Question 1

Existence check with H.C.F and L.C.M: Consider pairs of integers that have H.C.F = 16 and L.C.M = 136. What can we definitely say about such pairs?

Accepted Answer

Introduction / Context:This question asks whether any pair of integers can have the stated HCF and LCM. A necessary condition for two integers a and b is a * b = HCF * LCM. Additionally, the LCM must be a multiple of the HCF. Given Data / Assumptions: HCF = 16 LCM = 136 Integers a, b > 0 assumed. Concept / Approach:Write a = 16x and b = 16y with HCF(x, y) = 1. Then LCM(a, b) = 16xy. For LCM to be 136, we need 16xy = 136 ⇒ xy = 136/16 = 8.5, which is not an integer. Hence impossible. Step-by-Step Solution: If HCF = 16, then a = 16x, b = 16y with HCF(x, y) = 1LCM(a, b) = 16xy must equal 136Thus xy = 136 / 16 = 8.5 ⇒ not an integer ⇒ contradictionTherefore, there are no such integer pairs. Verification / Alternative check:Check divisibility: LCM must be a multiple of HCF. 136 is not divisible by 16 (136/16 = 8.5), so it cannot be an LCM when HCF is 16. Why Other Options Are Wrong:If no pair exists, counts like “one” or “two” or “many” are impossible. Common Pitfalls:Forgetting the requirement that LCM must be a multiple of HCF, or ignoring the integral nature of xy in the representation a = 16x, b = 16y. Final Answer:No such pair exists

Question 2

Smallest number with specified remainders: Find the least positive number that leaves remainders 3, 5, and 8 when divided by 16, 18, and 21 respectively.

Accepted Answer

Introduction / Context:This is a Chinese Remainder style problem with a useful observation: the three remainders (3, 5, 8) are all congruent to −13 with respect to their moduli (16, 18, 21). That means the sought number differs by the same constant from a common multiple of the moduli. Given Data / Assumptions: N ≡ 3 (mod 16) N ≡ 5 (mod 18) N ≡ 8 (mod 21) Concept / Approach:Note that 3 = −13 mod 16, 5 = −13 mod 18, 8 = −13 mod 21. Therefore, N ≡ −13 modulo LCM(16, 18, 21). Compute the LCM and then find the least positive N of the form LCM − 13. Step-by-Step Solution: LCM(16, 18, 21): 16 = 2^4, 18 = 2 * 3^2, 21 = 3 * 7 ⇒ LCM = 2^4 * 3^2 * 7 = 1008Thus N ≡ −13 (mod 1008) ⇒ least positive N = 1008 − 13 = 995Hence, the smallest such number is 995. Verification / Alternative check:Check: 995 mod 16 = 3, 995 mod 18 = 5, 995 mod 21 = 8. All conditions satisfied. Why Other Options Are Wrong:893, 992, 1024, and 1005 do not satisfy all three remainder conditions simultaneously. Common Pitfalls:Attempting to solve each congruence independently without noticing the convenient common shift of −13, or computing LCM incorrectly. Final Answer:995

Question 3

Greatest four-digit multiple: Find the greatest four-digit number that is divisible by each of 12, 18, 21, and 28.

Accepted Answer

Introduction / Context:The largest four-digit number divisible by several integers must be a multiple of their least common multiple (LCM). Compute the LCM, then take the largest multiple not exceeding 9999. Given Data / Assumptions: Divisors: 12, 18, 21, 28 Upper limit: 9999 (largest four-digit number) Concept / Approach:LCM via prime powers: choose the highest power of each prime appearing in the prime factorizations. Then compute floor(9999 / LCM) * LCM. Step-by-Step Solution: Prime factors: 12 = 2^2 * 3; 18 = 2 * 3^2; 21 = 3 * 7; 28 = 2^2 * 7LCM = 2^2 * 3^2 * 7 = 4 * 9 * 7 = 252Largest multiple ≤ 9999: floor(9999 / 252) = 39 ⇒ 39 * 252 = 9828Therefore, the greatest four-digit number divisible by all four is 9828. Verification / Alternative check:Check divisibility: 9828 / 12 = 819, /18 = 546, /21 = 468, /28 = 351 — all integers. Why Other Options Are Wrong:9864 and 9848 are not multiples of 252; 9636 is a multiple of 252 but smaller than 9828; 9996 is not divisible by all listed numbers. Common Pitfalls:Computing LCM incorrectly (e.g., missing 3^2) or multiplying by 40 (which gives 10080, exceeding four digits). Final Answer:9828

Question 4

Least number with uniform remainder: Find the least positive number which, when divided by 15, 27, 35, and 42, leaves a remainder of 7 in each case.

Accepted Answer

Introduction / Context:When a number leaves the same remainder r upon division by several moduli, then that number minus r is a common multiple of the moduli. This reduces directly to computing an LCM and then adding the remainder. Given Data / Assumptions: N ≡ 7 (mod 15) N ≡ 7 (mod 27) N ≡ 7 (mod 35) N ≡ 7 (mod 42) Concept / Approach:Let M be LCM(15, 27, 35, 42). Then the least positive N is M + 7. Compute M using prime powers: take the highest power of each prime across the factorizations. Step-by-Step Solution: Prime factors: 15 = 3 * 5; 27 = 3^3; 35 = 5 * 7; 42 = 2 * 3 * 7LCM = 2^1 * 3^3 * 5^1 * 7^1 = 2 * 27 * 5 * 7 = 1890Therefore, least N = 1890 + 7 = 1897 Verification / Alternative check:Check: 1897 mod 15 = 7; mod 27 = 7; mod 35 = 7; mod 42 = 7. All satisfied. Why Other Options Are Wrong:1883, 1987, and 2007 do not satisfy all four congruences. 1890 is LCM itself and would leave remainder 0, not 7. Common Pitfalls:Leaving out the factor 2 from 42 when computing LCM, or forgetting to add the remainder after finding the LCM. Final Answer:1897

Question 5

Measuring rods problem — find the shortest cloth length measured exactly by each Three measuring rods are 64 cm, 80 cm, and 96 cm long. What is the least length of cloth that can be measured an exact number of times using any one of these rods (i.e.…

Accepted Answer

Introduction / Context:This is a classic least common multiple (LCM) application. The “least length of cloth measured exactly by each rod” means we need a length that is simultaneously a multiple of 64 cm, 80 cm, and 96 cm, and as small as possible. Converting to meters at the end makes the answer presentation consistent with the options. Given Data / Assumptions: Rod lengths: 64 cm, 80 cm, 96 cm. We want the least positive length that is a multiple of all three. Final answer must be shown in meters. Concept / Approach:The LCM of integers a, b, c can be found by prime factorization or successive LCM calculations. LCM captures the smallest number divisible by all inputs. Step-by-Step Solution:64 = 2^6.80 = 2^4 * 5.96 = 2^5 * 3.Take the highest power of each prime across all: for 2 → 2^6; for 3 → 3^1; for 5 → 5^1.LCM = 2^6 * 3 * 5 = 64 * 15 = 960 cm.Convert to meters: 960 cm = 9.60 m. Verification / Alternative check:9.60 m = 960 cm. Now 960 ÷ 64 = 15, 960 ÷ 80 = 12, 960 ÷ 96 = 10, all integers, so 960 cm works. Any smaller number would miss at least one divisibility requirement. Why Other Options Are Wrong:0.96 m is only 96 cm, not divisible by 80. 19.20 m and 96.00 m are multiples of 9.60 m but not the least. 6.40 m (640 cm) fails divisibility by 96. Common Pitfalls:Taking the greatest common divisor (GCD) by mistake, or converting to meters too early and introducing rounding errors. Always compute LCM in centimeters first. Final Answer:9.60 m

Question 6

Count prime factors with multiplicity How many prime factors (counting multiplicity) are there in (6)^10 * (7)^17 * (11)^27?

Accepted Answer

Introduction / Context:This question checks understanding of “number of prime factors with multiplicity.” For a prime power p^k, there are exactly k prime factors counted with multiplicity. For a product of prime powers, add the exponents. Given Data / Assumptions: (6)^10 * (7)^17 * (11)^27. 6 is not prime, but its prime factorization is 2 * 3. However, the expression presents 6 as a base; we are asked to count prime factors of the whole product with multiplicity. Concept / Approach:Rewrite (6)^10 as (2 * 3)^10 = 2^10 * 3^10. Then the whole product becomes 2^10 * 3^10 * 7^17 * 11^27. The total number of prime factors with multiplicity is the sum of all exponents. Step-by-Step Solution:(6)^10 contributes 10 twos and 10 threes.(7)^17 contributes 17 sevens.(11)^27 contributes 27 elevens.Total = 10 + 10 + 17 + 27 = 54. Verification / Alternative check:You can explicitly state the prime-power breakdown: 2^10, 3^10, 7^17, 11^27. Summing exponents is the standard method and is unambiguous. Why Other Options Are Wrong:64, 71, 81, and 44 are simple mis-sums or include an error from misinterpreting 6 as a prime power without expanding it. Common Pitfalls:Forgetting to expand 6 = 2 * 3, or mistakenly adding 10 twice to get 20 but then also miscounting the other exponents. Final Answer:54

Question 7

Find the least subtraction for a uniform remainder condition What least number must be subtracted from 1294 so that when the result is divided by 9, 11, and 13 it leaves remainder 6 in each case?

Accepted Answer

Introduction / Context:We need the smallest nonnegative integer x such that N' = 1294 - x leaves the same remainder 6 upon division by 9, 11, and 13. This is a simultaneous congruence question that reduces to using the least common multiple (LCM) of the moduli. Given Data / Assumptions: N' mod 9 = 6, N' mod 11 = 6, N' mod 13 = 6. All moduli 9, 11, 13 are pairwise coprime. Concept / Approach:If N' leaves remainder 6 with each of 9, 11, 13, then N' - 6 is divisible by all three. Hence N' - 6 is a multiple of LCM(9, 11, 13). Because the moduli are pairwise coprime, the LCM is their product. Step-by-Step Solution:LCM(9, 11, 13) = 9 * 11 * 13 = 1287.So N' ≡ 6 (mod 1287). The closest number to 1294 that is congruent to 6 mod 1287 is 1287 + 6 = 1293.Thus 1294 - x = 1293 ⇒ x = 1. Verification / Alternative check:1293 ÷ 9 leaves 6; 1293 ÷ 11 leaves 6; 1293 ÷ 13 leaves 6. All checks pass. Why Other Options Are Wrong:0 gives 1294, which does not leave remainder 6 for all three. 2 or 3 land at 1292 or 1291, which fail at least one modulus. 7 similarly fails. Common Pitfalls:Forgetting to convert “same remainder 6” into the condition N' - 6 is divisible by each modulus, or using gcd instead of LCM. Final Answer:1

Question 8

Find the least number with mixed remainder conditions The smallest number leaves remainder 3 when divided by 5, 6, 7, and 8, but leaves no remainder when divided by 9. What is that number?

Accepted Answer

Introduction / Context:This is a Chinese Remainder Theorem style problem in an elementary form. We use LCM to handle the common remainder part, then apply the additional divisibility condition with 9 to select the correct instance. Given Data / Assumptions: N ≡ 3 (mod 5, 6, 7, 8). N ≡ 0 (mod 9). Concept / Approach:Numbers satisfying the first set can be written as N = LCM(5, 6, 7, 8) * k + 3. Then we search for k such that N is divisible by 9. Step-by-Step Solution:LCM(5, 6, 7, 8) = 840.Hence N = 840k + 3.Impose divisibility by 9: 840k + 3 ≡ 0 (mod 9). Since 840 ≡ 3 (mod 9), we have 3k + 3 ≡ 0 ⇒ k + 1 ≡ 0 (mod 3) ⇒ k ≡ 2 (mod 3).Smallest k = 2 ⇒ N = 840 * 2 + 3 = 1683. Verification / Alternative check:1683 ÷ 5 leaves 3; ÷ 6 leaves 3; ÷ 7 leaves 3; ÷ 8 leaves 3; and 1683 is divisible by 9 because 1 + 6 + 8 + 3 = 18, a multiple of 9. Why Other Options Are Wrong:1677, 2523, 3363, and 8403 violate at least one of the remainder conditions or the divisibility by 9 condition. Common Pitfalls:Using the sum 5 + 6 + 7 + 8 instead of LCM, or forgetting to test the added divisibility requirement by 9. Final Answer:1683

Question 9

Find the greatest divisor from non-equal remainders The greatest number that divides 1657 and 2037 leaving remainders 6 and 5 respectively is:

Accepted Answer

Introduction / Context:When a number d divides two numbers leaving remainders r1 and r2, it divides the corresponding differences (N1 - r1) and (N2 - r2) exactly. The required greatest divisor is then gcd(N1 - r1, N2 - r2). Given Data / Assumptions: N1 = 1657 leaves remainder 6. N2 = 2037 leaves remainder 5. d | (N1 - 6) and d | (N2 - 5). Concept / Approach:Compute A = 1657 - 6 and B = 2037 - 5, then find gcd(A, B) using the Euclidean algorithm. Step-by-Step Solution:A = 1651, B = 2032.B - A = 381.Compute gcd(1651, 2032): gcd(1651, 381) since 2032 - 1651 = 381.1651 = 381 * 4 + 127.381 = 127 * 3 + 0 ⇒ gcd = 127. Verification / Alternative check:Check divisibility: 1651 ÷ 127 = 13; 2032 ÷ 127 = 16; hence 127 divides both differences and works. Why Other Options Are Wrong:235, 260, 305, 381 do not evenly divide both 1651 and 2032, or are not the greatest common divisor. Common Pitfalls:Taking gcd of 1657 and 2037 directly (ignoring remainders), or subtracting the wrong remainders. Final Answer:127

Question 10

Six bells toll together — repaired data and count Six bells start together and toll at intervals of 5 s, 10 s, 15 s, 20 s, 25 s, and 30 s, respectively. In 60 minutes, how many times do they toll together (including the start)?

Accepted Answer

Introduction / Context:The original stem listed “six bells” but only five intervals. Using the Recovery-First Policy, we infer the missing interval to be 25 s so that six intervals are present: 5, 10, 15, 20, 25, and 30 seconds. We must find how many times all bells toll together in 60 minutes, counting the starting instant. Given Data / Assumptions: Intervals (seconds): 5, 10, 15, 20, 25, 30. Total time: 60 minutes = 3600 seconds. All bells toll together whenever time is a common multiple of all intervals. Concept / Approach:The bells toll together at times that are multiples of LCM of the six intervals. The count within T seconds equals floor(T / LCM) + 1 (the “+1” for time 0 at the start). Step-by-Step Solution:Compute LCM(5, 10, 15, 20, 25, 30).Prime powers: 5 = 5; 10 = 2 * 5; 15 = 3 * 5; 20 = 2^2 * 5; 25 = 5^2; 30 = 2 * 3 * 5.LCM = 2^2 * 3 * 5^2 = 4 * 3 * 25 = 300 seconds.Total occurrences in 3600 seconds: floor(3600 / 300) + 1 = 12 + 1 = 13. Verification / Alternative check:Every 300 s (5 min), all bells coincide: at 0, 300, 600, …, 3600 s (13 instants). Why Other Options Are Wrong:20, 15, 11 do not match the LCM-based count. 61 corresponds to an LCM of 60 s, which is not correct for the repaired set including 25 s. Common Pitfalls:Forgetting to include the start time, or computing with only the five originally typed intervals, which makes the problem inconsistent with the given options. Final Answer:13

Question 11

LCM feasibility given a fixed HCF If the HCF (gcd) of two positive integers is 26, which of the following cannot be their LCM?

Accepted Answer

Introduction / Context:For two positive integers a and b, the fundamental identity is: a * b = HCF(a, b) * LCM(a, b). If HCF = 26, then both a and b are multiples of 26, say a = 26m and b = 26n with gcd(m, n) = 1. Consequently, LCM = 26 * m * n. Given Data / Assumptions: HCF = 26. a = 26m, b = 26n, gcd(m, n) = 1. Therefore LCM must be 26 times an integer product of two coprime integers. Concept / Approach:Check each candidate whether it is a multiple of 26. If it is not divisible by 26, it cannot be the LCM. If it is divisible by 26, verify that the quotient can be written as a product of two coprime integers (always possible with 1 * k). Step-by-Step Solution:78 = 26 * 3 ⇒ possible (m, n) = (1, 3).104 = 26 * 4 ⇒ possible (m, n) = (1, 4).144 ÷ 26 is not an integer ⇒ impossible.234 = 26 * 9 ⇒ possible (m, n) = (1, 9).260 = 26 * 10 ⇒ possible (m, n) = (2, 5) with gcd 1. Verification / Alternative check:Because HCF divides both numbers, it must also divide their LCM. Any candidate LCM not divisible by 26 is invalid immediately. Why Other Options Are Wrong:All other options are multiples of 26 and are feasible. Common Pitfalls:Forgetting the divisibility requirement of LCM by the HCF, or believing m and n must both exceed 1; m = 1 is allowed. Final Answer:144

Question 12

Coprime property transfer If HCF(m, n) = 1 with m > n, then what are HCF(m + n, m) and HCF(m - n, n), respectively?

Accepted Answer

Introduction / Context:This problem uses standard gcd identities. If gcd(m, n) = 1 (m, n are coprime), then adding or subtracting multiples does not introduce a common factor. In particular, gcd(m, m ± n) = gcd(m, n) = 1, and similarly for gcd(n, m ± n). Given Data / Assumptions: gcd(m, n) = 1 and m > n. We need gcd(m + n, m) and gcd(m - n, n). Concept / Approach:Use the identity gcd(a, b) = gcd(a, b ± a). Then gcd(m + n, m) = gcd(m + n - m, m) = gcd(n, m) = 1. Likewise, gcd(m - n, n) = gcd(m, n) when adjusting with multiples. Step-by-Step Solution:gcd(m + n, m) = gcd(n, m) = 1.gcd(m - n, n) = gcd(m, n) = 1. Verification / Alternative check:Try small coprime examples, e.g., m = 5, n = 2. Then gcd(7, 5) = 1 and gcd(3, 2) = 1. The property holds in general. Why Other Options Are Wrong:Options involving 2 assume parity constraints; however, even if m and n share parity conditions, overall gcd remains 1 by the identities. Common Pitfalls:Assuming that adding or subtracting numbers may introduce a factor like 2; this does not happen in gcd algebra unless parity forces both to be even, which cannot occur for coprime m and n simultaneously. Final Answer:1 and 1

Question 13

Recover two numbers from LCM, HCF, and ratio The LCM and HCF of two numbers are 240 and 16, respectively. If their ratio is 3 : 5, find the two numbers.

Accepted Answer

Introduction / Context:Two unknown numbers are given indirectly via their LCM, HCF, and ratio. When the ratio is a : b with gcd(a, b) = 1, the actual numbers are a * HCF and b * HCF. We can verify the LCM at the end. Given Data / Assumptions: LCM = 240. HCF = 16. Ratio = 3 : 5 (coprime pair). Concept / Approach:Let numbers be 3x and 5x. Their gcd equals x because 3 and 5 are coprime. Given HCF = 16 ⇒ x = 16, so numbers are 48 and 80. Verify LCM using the identity LCM * HCF = product. Step-by-Step Solution:Assume numbers = 3x and 5x; gcd(3x, 5x) = x.Given HCF = 16 ⇒ x = 16.Numbers: 48 and 80.Check LCM: (48 * 80) / gcd(48, 80) = (3840) / 16 = 240, matches. Verification / Alternative check:Prime factors: 48 = 2^4 * 3; 80 = 2^4 * 5. LCM takes 2^4 * 3 * 5 = 240; gcd takes 2^4 = 16. Why Other Options Are Wrong:Other pairs either do not have gcd 16 or their LCM is not 240 when checked. Common Pitfalls:Forgetting that the given ratio terms must be coprime to apply the simple multiplication by HCF, or miscomputing the LCM check. Final Answer:48, 80

Question 14

Common linear factor condition — repaired for single-correct option For polynomials f(x) = x^3 + c x^2 - x + 2c and g(x) = x^2 + c x - 2 over the rationals, which one of the following values of c makes their HCF (gcd) a linear polynomial (degree 1)?

Accepted Answer

Introduction / Context:We want the greatest common divisor (over the rationals) of two polynomials to have degree 1. That means the polynomials share a common linear factor (i.e., they have at least one common rational/real root), but do not share a quadratic factor. We will derive c-values by eliminating c using a common root r. Given Data / Assumptions: f(x) = x^3 + c x^2 - x + 2c. g(x) = x^2 + c x - 2. Degree-1 gcd ⇒ f and g have a common linear factor (a common root r). Concept / Approach:If r is a common root, g(r) = 0 ⇒ c = (2 - r^2) / r (r ≠ 0). Substitute this c into f(r) = 0 and solve for r, then back-calculate c. This is an elimination technique often quicker than computing a full resultant in exams. Step-by-Step Solution:From g(r) = 0: c = (2 - r^2) / r.Compute f(r) = r^3 + c r^2 - r + 2c. Substitute c:f(r) = r^3 + ((2 - r^2)/r) * r^2 - r + 2 * (2 - r^2)/r.Simplify: r^3 + (2 - r^2) r - r + (4 - 2 r^2)/r = r + (4 - 2 r^2)/r.Multiply by r: r^2 + 4 - 2 r^2 = 0 ⇒ 4 - r^2 = 0 ⇒ r = ±2.For r = 2: c = (2 - 4)/2 = -1. For r = -2: c = (2 - 4)/(-2) = +1. Verification / Alternative check:For c = 1, g(x) = x^2 + x - 2 = (x + 2)(x - 1) and f(x) = x^3 + x^2 - x + 2 = (x + 2)(x^2 - 1) = (x + 2)(x - 1)(x + 1). The common factor is linear (x + 2). For c = -1, g(x) = x^2 - x - 2 = (x - 2)(x + 1) and f(x) = x^3 - x^2 - x - 2 = (x - 2)(x^2 + 1); again a linear common factor exists. The question asks for a single value; choosing +1 suffices. Why Other Options Are Wrong:c = +2 or +4 do not create a shared linear factor (quick factor checks fail). c = 0 does not yield a common linear factor either. The repaired single-correct framing accepts +1; -1 is also a valid value but not among the intended single-choice answers originally due to a duplicate option issue in the source. Common Pitfalls:Attempting long division repeatedly instead of eliminating c via a common root. Also, forgetting to verify that the gcd is degree 1, not higher. Final Answer:+1

Question 15

Find the greatest integer that divides each of the numbers 263, 935, and 1383 and leaves a remainder of 7 in every case.

Accepted Answer

Introduction / Context:This divisibility problem asks for the largest integer d that, when dividing 263, 935, and 1383, leaves the same remainder 7 in each case. Such questions reduce to computing a greatest common divisor (GCD) after adjusting for the common remainder. Given Data / Assumptions: The three integers are 263, 935, and 1383. The common remainder on division by the required number is 7. d is a positive integer and must be as large as possible. Concept / Approach:If each number leaves remainder 7 upon division by d, then subtracting 7 from each number yields exact multiples of d. Therefore, d must divide 263 − 7, 935 − 7, and 1383 − 7. Hence, d = gcd(256, 928, 1376). Compute the GCD stepwise using the Euclidean algorithm. Step-by-Step Solution: Transform numbers: 263 − 7 = 256; 935 − 7 = 928; 1383 − 7 = 1376.Compute gcd(256, 928): 928 mod 256 = 160; 256 mod 160 = 96; 160 mod 96 = 64; 96 mod 64 = 32; 64 mod 32 = 0 ⇒ gcd = 32.Now gcd(32, 1376): 1376 mod 32 = 0 ⇒ gcd = 32.Therefore, the greatest required divisor is 32. Verification / Alternative check:Direct check: 256/32, 928/32, and 1376/32 are integers (8, 29, and 43). This confirms 32 divides all adjusted values, and no larger divisor can exceed the GCD. Why Other Options Are Wrong: 31, 35, 30, 28: None divides all of 256, 928, and 1376 exactly, so they cannot be the common divisor after adjusting the remainder. Common Pitfalls: Forgetting to subtract the common remainder before computing the GCD. Stopping the Euclidean algorithm too early and reporting a non-maximum common divisor. Final Answer:32

Question 16

Find the greatest number that, when dividing 390, 480, and 620, leaves the same remainder in each case.

Accepted Answer

Introduction / Context:When a single divisor leaves the same remainder upon division of several numbers, the differences between the numbers are exact multiples of that divisor. The largest such divisor equals the greatest common divisor (GCD) of all pairwise differences. Given Data / Assumptions: Numbers: 390, 480, 620. One unknown divisor d leaves the same remainder on each division. Concept / Approach:If 390 ≡ 480 ≡ 620 (mod d), then d divides 480 − 390, 620 − 480, and 620 − 390. So compute d = gcd(90, 140, 230). The largest such d is the answer. Step-by-Step Solution: Compute differences: 480 − 390 = 90; 620 − 480 = 140; 620 − 390 = 230.gcd(90, 140): 140 mod 90 = 50; 90 mod 50 = 40; 50 mod 40 = 10; 40 mod 10 = 0 ⇒ gcd = 10.gcd(10, 230) = 10 (since 230 mod 10 = 0).Hence the greatest possible divisor is 10. Verification / Alternative check:Because 10 divides each difference, any common remainder r (0 ≤ r < 10) will be the same for all; no larger number divides all the differences simultaneously. Why Other Options Are Wrong: 15, 20, 19, 30: These do not divide all of 90, 140, and 230 together; hence they cannot produce the same remainder for all three original numbers. Common Pitfalls: Mistaking the GCD of the original numbers for the required divisor. Not considering all pairwise differences. Final Answer:10

Question 17

Find all numbers between 300 and 400 that leave a remainder of 4 when divided by 12, 18, and 36. Then compute the sum of those numbers.

Accepted Answer

Introduction / Context:We are told of a number N between 300 and 400 that leaves remainder 4 upon division by 12, 18, and 36. Because 36 is a multiple of both 12 and 18, the single congruence modulo 36 controls the condition. The task then asks for the sum of all such numbers in the interval (there are two), not just one value. Given Data / Assumptions: N is strictly between 300 and 400. N ≡ 4 (mod 12), N ≡ 4 (mod 18), and N ≡ 4 (mod 36). LCM(12, 18, 36) = 36, so it suffices that N ≡ 4 (mod 36). Concept / Approach:All solutions are of the form N = 4 + 36k. Choose integers k such that 300 < N < 400. List the valid N and then add them. Ensure 400 is excluded if the wording requires strictly between 300 and 400. Step-by-Step Solution: General form: N = 4 + 36k.Find k with 300 < 4 + 36k < 400 ⇒ 296 < 36k < 396.Divide by 36: 8.22… < k < 11.Integer k values: 9 and 10.Corresponding N: 4 + 36*9 = 328 and 4 + 36*10 = 364.Sum = 328 + 364 = 692. Verification / Alternative check:Check remainders: 328 mod 36 = 4; 364 mod 36 = 4. Both also leave remainder 4 modulo 12 and 18 since 36 encompasses those moduli. 400 is excluded as it equals the upper boundary; while 400 mod 36 = 4, it is not strictly between 300 and 400. Why Other Options Are Wrong: 840, 324, 1024, 728: Do not match the computed sum of valid values in the specified interval. Common Pitfalls: Including 400 in the sum despite the phrase “between 300 and 400.” Using LCM(12, 18) instead of LCM(12, 18, 36) without noticing 36 already governs both. Final Answer:692

Question 18

Compute the H.C.F. (greatest common divisor) of the integers 132, 204, and 228.

Accepted Answer

Introduction / Context:We seek the highest common factor (HCF) of three integers: 132, 204, and 228. The HCF is the largest positive integer that divides each of the numbers exactly, leaving no remainder. Given Data / Assumptions: Numbers: 132, 204, 228. Standard divisibility and Euclidean algorithm apply. Concept / Approach:Compute HCF stepwise: gcd(132, 204) first, then gcd(result, 228). The Euclidean algorithm is efficient and reliable for such calculations. Step-by-Step Solution: gcd(132, 204): 204 mod 132 = 72; 132 mod 72 = 60; 72 mod 60 = 12; 60 mod 12 = 0 ⇒ gcd = 12.gcd(12, 228): 228 mod 12 = 0 ⇒ gcd = 12.Therefore, HCF(132, 204, 228) = 12. Verification / Alternative check:Prime factors: 132 = 2^2*3*11; 204 = 2^2*3*17; 228 = 2^2*3*19. The common prime product is 2^2*3 = 12, confirming the result. Why Other Options Are Wrong: 18, 6, 21, 24: These either do not divide all three numbers or are not the greatest common factor. Only 12 divides all and is maximal. Common Pitfalls: Selecting a common divisor (like 6) and overlooking a larger common divisor (12). Final Answer:12

Question 19

Find the H.C.F. (greatest common divisor) of the fractions 4/5 and 7/15.

Accepted Answer

Introduction / Context:For fractions, the HCF (also called GCD) is computed using a specific rule that involves the numerators and denominators separately. This problem checks knowledge of that rule and careful simplification. Given Data / Assumptions: Fractions: 4/5 and 7/15. All values are positive rational numbers. Concept / Approach:The HCF of fractions a/b and c/d is given by HCF(numerators) / LCM(denominators). Thus compute HCF(4, 7) and LCM(5, 15). Since 4 and 7 are coprime, their HCF is 1. For 5 and 15, LCM is 15. Step-by-Step Solution: HCF of numerators: HCF(4, 7) = 1.LCM of denominators: LCM(5, 15) = 15.Therefore, HCF of the fractions = 1 / 15. Verification / Alternative check:Any common fractional divisor must be at most 1/15 because 1/15 times an integer must divide both fractions cleanly; testing confirms 1/15 is feasible and larger fractional candidates fail simultaneously for both fractions. Why Other Options Are Wrong: 1/5 and 1/25: Do not satisfy the fraction HCF rule here. 1/13 and 2/15: Not derived from HCF(4, 7) and LCM(5, 15); these are distractors. Common Pitfalls: Using LCM of numerators over HCF of denominators (that is for LCM of fractions, not HCF). Final Answer:1/15

Question 20

Find the H.C.F. (greatest common divisor) of the fractions 1/2, 3/4, and 4/5.

Accepted Answer

Introduction / Context:The HCF of multiple fractions is computed via the same rule as for two fractions: HCF of numerators divided by LCM of denominators. This problem applies that rule to three fractions simultaneously. Given Data / Assumptions: Fractions: 1/2, 3/4, 4/5. All are positive rationals. Concept / Approach:For fractions a/b, c/d, e/f: HCF = HCF(a, c, e) / LCM(b, d, f). Compute the integer HCF of the numerators and the integer LCM of the denominators, then form the fraction. Step-by-Step Solution: Numerator HCF: HCF(1, 3, 4) = 1.Denominator LCM: LCM(2, 4, 5) = 20.Therefore, HCF of the fractions = 1 / 20. Verification / Alternative check:Check that 1/20 is a common fractional divisor: (1/2) ÷ (1/20) = 10; (3/4) ÷ (1/20) = 15; (4/5) ÷ (1/20) = 16. All quotients are integers, confirming 1/20 works; any larger fractional divisor would require a smaller denominator than 20, which would fail at least one fraction. Why Other Options Are Wrong: 1/40: Too small; not the greatest possible. 20, 15: These are integers, not fractional HCFs here. 1/10: Larger than the true HCF; it does not divide all three fractions exactly. Common Pitfalls: Confusing the rules for HCF and LCM of fractions. Forgetting to take HCF of numerators and LCM of denominators in the correct places. Final Answer:1/20

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