:20:00 Concept / Approach: The next simultaneous change occurs after LCM(48, 72, 108) seconds. Convert the LCM to minutes and seconds and add to the clock time. Step-by-Step Solution: Prime factors: 48 = 2^4 * 3; 72 = 2^3 * 3^2; 108 = 2^2 * 3^3.LCM = 2^4 * 3^3 = 16 * 27 = 432 seconds.432 s = 7 minutes 12 seconds.08:20:00 + 00:07:12 = 08:27:12. Verification / Alternative check: Check 432 is a multiple of each interval: 432/48 = 9, 432/72 = 6, 432/108 = 4, all integers. Why Other Options Are Wrong: Other times correspond to adding 444 s, 456 s, etc., which are not the LCM or not multiples of all three intervals. Common Pitfalls: Mistaking gcd for LCM, or making a clock arithmetic error when adding minutes and seconds. Final Answer: 8 : 27 : 12 hrs.

Question 5

Four metal rods have lengths 78 cm, 104 cm, 117 cm, and 169 cm. They must be cut into equal-length pieces with no wastage. What is the maximum number of pieces that can be obtained in total?

Accepted Answer

Introduction: In problems about cutting objects into equal parts without wastage, the key idea is to use the highest common factor, also called greatest common divisor, of the given lengths. This ensures each piece length divides every original length exactly, creating the maximum possible number of pieces. Given Data / Assumptions: Rod lengths: 78 cm, 104 cm, 117 cm, 169 cm Pieces must have equal length No material wastage is allowed Concept / Approach: The longest feasible piece length is HCF(78, 104, 117, 169). Once the piece length is known, the total number of pieces equals the sum of (length / piece length) over all rods. Step-by-Step Solution: HCF(78, 104) = 26 HCF(26, 117) = 13 HCF(13, 169) = 13 So, piece length = 13 cm Pieces from 78 cm = 78 / 13 = 6 Pieces from 104 cm = 104 / 13 = 8 Pieces from 117 cm = 117 / 13 = 9 Pieces from 169 cm = 169 / 13 = 13 Total pieces = 6 + 8 + 9 + 13 = 36 Verification / Alternative check: If we chose any piece length greater than 13, at least one rod length would not be divisible, causing wastage or inequality. Therefore 13 cm is maximal. Why Other Options Are Wrong: 27, 43, 48, or 26 do not match the sum of quotients when using the correct HCF, nor do they arise from any valid common divisor analysis. Common Pitfalls: Confusing HCF with LCM, or summing individual HCFs pairwise instead of computing the HCF of all lengths together. Final Answer: 36

Question 6

Five college hobby clubs meet periodically: gardening every 2 days, electronics every 3 days, chess every 4 days, yachting every 5 days, and photography every 6 days. Within a span of 180 days, how many times do all five clubs meet on the same day (…

Accepted Answer

Introduction: When periodic events coincide, the interval between coincidences equals the least common multiple of their individual periods. We apply that idea to five meeting schedules measured in days. Given Data / Assumptions: Periods: 2, 3, 4, 5, and 6 days Window: 180 days We count coincidences after day 0, excluding the starting day meeting, which is a standard counting convention for such questions. Concept / Approach: Find LCM(2, 3, 4, 5, 6). The groups meet together every LCM days. The number of such meetings within 180 days, excluding the start, is 180 / LCM. Step-by-Step Solution: LCM(2, 3, 4, 5, 6) = 60 Coincidences after the start within 180 days: 180 / 60 = 3 Verification / Alternative check: The common meeting days fall on day 60, day 120, and day 180 when we exclude day 0. Counting day 0 would give 4, but the problem asks within 180 days excluding the starting day by our stated convention. Why Other Options Are Wrong: 5, 10, and 18 overcount; 4 would include the starting day, which the present interpretation excludes. Common Pitfalls: Including the starting day when the phrase within 180 days is interpreted as after the start. Always clarify the counting convention. Final Answer: 3

Question 7

Three runners A, B, and C start together from the same point around a circular track. Their lap times are A: 252 s, B: 308 s, and C: 198 s. After how much time will they next be together at the starting point?

Accepted Answer

Introduction: When several runners start together and repeat laps with different lap times, their next common meeting at the start occurs after a time equal to the least common multiple of their lap times. Given Data / Assumptions: A lap: 252 s B lap: 308 s C lap: 198 s All start together at the same point Concept / Approach: Compute LCM(252, 308, 198). Convert the resulting seconds to minutes and seconds for a readable time format. Step-by-Step Solution: LCM(252, 308, 198) = 2772 s 2772 s = 46 min and 12 s Verification / Alternative check: Prime factorizations confirm that 2772 is divisible by 252, 308, and 198, and there is no smaller positive time that satisfies all three simultaneously. Why Other Options Are Wrong: 26 min 18 s, 42 min 36 s, 45 min, and 44 min are not common multiples of all three lap times. Common Pitfalls: Taking a simple sum or pairwise LCM instead of the LCM of all three values. Always combine all periods. Final Answer: 46 min and 12 s

Question 8

Find the LCM of the expressions: (x^2 - y^2 - z^2 - 2yz), (x^2 - y^2 + z^2 + 2xz), and (x^2 + y^2 - z^2 - 2xy). Express the LCM as a product of linear factors.

Accepted Answer

Introduction: The task is to find the least common multiple of three quadratic forms by factoring each into linear factors and then collecting distinct factors with the highest needed multiplicities. Given Data / Assumptions: f1 = x^2 - y^2 - z^2 - 2yz f2 = x^2 - y^2 + z^2 + 2xz f3 = x^2 + y^2 - z^2 - 2xy Concept / Approach: Rewrite each quadratic as a difference of squares by grouping. Factor each to identify the linear factors present across the set. The LCM collects each distinct factor at the maximum multiplicity it occurs in any single polynomial. Step-by-Step Solution: f1 = x^2 - (y + z)^2 = (x - y - z)(x + y + z) f2 = (x + z)^2 - y^2 = (x + z - y)(x + z + y) = (x - y + z)(x + y + z) f3 = (x - y)^2 - z^2 = (x - y - z)(x - y + z) Distinct factors: (x + y + z), (x - y - z), (x - y + z) Therefore LCM = (x + y + z)(x - y - z)(x - y + z) Verification / Alternative check: Each of f1, f2, and f3 divides the product (x + y + z)(x - y - z)(x - y + z). No other linear factor is required. Why Other Options Are Wrong: Options that replace (x - y - z) with (x + y - z) or omit factors do not contain all necessary linear factors to cover each polynomial. Common Pitfalls: Sign errors when grouping terms to form a difference of squares, which can flip the identification of factors. Final Answer: (x + y + z)(x - y - z)(x - y + z)

Question 9

Four copper rods have lengths 52 m, 65 m, 78 m, and 91 m. They are to be cut into equal-length pieces with no wastage. What is the minimum total number of pieces obtained?

Accepted Answer

Introduction: To cut rods into equal parts without any waste, we choose the piece length as the highest common factor of all given lengths. This maximizes piece length and therefore minimizes the total number of pieces. Given Data / Assumptions: Rod lengths: 52 m, 65 m, 78 m, 91 m Equal piece length No wastage Concept / Approach: Piece length = HCF(52, 65, 78, 91). Then number of pieces = sum of (length / HCF). Step-by-Step Solution: HCF(52, 65) = 13 HCF(13, 78) = 13 HCF(13, 91) = 13 Piece length = 13 m Pieces: 52 / 13 = 4, 65 / 13 = 5, 78 / 13 = 6, 91 / 13 = 7 Total pieces = 4 + 5 + 6 + 7 = 22 Verification / Alternative check: Any greater common length does not exist, so 13 m is optimal, giving the minimum number of pieces. Why Other Options Are Wrong: 20, 21, 23, and 24 do not match the arithmetic based on the correct HCF and quotients. Common Pitfalls: Using LCM instead of HCF or adding lengths before dividing. Always divide each length by the HCF and sum the quotients. Final Answer: 22

Question 10

Three runners A, B, and C run around a circular stadium and start together from the same point. Their lap times are A: 252 s, B: 308 s, and C: 198 s. After how much time will they all meet again at the starting point?

Accepted Answer

Introduction: This is the same common meeting concept as earlier lap problems. The time to meet again equals the least common multiple of their lap times because each runner must complete an integer number of laps by that moment. Given Data / Assumptions: A: 252 s per lap B: 308 s per lap C: 198 s per lap Concept / Approach: Compute LCM(252, 308, 198) and convert to minutes and seconds. Step-by-Step Solution: LCM = 2772 s 2772 s = 46 min and 12 s Verification / Alternative check: Each of 252, 308, and 198 divides 2772, so they are together again at that time. No smaller positive time works for all three. Why Other Options Are Wrong: The other values are not common multiples of all three lap times. Common Pitfalls: Forgetting to include the third runner when taking the LCM, which yields an incorrect earlier time. Final Answer: 46 min and 12 s

Question 11

Three liquids of 403 L petrol, 456 L diesel, and 496 L motor oil are to be filled into bottles of equal capacity, with each bottle completely full and no mixing across types. What is the least possible number of bottles required?

Accepted Answer

Introduction: Packing different liquids into equal-sized bottles without mixing requires that bottle capacity be a common divisor of each total volume. To minimize the number of bottles, we maximize the bottle capacity, which equals the highest common factor of the volumes. Given Data / Assumptions: Volumes: 403 L, 456 L, 496 L Equal bottle capacity Each bottle filled completely, no mixing of liquids Concept / Approach: Bottle size = HCF(403, 456, 496). Total bottles = 403 / size + 456 / size + 496 / size. Step-by-Step Solution: HCF(403, 456, 496) = 1 Bottle size = 1 L Total bottles = 403 + 456 + 496 = 1355 Verification / Alternative check: Since the greatest common divisor is 1, no larger equal capacity is possible that divides all three volumes exactly. Therefore 1 L is forced, giving 1355 bottles. Why Other Options Are Wrong: 34, 44, 46, and 54 do not equal the computed minimum 1355, so the correct choice is the catchall None of these. Common Pitfalls: Attempting to use pairwise common divisors that do not divide the third volume, which would break the equal capacity rule. Final Answer: None of these

Question 12

There are 30 pineapple trees, 45 orange trees, and 60 mango trees. They are to be arranged in rows so that each row contains the same number of trees and each row uses only one variety. What is the minimum number of rows required?

Accepted Answer

Introduction: To minimize the number of rows while keeping an equal number of trees per row for each variety, we choose the count per row as the highest common factor of the three totals. That makes each row as full as possible and reduces the number of rows. Given Data / Assumptions: Pineapple: 30 Orange: 45 Mango: 60 Each row contains only one variety Each row contains the same number of trees Concept / Approach: Trees per row = HCF(30, 45, 60). Number of rows = (30 + 45 + 60) / trees per row. Step-by-Step Solution: HCF(30, 45, 60) = 15 Total trees = 135 Rows required = 135 / 15 = 9 Verification / Alternative check: Using any smaller common factor would result in fewer trees per row and more rows, so 15 per row is optimal for minimizing rows. Why Other Options Are Wrong: 10, 12, 15, or 25 do not match the computed minimum number of rows from the correct common divisor approach. Common Pitfalls: Using the least common multiple or dividing each count differently, which violates the equal trees per row requirement. Final Answer: 9

Question 13

If HCF of x 3 - 10m 2 + 31x - 30m and x 2 - 8mx + 15 is a linear polynomial, then what is the value of m?

Accepted Answer

Go through options. Put m = 1 in two given expression,we have x3 - 10x2 + 31x - 30 and x2 - 8x + 15 x2 - 8x + 15 = (x - 5) (x - 3) x3 - 10x2 + 31x - 30 = (x - 2) (x - 5) (x - 3) ∴ (x - 5) (x - 3) = x2 - 8x + 15 is the HCF of both expressions. ∴ m = 1

Question 14

Find the sum of digits of the least positive number that leaves remainders 35 when divided by 54, 61 when divided by 80, and 100 when divided by 119.

Accepted Answer

Introduction: This is a simultaneous congruence problem. When all remainders align as the same negative offset from each modulus, the least positive solution is the least common multiple of the moduli minus that common offset. We then sum the digits of the resulting least number. Given Data / Assumptions: N mod 54 = 35 N mod 80 = 61 N mod 119 = 100 Concept / Approach: Recognize that 35 = 54 - 19, 61 = 80 - 19, 100 = 119 - 19, so N is congruent to -19 modulo each modulus. Therefore N = LCM(54, 80, 119) - 19 gives the least positive solution. Step-by-Step Solution: LCM(54, 80, 119) = 257040 Least N = 257040 - 19 = 257021 Sum of digits = 2 + 5 + 7 + 0 + 2 + 1 = 17 Verification / Alternative check: Check: 257021 mod 54 = 35, mod 80 = 61, mod 119 = 100, confirming correctness. Why Other Options Are Wrong: 19, 23, 26, and 28 are plausible digit sums but do not match the sum of digits of the true least solution 257021. Common Pitfalls: Solving each congruence separately without noticing the common negative offset, which significantly simplifies the Chinese remainder reasoning. Final Answer: 17

Question 15

Find the sum of digits of the least positive number that leaves remainders 33 when divided by 52, 59 when divided by 78, and 98 when divided by 117.

Accepted Answer

Introduction: This follows the same remainder alignment idea. When each remainder equals modulus minus the same offset, the least solution is the common period minus that offset. We then sum the digits of the resulting number. Given Data / Assumptions: N mod 52 = 33 N mod 78 = 59 N mod 117 = 98 Concept / Approach: Note that 33 = 52 - 19, 59 = 78 - 19, 98 = 117 - 19, so N is congruent to -19 modulo each modulus. Compute LCM and then subtract 19 to get the least positive N. Step-by-Step Solution: LCM(52, 78, 117) = 468 Least N = 468 - 19 = 449 Sum of digits = 4 + 4 + 9 = 17 Verification / Alternative check: Check: 449 mod 52 = 33, 449 mod 78 = 59, 449 mod 117 = 98. All remainders match. Why Other Options Are Wrong: 19, 21, 27, and 36 are not the digit sum of the least valid N, which equals 449. Common Pitfalls: Missing the shared offset or taking a non-minimal solution by adding an extra multiple of the LCM. Final Answer: 17

Question 16

Two numbers are in the ratio 3 : 4 and their HCF is 4. What is their LCM?

Accepted Answer

Introduction: If two numbers are 3k and 4k and their highest common factor is k, then using the product relation HCF * LCM = product of the numbers gives the LCM directly. Given Data / Assumptions: Numbers in ratio 3 : 4 HCF = 4 Concept / Approach: Let the numbers be 3k and 4k with HCF = k. Given k = 4, so numbers are 12 and 16. Then LCM = (12 * 16) / HCF(12, 16) or more simply LCM = product / HCF. Step-by-Step Solution: Numbers: 12 and 16 Product = 12 * 16 = 192 HCF(12, 16) = 4 LCM = 192 / 4 = 48 Verification / Alternative check: Multiples of 16 that are also multiples of 12: 48 is the least such value. Why Other Options Are Wrong: 12 and 16 are too small, 24 is a multiple of 12 but not 16, and 64 is larger than necessary. Common Pitfalls: Confusing HCF with the common multiplier k in the ratio without following through to actual numbers. Final Answer: 48

Question 17

The sum of two numbers is 1056 and their HCF is 66. How many unordered pairs of such numbers exist?

Accepted Answer

Introduction: Represent numbers with a common HCF using coprime multipliers. Then convert the sum condition into a constraint on those multipliers, and count the coprime solutions accordingly. Given Data / Assumptions: Let the numbers be 66a and 66b HCF(66a, 66b) = 66 implies HCF(a, b) = 1 66a + 66b = 1056 Concept / Approach: From the sum we get a + b = 1056 / 66 = 16. Count coprime positive pairs (a, b) with a + b = 16. For unordered pairs, count each pair once with a < b. Step-by-Step Solution: a + b = 16, HCF(a, b) = 1 Valid unordered coprime pairs: (1, 15), (3, 13), (5, 11), (7, 9) Count = 4 Verification / Alternative check: These correspond to numbers (66, 990), (198, 858), (330, 726), and (462, 594), each with HCF 66 and sum 1056. Why Other Options Are Wrong: 2 and 3 undercount; 6 and 8 double count ordered pairs or include noncoprime pairs. Common Pitfalls: Counting ordered pairs separately or including pairs like (2, 14) where the multipliers are not coprime. Final Answer: 4

Question 18

Two natural numbers have HCF 12 and LCM 72. If one number is 24, what is the difference between the two numbers?

Accepted Answer

Introduction: The product of two numbers equals the product of their HCF and LCM. Knowing one number allows us to find the other directly, then compute the difference. Given Data / Assumptions: HCF = 12 LCM = 72 One number = 24 Concept / Approach: a * b = HCF * LCM. Solve for the unknown and subtract to find the difference. Step-by-Step Solution: a * b = 12 * 72 = 864 If a = 24, then b = 864 / 24 = 36 Difference = 36 - 24 = 12 Verification / Alternative check: HCF(24, 36) = 12 and LCM(24, 36) = 72, confirming consistency. Why Other Options Are Wrong: 18, 21, and 24 do not match the computed difference for the consistent pair. Common Pitfalls: Using sum instead of product relation, or forgetting to check that the derived pair satisfies both HCF and LCM values. Final Answer: 12

Question 19

How many integers between 4000 and 6000 are divisible by 32, 40, 48, and 60?

Accepted Answer

Introduction: An integer divisible by multiple numbers must be a multiple of their least common multiple. We count how many multiples of that LCM lie in the specified interval. Given Data / Assumptions: Range: 4000 to 6000 inclusive Divisors: 32, 40, 48, 60 Concept / Approach: Compute LCM(32, 40, 48, 60). Then count multiples in the range using ceiling and floor division. Step-by-Step Solution: LCM = 480 Smallest multiple ≥ 4000: 9 * 480 = 4320 Largest multiple ≤ 6000: 12 * 480 = 5760 Values: 4320, 4800, 5280, 5760 Count = 4 Verification / Alternative check: Each listed value is divisible by all four divisors by construction through the LCM. Why Other Options Are Wrong: 2, 3, and 5 do not match the exact count of LCM multiples in the interval. Common Pitfalls: Taking pairwise divisibility instead of using the LCM, which can overcount or undercount. Final Answer: 4

Question 20

For any integer n, evaluate HCF(22n + 7, 33n + 10).

Accepted Answer

Introduction: The HCF of linear expressions in n can often be simplified using the Euclidean algorithm by forming integer combinations that remove n, possibly yielding a constant HCF. Given Data / Assumptions: Expressions: 22n + 7 and 33n + 10 n is any integer Concept / Approach: Apply the Euclidean algorithm to reduce the pair step by step until reaching a constant. If that constant is 1, the HCF is 1 for all integers n. Step-by-Step Solution: (33n + 10) - (22n + 7) = 11n + 3 (22n + 7) - 2 * (11n + 3) = 1 Therefore, HCF(22n + 7, 33n + 10) = HCF(11n + 3, 1) = 1 Verification / Alternative check: Since an integer combination of the two gives 1, they are coprime for all integer n. Why Other Options Are Wrong: 0 is not a valid HCF here, 11 is not a universal divisor of both expressions for all n, and None of these does not apply because 1 is correct. Common Pitfalls: Stopping the Euclidean steps early and missing the reduction to 1. Final Answer: 1

Problems on H.C.F and L.C.M Questions