Question 1

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

Accepted Answer

Introduction / Context:This number theory question combines the concept of least common multiple (L.C.M.) with congruences and an additional condition that the resulting number must be a multiple of 7. It tests your ability to interpret and manipulate modular arithmetic conditions in a structured way. Given Data / Assumptions: We seek a number N that is a multiple of 7. N leaves a remainder of 4 when divided by 6, 9, 15, and 18. We must find the least such multiple of 7. Concept / Approach:If N gives remainder 4 when divided by each of 6, 9, 15, and 18, then N - 4 is exactly divisible by those numbers. Therefore, N - 4 must be a common multiple of 6, 9, 15, and 18. The smallest such value is their L.C.M. Let that L.C.M. be L. Then N can be written as N = 4 + kL. We must choose the smallest such N that is also a multiple of 7, that is, 4 + kL must be divisible by 7 for some positive integer k. Step-by-Step Solution:Step 1: Let N be the required number.Step 2: N leaves remainder 4 when divided by 6, 9, 15, and 18, so N - 4 is divisible by all four numbers.Step 3: Compute L = L.C.M.(6, 9, 15, 18).Step 4: Factorize: 6 = 2 * 3, 9 = 3^2, 15 = 3 * 5, 18 = 2 * 3^2.Step 5: Take highest powers: 2^1, 3^2, and 5^1.Step 6: L = 2 * 9 * 5 = 90.Step 7: So N - 4 = 90k, and N = 4 + 90k for some integer k.Step 8: We also need N to be a multiple of 7, so 4 + 90k must be divisible by 7.Step 9: Work modulo 7: 90 ≡ 6 (mod 7), so 4 + 90k ≡ 4 + 6k (mod 7).Step 10: We require 4 + 6k ≡ 0 (mod 7), or 6k ≡ -4 ≡ 3 (mod 7).Step 11: The inverse of 6 mod 7 is 6, so k ≡ 3 * 6 ≡ 18 ≡ 4 (mod 7).Step 12: The smallest positive k that satisfies this is k = 4.Step 13: Therefore N = 4 + 90 * 4 = 4 + 360 = 364. Verification / Alternative check:Check divisibility: 364 is a multiple of 7 (7 * 52 = 364).Check remainders: 364 ÷ 6 = 60 remainder 4, 364 ÷ 9 = 40 remainder 4, 364 ÷ 15 = 24 remainder 4, 364 ÷ 18 = 20 remainder 4. All conditions are satisfied, and because we used the smallest k compatible with the modular condition, 364 is the least such multiple. Why Other Options Are Wrong:Options a (74), b (94), and c (184) either do not satisfy the remainder condition for all four divisors or are not multiples of 7. A quick check reveals that at least one of the specified divisibility conditions fails for each of these numbers. Common Pitfalls:Many students try random trial and error with the answer choices rather than using a systematic L.C.M. and modular arithmetic approach. Others may forget that N must also be a multiple of 7 or mis-handle the modulo calculations. Carefully translating the conditions into equations and congruences helps avoid these mistakes. Final Answer:The least multiple of 7 that satisfies the given conditions is 364.

Question 2

The sum of two numbers is 55 and their H.C.F. and L.C.M. are 5 and 120 respectively. What is the sum of the reciprocals of these two numbers?

Accepted Answer

Introduction / Context:This problem again uses the relationship between the H.C.F., L.C.M., and the product of two numbers, but it adds an extra twist by asking for the sum of reciprocals. It tests your ability to manipulate algebraic expressions involving sums, products, and reciprocals, which is a common theme in competitive examinations. Given Data / Assumptions: Let the two positive integers be x and y. x + y = 55. H.C.F. of x and y is 5. L.C.M. of x and y is 120. We must find 1/x + 1/y. Concept / Approach:We use the key identity x * y = H.C.F. * L.C.M. to find the product x * y. Once we have the sum and product of the two numbers, the sum of the reciprocals can be written as (x + y) / (x * y). This avoids the need to find the numbers explicitly, although we can still find them as a check. Step-by-Step Solution:Step 1: Let the numbers be x and y.Step 2: Given x + y = 55.Step 3: Given H.C.F. = 5 and L.C.M. = 120.Step 4: Use the relation x * y = H.C.F. * L.C.M. = 5 * 120 = 600.Step 5: The sum of reciprocals is 1/x + 1/y = (x + y) / (x * y).Step 6: Substitute values: (x + y) / (x * y) = 55 / 600.Step 7: Simplify the fraction 55 / 600 by dividing numerator and denominator by 5.Step 8: 55 ÷ 5 = 11 and 600 ÷ 5 = 120, so the result is 11/120. Verification / Alternative check:We can also compute x and y explicitly by solving t^2 - 55t + 600 = 0.The roots are t = 40 and t = 15.Then 1/40 + 1/15 = (15 + 40) / (40 * 15) = 55 / 600 = 11/120. This matches our earlier result, confirming that the sum of the reciprocals is 11/120. Why Other Options Are Wrong:Options a (55/601) and b (601/55) are unrelated to the derived formula and represent incorrect manipulations of the sum and product. Option d (120/11) is the reciprocal of the correct answer and corresponds to a common mistake of inverting the fraction incorrectly. The correct direction of the reciprocal transformation must be maintained. Common Pitfalls:Some students mistakenly try to take reciprocals of individual numbers guessed from the answer choices rather than using the formula (x + y) / (x * y). Others invert the final fraction by mistake or forget to simplify it. Remember that the sum of reciprocals of two numbers with known sum and product has a direct and simple expression, which avoids unnecessary work. Final Answer:The sum of the reciprocals of the two numbers is 11/120.

Question 3

The ratio of two numbers is 3 : 4 and their H.C.F. is 4. What is the L.C.M. of these two numbers?

Accepted Answer

Introduction / Context:This is a straightforward question that combines ratio and highest common factor (H.C.F.) to determine the actual numbers and then asks for their least common multiple (L.C.M.). It is a common pattern in aptitude tests because it connects basic ratio knowledge with H.C.F. and L.C.M. properties. Given Data / Assumptions: The ratio of two numbers is 3 : 4. Their H.C.F. is 4. We must find the L.C.M. of the two numbers. Concept / Approach:If two numbers are in the ratio 3 : 4, we can write them as 3k and 4k for some positive integer k. The H.C.F. of 3k and 4k is k times the H.C.F. of 3 and 4. Since 3 and 4 are co-prime, their H.C.F. is 1, so the overall H.C.F. is k. We are given that this H.C.F. is 4, so k = 4. Once we know the numbers, we can compute their L.C.M. using either prime factorization or the formula involving product and H.C.F. Step-by-Step Solution:Step 1: Let the two numbers be 3k and 4k.Step 2: The H.C.F. of 3k and 4k is k * H.C.F.(3, 4) = k * 1 = k.Step 3: Given H.C.F. = 4, so k = 4.Step 4: Therefore, the actual numbers are 3 * 4 = 12 and 4 * 4 = 16.Step 5: Now find L.C.M.(12, 16).Step 6: Prime factors: 12 = 2^2 * 3, 16 = 2^4.Step 7: Take highest powers: 2^4 and 3^1.Step 8: L.C.M. = 2^4 * 3 = 16 * 3 = 48. Verification / Alternative check:We can also use the relation L.C.M. = (product) / H.C.F.Product of numbers = 12 * 16 = 192.Given H.C.F. = 4, so L.C.M. = 192 / 4 = 48. Both methods give the same result, confirming that the L.C.M. is 48. Why Other Options Are Wrong:Option a (12) and option b (16) are the smaller factors and cannot be the least common multiple since they do not include all prime powers needed to cover both 12 and 16. Option c (24) is a common multiple but not the least one that covers the largest factor 16 multiplied by 3. Only 48 satisfies the condition of being divisible by both 12 and 16 and being the least such positive integer. Common Pitfalls:Some students mistakenly treat the ratio numbers (3 and 4) themselves as the original numbers or misinterpret H.C.F. in that context. Others miscompute the L.C.M. of 12 and 16 by not taking the highest power of 2. Always reconstruct the original numbers carefully from the ratio before calculating H.C.F. and L.C.M. Final Answer:The L.C.M. of the two numbers is 48.

Question 4

The H.C.F. and L.C.M. of two numbers are 11 and 385 respectively. If one of the numbers lies between 75 and 125, what is that number?

Accepted Answer

Introduction / Context:This question requires you to use the relation between H.C.F., L.C.M., and the original numbers to find specific values under a given range constraint. Such problems are standard in aptitude tests to evaluate your ability to connect number theoretic properties with inequality conditions. Given Data / Assumptions: H.C.F. of the two numbers is 11. L.C.M. of the two numbers is 385. One of the numbers lies between 75 and 125. We are asked to identify that number. Concept / Approach:Let the two numbers be 11a and 11b, where a and b are co-prime integers. Then their L.C.M. is 11ab. We are told the L.C.M. is 385, so 11ab = 385. From this we find ab. The possible co-prime pairs (a, b) that satisfy this equation lead to the actual numbers. Finally, we check which of these numbers lies within the given range 75 to 125. Step-by-Step Solution:Step 1: Let the numbers be 11a and 11b with H.C.F.(a, b) = 1.Step 2: L.C.M.(11a, 11b) = 11ab.Step 3: Given that L.C.M. = 385, so 11ab = 385.Step 4: Divide both sides by 11: ab = 385 / 11 = 35.Step 5: The factor pairs of 35 are (1, 35) and (5, 7), and their reverses.Step 6: Both pairs (1, 35) and (5, 7) have co-prime components.Step 7: Corresponding numbers are: For (1, 35) we get 11 and 385. For (5, 7) we get 55 and 77.Step 8: Among 11, 385, 55, and 77, the numbers lying between 75 and 125 are 77 only. Verification / Alternative check:Check 55 and 77 as a possible pair: H.C.F.(55, 77) = 11 and L.C.M. = (55 * 77) / 11 = 385. This confirms that 55 and 77 are valid numbers for the given H.C.F. and L.C.M. Only 77 lies in the required range, so it must be the answer. Why Other Options Are Wrong:Option b (88), option c (99), and option d (110) do not pair with any integer in such a way that H.C.F. becomes 11 and L.C.M. remains 385. They either produce a different product or lead to the wrong L.C.M. when combined with any compatible pair. A quick check using the formula L.C.M. = (product) / H.C.F. shows that they do not yield 385. Common Pitfalls:Students sometimes attempt trial and error directly on answer choices, which can be time-consuming and error-prone. Others forget that the numbers must be multiples of 11 because the H.C.F. is 11. Working with the simplified variables a and b and using ab = 35 provides a structured path to the solution. Final Answer:The required number between 75 and 125 is 77.

Question 5

Three numbers are in the ratio 1 : 2 : 3 and their H.C.F. is 12. What are the three numbers?

Accepted Answer

Introduction / Context:This is a straightforward ratio and H.C.F. question. It tests whether you can correctly scale a given ratio by an appropriate factor derived from the highest common factor to reconstruct the actual numbers. Such questions are common in basic arithmetic sections of aptitude exams. Given Data / Assumptions: The ratio of three numbers is 1 : 2 : 3. The H.C.F. of the three numbers is 12. We must determine the actual three numbers. Concept / Approach:If three numbers are in the ratio 1 : 2 : 3, we can represent them as k, 2k, and 3k for some positive integer k. The H.C.F. of k, 2k, and 3k is k times the H.C.F. of 1, 2, and 3. Since H.C.F.(1, 2, 3) is 1, the overall H.C.F. is k. We are told that the H.C.F. is 12, so we can directly identify k as 12 and then compute the three numbers. Step-by-Step Solution:Step 1: Let the three numbers be k, 2k, and 3k.Step 2: The H.C.F. of k, 2k, and 3k is H.C.F.(1, 2, 3) * k = 1 * k = k.Step 3: Given that the H.C.F. is 12, we have k = 12.Step 4: Substitute k = 12 into the expressions for the numbers.Step 5: First number = k = 12.Step 6: Second number = 2k = 24.Step 7: Third number = 3k = 36. Verification / Alternative check:Check the ratio: 12 : 24 : 36 simplifies to 1 : 2 : 3 when each term is divided by 12.Check the H.C.F.: H.C.F.(12, 24, 36) is 12, since 12 divides all three and no larger number does. Both conditions are satisfied, so the numbers 12, 24, 36 are correct. Why Other Options Are Wrong:Option a (4, 8, 12) has H.C.F. 4, not 12. Option b (5, 10, 15) has H.C.F. 5. Option c (10, 20, 30) has H.C.F. 10. None of these match the required H.C.F. of 12, even though they may share the same ratio pattern. Only option d satisfies both the ratio and H.C.F. conditions. Common Pitfalls:Students sometimes pick numbers that preserve the ratio but do not check the H.C.F. explicitly. It is essential to verify both given conditions. Another typical error is scaling by the wrong factor or miscomputing the H.C.F. Always verify the ratio and the H.C.F. at the end. Final Answer:The three numbers are 12, 24, 36.

Question 6

Find the H.C.F. of the three numbers 4 × 27 × 3125, 8 × 9 × 25 × 7 and 16 × 81 × 5 × 11 × 49.

Accepted Answer

Introduction / Context:This question assesses your skill in working with composite numbers given in factored form and finding their highest common factor (H.C.F.). It is a typical number theory problem where prime factorization plays a major role, and it helps reinforce the method of extracting common prime powers from several complex products. Given Data / Assumptions: First number: 4 × 27 × 3125. Second number: 8 × 9 × 25 × 7. Third number: 16 × 81 × 5 × 11 × 49. We must find the H.C.F. of these three numbers. Concept / Approach:The H.C.F. of a set of numbers is obtained by taking each prime factor that is common to all the numbers with the smallest exponent appearing in their prime factorizations. Since the numbers are given as products, our first step is to fully factor each component into primes. After that, we identify the primes common to all three numbers and multiply them with the lowest powers to obtain the H.C.F. Step-by-Step Solution:Step 1: Prime factorize each component.4 = 2^2, 27 = 3^3, 3125 = 5^5.Thus first number = 2^2 * 3^3 * 5^5.8 = 2^3, 9 = 3^2, 25 = 5^2, 7 = 7^1.Thus second number = 2^3 * 3^2 * 5^2 * 7.16 = 2^4, 81 = 3^4, 5 = 5^1, 11 = 11^1, 49 = 7^2.Thus third number = 2^4 * 3^4 * 5 * 7^2 * 11.Step 2: Identify primes common to all three numbers: 2, 3, and 5.Step 3: For prime 2, the powers are 2^2, 2^3, and 2^4, so smallest power is 2^2.Step 4: For prime 3, the powers are 3^3, 3^2, and 3^4, so smallest power is 3^2.Step 5: For prime 5, the powers are 5^5, 5^2, and 5^1, so smallest power is 5^1.Step 6: The H.C.F. is therefore 2^2 * 3^2 * 5.Step 7: Compute: 2^2 = 4, 3^2 = 9, so H.C.F. = 4 * 9 * 5 = 36 * 5 = 180. Verification / Alternative check:Check that 180 divides each number: 180 = 2^2 * 3^2 * 5.First number has 2^2, 3^3, 5^5 so it contains at least these prime powers.Second number has 2^3, 3^2, 5^2 so it also contains them.Third number has 2^4, 3^4, 5^1 so it contains them as well. No higher power of any prime is common to all three, so 180 is indeed the highest common factor. Why Other Options Are Wrong:Options b (360) and c (540) include higher powers of some primes, making them larger than the true H.C.F. and no longer dividing at least one of the numbers exactly. Option d (1260) includes extra prime factors (such as 7) with unnecessary powers and will not divide all three numbers without remainder. Only 180 satisfies the divisibility condition for all three and is maximal. Common Pitfalls:Students may overlook that only common primes with their smallest exponents contribute to the H.C.F. It is also easy to make arithmetic errors when expanding the products or to mix up H.C.F. with L.C.M. Staying organised with prime factorization and clearly tracking exponents prevents these mistakes. Final Answer:The H.C.F. of the three given numbers is 180.

Question 7

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, what is the greater of the two numbers?

Accepted Answer

Introduction / Context:This is a direct application of the relationship between the product of two numbers and their highest common factor (H.C.F.) when you are asked to determine the individual numbers. It is a standard type of question in quantitative aptitude, and knowing how to factor correctly is key. Given Data / Assumptions: The product of the two numbers is 4107. Their H.C.F. is 37. We must find the greater of the two numbers. Concept / Approach:If two numbers have H.C.F. 37, we can express them as 37a and 37b, where a and b are co-prime positive integers. Then their product is (37a) * (37b) = 37^2 * ab. Since the product is given, we can solve for ab and then find co-prime pairs (a, b) whose product matches this value. This will lead us to the actual numbers, from which we choose the larger. Step-by-Step Solution:Step 1: Let the numbers be 37a and 37b, with H.C.F.(a, b) = 1.Step 2: Product = 37^2 * ab.Step 3: Given product is 4107, so 37^2 * ab = 4107.Step 4: Calculate 37^2 = 1369.Step 5: So 1369 * ab = 4107.Step 6: Divide to find ab: ab = 4107 / 1369 = 3.Step 7: The co-prime factor pairs of 3 are (1, 3) and (3, 1).Step 8: The corresponding numbers are 37 * 1 = 37 and 37 * 3 = 111.Step 9: Of these, the greater number is 111. Verification / Alternative check:Check product: 37 * 111 = 4107, which matches the given product.Check H.C.F.: gcd(37, 111) is 37 because 111 = 3 * 37. Both conditions are satisfied, confirming that the numbers are 37 and 111 and the greater one is 111. Why Other Options Are Wrong:Options a (101), b (107), and d (185) either do not produce the correct product with a compatible partner or do not have H.C.F. 37 with any other integer such that the product is 4107. For example, 101 is prime and 4107 / 101 is not an integer, so it cannot be one of the numbers. Common Pitfalls:Students may forget to represent the numbers in the form 37a and 37b and instead try random factorizations of 4107. Others may miscalculate 37^2 or the division 4107 / 1369. Writing the steps cleanly and doing the arithmetic carefully makes this question very manageable. Final Answer:The greater of the two numbers is 111.

Question 8

The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, what is the other number?

Accepted Answer

Introduction / Context:This question uses the fundamental relationship between the product of two numbers, their highest common factor (H.C.F.), and their least common multiple (L.C.M.) to find an unknown number. It is a very standard pattern in aptitude exams and can be solved quickly if you remember the key identity. Given Data / Assumptions: H.C.F. of the two numbers is 11. L.C.M. of the two numbers is 7700. One of the numbers is 275. We must find the other number. Concept / Approach:For any two positive integers a and b, their product satisfies the identity a * b = H.C.F. * L.C.M. When we know one of the numbers and both the H.C.F. and L.C.M., we can find the other number by rearranging this identity. This avoids the need for any complicated factorization. Step-by-Step Solution:Step 1: Let the unknown number be x and the known number be 275.Step 2: Given H.C.F. = 11 and L.C.M. = 7700.Step 3: Use the relation 275 * x = H.C.F. * L.C.M. = 11 * 7700.Step 4: Compute the right side: 11 * 7700 = 84700.Step 5: So 275 * x = 84700.Step 6: Solve for x: x = 84700 / 275.Step 7: Perform the division: 84700 ÷ 275 = 308.Step 8: Therefore, the other number is 308. Verification / Alternative check:Check product: 275 * 308 = 84700.Check H.C.F. of 275 and 308: gcd(275, 308) = 11.Check L.C.M.: (product) / H.C.F. = 84700 / 11 = 7700. All conditions are met, so 308 is consistent with the given H.C.F. and L.C.M. Why Other Options Are Wrong:Options a (279), b (283), and d (318) do not satisfy the identity a * b = 11 * 7700 when paired with 275. Either their product with 275 is not 84700, or their greatest common divisor with 275 is not 11, so they cannot be correct answers. Common Pitfalls:Some students misremember the formula and incorrectly use H.C.F. + L.C.M. or some other operation instead of the product. Others make arithmetic mistakes while multiplying 11 by 7700 or when dividing by 275. Carefully applying the correct identity and doing the arithmetic step by step ensures an accurate result. Final Answer:The other number is 308.

Question 9

If the HCF (GCD) of two positive integers is 8, then which of the following values can NEVER be their LCM?  (Choose the one option that is impossible.)

Accepted Answer

Introduction / Context:This question tests the relationship between HCF (GCD) and LCM for two numbers. If two numbers have HCF = 8, both numbers must be multiples of 8. A key fact is: for numbers x and y, x * y = HCF(x,y) * LCM(x,y). Also, if we write the numbers in reduced form, the LCM becomes tightly constrained. Given Data / Assumptions: HCF of the two numbers is 8 We must decide which candidate LCM value is impossible Concept / Approach:Let the two numbers be 8a and 8b, where gcd(a,b) = 1 (after factoring out the HCF, the remaining parts are coprime). Then:LCM(8a, 8b) = 8 * a * bBecause a and b are integers, the LCM must be a multiple of 8. A necessary condition is that any LCM value must be divisible by 8. Step-by-Step Solution:Step 1: If HCF = 8, write the numbers as 8a and 8b with gcd(a,b) = 1.Step 2: LCM becomes 8ab, so the LCM must be divisible by 8.Step 3: Check each option for divisibility by 8:32 ÷ 8 = 4 (divisible)48 ÷ 8 = 6 (divisible)60 ÷ 8 is not an integer (not divisible)152 ÷ 8 = 19 (divisible)120 ÷ 8 = 15 (divisible)Step 4: The only value that is not a multiple of 8 is 60, so it can never be the LCM. Verification / Alternative check:Construct examples: (8, 32) gives HCF 8 and LCM 32; (8, 48) gives LCM 48; (8, 152) gives LCM 152; (8, 120) gives LCM 120. No pair of numbers with HCF 8 can have LCM 60, because 60 is not a multiple of 8. Why Other Options Are Wrong:32: Possible with numbers such as 8 and 32.48: Possible with numbers such as 8 and 48.152: Possible with numbers such as 8 and 152.120: Possible with numbers such as 8 and 120. Common Pitfalls:Forgetting that both numbers must be multiples of the HCF.Not checking divisibility of the candidate LCM by the HCF.Confusing the roles of HCF and LCM in the product formula. Final Answer:60

Question 10

Find the least positive integer N such that:  • When N is divided by 5, 6, 7, and 8, it leaves a remainder of 3 each time, and • When N is divided by 9, it leaves no remainder.  What is the value of N?

Accepted Answer

Introduction / Context:This problem involves modular arithmetic and least common multiples. When a number leaves the same remainder upon division by several divisors, we can convert that statement into one involving the LCM of those divisors. The added condition that the number is divisible by 9 allows us to select the correct multiple. Given Data / Assumptions: N mod 5 = 3 N mod 6 = 3 N mod 7 = 3 N mod 8 = 3 N mod 9 = 0 Concept / Approach:If N leaves remainder 3 when divided by 5, 6, 7, and 8, then N - 3 is divisible by all of these numbers. Therefore:N - 3 is a multiple of LCM(5, 6, 7, 8)So we can write:N = 840k + 3for some positive integer k, where 840 is the LCM. Then we use the condition N mod 9 = 0 to determine the smallest suitable k. Step-by-Step Solution:Step 1: Compute LCM(5, 6, 7, 8).5 = 5, 6 = 2 * 3, 7 = 7, 8 = 2^3.LCM = 2^3 * 3 * 5 * 7 = 8 * 3 * 5 * 7 = 840.Step 2: Write N as N = 840k + 3.Step 3: Apply divisibility by 9: 840k + 3 ≡ 0 (mod 9).840 mod 9 = 3, so 840k + 3 ≡ 3k + 3 (mod 9).Thus 3k + 3 ≡ 0 (mod 9) ⇒ 3(k + 1) ≡ 0 (mod 9).Divide by 3: k + 1 ≡ 0 (mod 3) ⇒ k ≡ 2 (mod 3).Step 4: The smallest positive k that satisfies this is k = 2.Step 5: Therefore N = 840 * 2 + 3 = 1683. Verification / Alternative check:Check N = 1683: 1683 - 3 = 1680, which is divisible by 5, 6, 7, and 8. Also, 1 + 6 + 8 + 3 = 18, which is divisible by 9, so 1683 is divisible by 9. No smaller positive k gives a valid N, so 1683 is the least such number. Why Other Options Are Wrong:1677: 1677 - 3 = 1674 is not divisible by 8.2523 and 3363: They can be written in the form 840k + 3, but correspond to larger k values, so they are not the least.8403: Much larger and not the minimum solution. Common Pitfalls:Forgetting to transform N to N - 3 before applying the LCM idea.Confusing LCM and HCF when dealing with common divisibility.Not enforcing the condition that N must be the smallest positive integer that works. Final Answer:1683

Question 11

A heap of coconuts is divided into groups of 2, 3, and 5, and each time exactly 1 coconut is left over.  What is the least possible number of coconuts in the heap?

Accepted Answer

Introduction / Context:This question is another example of using LCM with a constant remainder. When a number leaves the same remainder when divided by several divisors, subtracting that remainder from the number gives a value that is divisible by all of those divisors. We then look for the smallest such number. Given Data / Assumptions: N mod 2 = 1 N mod 3 = 1 N mod 5 = 1 N is the least positive integer satisfying these conditions Concept / Approach:If N leaves remainder 1 in each case, then N - 1 is divisible by 2, 3, and 5. Therefore:N - 1 is a multiple of LCM(2, 3, 5)So we can write:N = LCM(2, 3, 5) * k + 1for some positive integer k, and we choose the smallest k that gives a positive N. Step-by-Step Solution:Step 1: Compute LCM(2, 3, 5).LCM(2, 3, 5) = 2 * 3 * 5 = 30.Step 2: Express N as N = 30k + 1.Step 3: For the least positive N, take k = 1.Then N = 30 * 1 + 1 = 31. Verification / Alternative check:Verify N = 31: 31 mod 2 = 1, 31 mod 3 = 1 (since 30 is divisible by 3), and 31 mod 5 = 1 (since 30 is divisible by 5). Any smaller positive candidate would not satisfy all three congruences simultaneously. Why Other Options Are Wrong:27: 27 mod 3 = 0, so it does not leave remainder 1 when divided by 3.16: 16 mod 2 = 0, so it fails for divisor 2.63: 63 mod 3 = 0, so it fails for divisor 3.61: 61 leaves remainder 1 for 2, 3, and 5 but it is not the least, because 31 is smaller. Common Pitfalls:Using HCF instead of LCM when considering several divisors at once.Forgetting to subtract the remainder (1) before applying the LCM idea.Not paying attention to the requirement that the number must be the smallest possible. Final Answer:31

Question 12

Find the LCM (Least Common Multiple) of the three numbers 80, 85, and 90.  Choose the correct LCM value.

Accepted Answer

Introduction / Context:This problem checks your understanding of LCM using prime factorization. The LCM of several integers is the smallest positive integer that is divisible by each of them. The standard method is to break each number into prime factors and then take each prime to its highest power that appears in any of the numbers. Given Data / Assumptions: Numbers: 80, 85, 90 Goal: LCM(80, 85, 90) Concept / Approach:Using prime factorization, write each number in the form of prime powers. Then:LCM = product of all primes appearing in any number, each raised to its maximum exponentThis guarantees that the result is divisible by all the numbers and is as small as possible. Step-by-Step Solution:Step 1: Factorize 80.80 = 2^4 * 5Step 2: Factorize 85.85 = 5 * 17Step 3: Factorize 90.90 = 2 * 3^2 * 5Step 4: List the primes and pick maximum exponents:2: max power = 2^43: max power = 3^25: max power = 5^117: max power = 17^1Step 5: LCM = 2^4 * 3^2 * 5 * 17 = 16 * 9 * 5 * 17.16 * 9 = 144, 144 * 5 = 720, 720 * 17 = 12240. Verification / Alternative check:Check divisibility: 12240 ÷ 80 = 153, 12240 ÷ 85 = 144, 12240 ÷ 90 = 136. All quotients are integers, so 12240 is a common multiple. Because we used the prime-power method, this common multiple is guaranteed to be the least. Why Other Options Are Wrong:11440: Does not include 3^2, so it is not divisible by 90.12880: Contains extra factors and is larger than necessary, so it is not the least common multiple.15300: Also a common multiple but not the smallest one.10200: Not divisible by 85, because it lacks the factor 17. Common Pitfalls:Forgetting to include the prime 17 from 85 in the LCM.Using addition instead of multiplication when combining prime powers.Confusing LCM with HCF, which uses minimum exponents instead of maximum ones. Final Answer:12240

Question 13

A drink vendor has 368 liters of Maaza, 80 liters of Pepsi, and 144 liters of Sprite. He wants to pack them in cans such that:  • Each can contains the same volume (in liters), • No can contains a mix of two different drinks, and • All the drink is …

Accepted Answer

Introduction / Context:This problem combines the idea of HCF with an optimization goal. To minimize the number of cans, each can should have the maximum possible capacity that divides each drink quantity exactly. Once the can size is fixed, we simply divide each total volume by that size and add them up to get the least number of cans. Given Data / Assumptions: Maaza = 368 liters Pepsi = 80 liters Sprite = 144 liters All cans must be of equal capacity No mixing of drinks in the same can and no leftover liquid Concept / Approach:If the capacity of each can is x liters, then x must be a common divisor of 368, 80, and 144. To minimize the number of cans, x should be as large as possible, which means:x = HCF(368, 80, 144)Once we find x, we compute the total number of cans needed for each drink separately and then add them. Step-by-Step Solution:Step 1: Find gcd(368, 80).368 mod 80 = 48.80 mod 48 = 32.48 mod 32 = 16.32 mod 16 = 0, so gcd(368, 80) = 16.Step 2: Find gcd(16, 144).144 mod 16 = 0, so gcd(16, 144) = 16.Therefore, can capacity x = 16 liters.Step 3: Number of cans for each drink:Maaza cans = 368 / 16 = 23.Pepsi cans = 80 / 16 = 5.Sprite cans = 144 / 16 = 9.Step 4: Total cans = 23 + 5 + 9 = 37. Verification / Alternative check:If we choose any smaller can size that also divides all three quantities, such as 8 liters, the total number of cans would be 368/8 + 80/8 + 144/8 = 46 + 10 + 18 = 74, which is larger than 37. Since 16 is the greatest common divisor, it guarantees the least possible number of cans. Why Other Options Are Wrong:35: Would correspond to a different and smaller can size that does not divide all three quantities exactly.46 and 47: These totals would occur only if the can size were smaller than 16 liters, which is not optimal.53: Too large and inconsistent with the HCF-based calculation. Common Pitfalls:Using the LCM instead of the HCF when choosing can capacity.Forgetting that each drink must be packed separately with no mixing.Not summing the cans across all types of drinks when computing the final count. Final Answer:37

Question 14

Find the maximum possible number of students among whom 1001 pens and 910 pencils can be distributed such that:  • Each student gets the same number of pens, and • Each student gets the same number of pencils.  What is the greatest possible number o…

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Introduction / Context:This problem is about equal distribution of two different items. If the same number of pens and the same number of pencils must be given to each student with no remainder, then the number of students must divide both totals exactly. The largest such number is the HCF (GCD) of the two quantities. Given Data / Assumptions: Total pens = 1001 Total pencils = 910 Each student must receive the same integer number of pens Each student must receive the same integer number of pencils Concept / Approach:If the number of students is S, then S must divide 1001 and 910. Therefore:S = gcd(1001, 910)This gcd gives the maximum possible number of students for which a perfect distribution is possible. Step-by-Step Solution:Step 1: Compute gcd(1001, 910).1001 - 910 = 91, so 1001 mod 910 = 91.910 mod 91 = 0, because 91 * 10 = 910.Therefore, gcd(1001, 910) = 91.Step 2: So the maximum possible number of students is S = 91. Verification / Alternative check:For 91 students: pens per student = 1001 / 91 = 11 and pencils per student = 910 / 91 = 10. Both are integers, so the distribution works perfectly. Any number greater than 91 would fail to divide at least one of the quantities. Why Other Options Are Wrong:101: 910 is not divisible by 101.130: 1001 is not divisible by 130.910: 1001 is not divisible by 910.1001: 910 is not divisible by 1001. Common Pitfalls:Choosing the smaller of the two numbers or their difference without computing the gcd.Confusing HCF with LCM in equal distribution problems.Not checking divisibility for both items (pens and pencils) simultaneously. Final Answer:91

Question 15

Find the greatest number that can divide 43, 91, and 183 in such a way that the remainder left in each case is the same.  What is that greatest number?

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Introduction / Context:This question explores an important property of equal remainders. If several numbers leave the same remainder when divided by a positive integer d, then the differences between those numbers are divisible by d. We can therefore convert the problem into finding the HCF of these differences. Given Data / Assumptions: Numbers: 43, 91, 183 Each leaves the same remainder when divided by d We need the greatest possible value of d Concept / Approach:If each number leaves the same remainder r when divided by d, then:43 = d * q1 + r91 = d * q2 + r183 = d * q3 + rSubtracting these equations pairwise removes r, so d divides the differences:d divides (91 - 43), (183 - 91), and (183 - 43)Thus, d must be a common divisor of these differences, and the greatest such d is their HCF. Step-by-Step Solution:Step 1: Compute the differences.91 - 43 = 48183 - 91 = 92183 - 43 = 140Step 2: Find gcd(48, 92, 140).First, gcd(48, 92): 92 mod 48 = 44, 48 mod 44 = 4, 44 mod 4 = 0, so gcd(48, 92) = 4.Next, gcd(4, 140) = 4, since 140 mod 4 = 0.Step 3: Therefore, the greatest possible d is 4. Verification / Alternative check:Check with d = 4: 43 mod 4 = 3, 91 mod 4 = 3, and 183 mod 4 = 3. All three numbers leave the same remainder, so d = 4 is valid. Any larger value would need to divide all the differences 48, 92, and 140, but the HCF of those differences is 4, so no larger value can work. Why Other Options Are Wrong:7: 48 is not divisible by 7.9: 92 is not divisible by 9.11: 48 is not divisible by 11.13: 92 is not divisible by 13. Common Pitfalls:Trying to take the HCF of the original numbers instead of their differences.Not realizing that equal remainders imply divisibility of pairwise differences.Checking only one difference instead of verifying divisibility for all differences. Final Answer:4

Question 16

Find the LCD (Least Common Denominator) of 12 and 18.  (Note: For positive integers, the LCD is the same as the LCM of the denominators.)

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Introduction / Context:The LCD (Least Common Denominator) of two fractions with denominators 12 and 18 is simply the LCM of 12 and 18. The LCD is important when adding or comparing fractions, because it is the smallest number that both denominators divide into exactly. Given Data / Assumptions: Denominators: 12 and 18 LCD(12,18) = LCM(12,18) Concept / Approach:Use prime factorization to find the LCM. Express each number as a product of prime powers, then take the highest power of each prime that appears in either factorization:LCM = product of primes with maximum exponents Step-by-Step Solution:Step 1: Factorize 12.12 = 2^2 * 3Step 2: Factorize 18.18 = 2 * 3^2Step 3: Combine primes using maximum exponents:Prime 2: max exponent = 2^2Prime 3: max exponent = 3^2Step 4: LCM = 2^2 * 3^2 = 4 * 9 = 36. Verification / Alternative check:36 ÷ 12 = 3 and 36 ÷ 18 = 2, both integers, so 36 is a common multiple. Any smaller number, such as 18 or 24, fails to be divisible by both 12 and 18, so 36 is indeed the least common denominator. Why Other Options Are Wrong:18: Divisible by 18 but not by 12.42: Not divisible by 12.12: Not divisible by 18.72: Is a common multiple, but it is not the least common multiple. Common Pitfalls:Confusing LCD with HCF.Multiplying 12 and 18 directly without reducing by gcd, which gives a non-minimal common multiple.Missing the correct exponent when writing prime factors (for example, forgetting that 18 contains 3^2). Final Answer:36

Question 17

Among the following numbers, which one has the greatest number of positive divisors?  (Count all factors, including 1 and the number itself.)

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Introduction / Context:This question tests your understanding of how to count the number of divisors of a number using prime factorization. When a number is expressed as a product of prime powers, there is a formula that gives the count of its positive divisors. We apply this to each option and then compare the results. Given Data / Assumptions: Numbers to compare: 99, 101, 176, 182 We need the number with the most positive divisors Concept / Approach:If n is written as:n = p1^a * p2^b * p3^c * ...then the total number of positive divisors of n is:(a + 1)(b + 1)(c + 1)...We factor each option, compute the divisor count using this formula, and then choose the largest. Step-by-Step Solution:Step 1: Factorize 99.99 = 9 * 11 = 3^2 * 11^1.Divisors of 99 = (2 + 1)(1 + 1) = 3 * 2 = 6.Step 2: Factorize 101.101 is prime.Divisors of 101 = 2 (1 and 101).Step 3: Factorize 176.176 = 16 * 11 = 2^4 * 11^1.Divisors of 176 = (4 + 1)(1 + 1) = 5 * 2 = 10.Step 4: Factorize 182.182 = 2 * 91 = 2^1 * 7^1 * 13^1.Divisors of 182 = (1 + 1)(1 + 1)(1 + 1) = 2 * 2 * 2 = 8.Step 5: Compare divisor counts: 99 has 6, 101 has 2, 176 has 10, and 182 has 8. The maximum is 10 for the number 176. Verification / Alternative check:You can confirm by listing divisors of 176: 1, 2, 4, 8, 16, 11, 22, 44, 88, 176. There are exactly 10, which matches the formula. The other numbers have fewer divisors when listed explicitly. Why Other Options Are Wrong:99: Only 6 divisors, fewer than 176.101: Prime, so it has only 2 divisors.182: Has 8 divisors, again fewer than 176.120 (extra option): It has many divisors, but it is not among the original four choices we are asked to compare. Common Pitfalls:Assuming that the largest number must have the most divisors.Miscomputing prime factorization and exponents, which leads to incorrect divisor counts.Forgetting to add 1 to each exponent before multiplying them to get the total number of divisors. Final Answer:176

Question 18

Find the LCM (Least Common Multiple) of 54 and 60.  Choose the correct value of LCM(54, 60).

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Introduction / Context:This question tests LCM calculation for two numbers using prime factorization or the gcd method. The LCM is the smallest number that both inputs divide exactly. It is a fundamental concept in arithmetic and number theory, especially useful in scheduling and fraction problems. Given Data / Assumptions: Numbers: 54 and 60 Goal: LCM(54, 60) Concept / Approach:We use prime factorization. Express each number in prime powers, then:LCM = product of each prime raised to its highest exponent across the numbersAn alternative method for verification is the identity:LCM(x, y) * gcd(x, y) = x * y Step-by-Step Solution:Step 1: Factorize 54.54 = 2 * 27 = 2 * 3^3.Step 2: Factorize 60.60 = 2^2 * 3 * 5.Step 3: Take maximum exponents of each prime:2: max exponent = 2^23: max exponent = 3^35: max exponent = 5^1Step 4: LCM = 2^2 * 3^3 * 5 = 4 * 27 * 5 = 540. Verification / Alternative check:Compute gcd(54, 60). The gcd is 6. Then:LCM = (54 * 60) / gcd(54, 60) = 3240 / 6 = 540This matches the prime factorization method, confirming that 540 is correct. Why Other Options Are Wrong:600: 600 / 54 is not an integer, so 600 is not a common multiple of both numbers.360: 360 is not divisible by 54.270: Not divisible by 60.720: Although it is a common multiple, it is larger than the least common multiple, which is 540. Common Pitfalls:Using HCF instead of LCM when reading the question quickly.Multiplying the numbers directly without dividing by their gcd, which gives a non-minimal common multiple.Missing a prime factor or using a lower exponent rather than the highest exponent across the factorizations. Final Answer:540

Question 19

If the HCF (GCD) of 210 and 55 can be expressed in the form:  HCF(210, 55) = 210 × 5 + 55 × P  then what is the value of P?

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Introduction / Context:This question checks your understanding of Bézout's identity and the Euclidean algorithm. The HCF of two integers can always be written as a linear combination of those integers. Here, the HCF of 210 and 55 is expressed as 210 * 5 + 55 * P, and we need to determine P. Given Data / Assumptions: Numbers: 210 and 55 HCF(210, 55) is expressed as 210 * 5 + 55 * P We need to find P Concept / Approach:First, compute the HCF of 210 and 55 using the Euclidean algorithm. Then set that value equal to the expression 210 * 5 + 55 * P and solve for P. Note that P can be negative, which is completely acceptable in a linear combination representation. Step-by-Step Solution:Step 1: Find gcd(210, 55).210 mod 55 = 45.55 mod 45 = 10.45 mod 10 = 5.10 mod 5 = 0, so gcd(210, 55) = 5.Step 2: Use the given expression:HCF(210, 55) = 210 * 5 + 55 * P.So 5 = 210 * 5 + 55 * P.Step 3: Compute 210 * 5 = 1050.Thus 5 = 1050 + 55P.Step 4: Rearrange to solve for P: 55P = 5 - 1050 = -1045.P = -1045 / 55 = -19. Verification / Alternative check:Substitute P = -19 back into the expression: 210 * 5 + 55 * (-19) = 1050 - 1045 = 5, which matches the HCF of 210 and 55. This confirms that P = -19 is correct. Why Other Options Are Wrong:-23: 210 * 5 + 55 * (-23) = 1050 - 1265 = -215, not 5.16: 210 * 5 + 55 * 16 = 1050 + 880 = 1930, not 5.27: 210 * 5 + 55 * 27 = 1050 + 1485 = 2535, not 5.-9: 210 * 5 + 55 * (-9) = 1050 - 495 = 555, not 5. Common Pitfalls:Failing to compute the HCF correctly with the Euclidean algorithm.Forgetting that P can be negative, which is allowed in linear combinations.Making sign errors when rearranging the equation 5 = 1050 + 55P. Final Answer:-19

Question 20

The difference of two positive integers is 14. Their LCM is 441 and their HCF (GCD) is 7.  Find the two numbers.

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Introduction / Context:This problem combines several important facts: the relationship between two numbers and their HCF and LCM, the idea of factoring out the HCF, and the constraint on the difference between the two numbers. It is a classic type of HCF and LCM reasoning question. Given Data / Assumptions: Difference between the numbers = 14 HCF = 7 LCM = 441 Numbers are positive integers Concept / Approach:Let the two numbers be 7a and 7b because 7 is the HCF. Then gcd(a, b) = 1, which means a and b are coprime. The LCM of 7a and 7b is 7ab. We are told:LCM = 7ab = 441 ⇒ ab = 63Also, the difference is 14:|7a - 7b| = 14 ⇒ |a - b| = 2We must find a and b such that a * b = 63, |a - b| = 2, and gcd(a, b) = 1. Step-by-Step Solution:Step 1: From 7ab = 441, divide by 7: ab = 63.Step 2: From 7|a - b| = 14, divide by 7: |a - b| = 2.Step 3: List factor pairs of 63: (1, 63), (3, 21), (7, 9).Step 4: Check each pair for difference 2 and gcd = 1.(1, 63): difference 62, so not suitable.(3, 21): difference 18 and gcd(3, 21) = 3, so not suitable.(7, 9): difference 2 and gcd(7, 9) = 1, so this pair works.Step 5: Thus a = 7, b = 9 (or vice versa). The numbers are 7a = 49 and 7b = 63. Verification / Alternative check:Check HCF and LCM: gcd(49, 63) = 7, and LCM(49, 63) = (49 * 63) / 7 = 441. The difference is |63 - 49| = 14. All conditions match the given data. Why Other Options Are Wrong:48 and 64: HCF is 16, and LCM is not 441.46 and 62: HCF is 2, not 7.56 and 70: Difference is 14, but HCF is 14 and LCM is 280.42 and 56: HCF is 14, again not equal to 7, and the LCM does not match 441. Common Pitfalls:Not factoring out the HCF first, which makes it harder to use the LCM relation.Ignoring the requirement that a and b must be coprime after factoring out the HCF.Selecting factor pairs of 63 that do not satisfy the difference constraint of 2. Final Answer:49 and 63

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