Number Theory, HCF and LCM

The number series follows the pattern where each term is obtained by multiplying the previous term by 3 and then subtracting 3. Specifically: 3 × 3 - 3 = 6, 6 × 3 - 3 = 15, 15 × 3 - 3 = 42, 42 × 3 - 3 = 123, and 123 × 3 - 3 = 366. Therefore, the missing number in the series is 42, which corresponds to option C.

To find the number to add to 11.138 to make it divisible by 77, first find the remainder when 11,138 is divided by 77. 11,138 ÷ 77 gives a remainder of 69. Since 77 - 69 = 8, adding 8 to 11,138 makes it divisible by 77 exactly.

XCIX in Roman numerals represents 99 because XC means 90 (100 - 10) and IX means 9 (10 - 1), so together they form 99.

The sequence consists mostly of perfect squares: 1 (1²), 4 (2²), 9 (3²), 16 (4²), and 36 (6²). The number 26 is not a perfect square, making it the odd one out.

The sequence given is 9, 16, x, 36. These numbers correspond to perfect squares of consecutive integers: 9 = 3², 16 = 4², x = 5², and 36 = 6². Therefore, the missing number x is 25 (5²). Since 25 corresponds to option D, the correct answer is D.

The number series follows the pattern where each term is obtained by multiplying the previous term by 3 and then subtracting 3. Specifically, aₙ₊₁ = 3 × aₙ - 3. Starting with 3: 3 × 3 - 3 = 6, 3 × 6 - 3 = 15, 3 × 15 - 3 = 42, 3 × 42 - 3 = 123, and 3 × 123 - 3 = 366. Therefore, the missing number in the series is 42 (option C).

The pattern in the series is that the differences between consecutive terms increase by 4 each time: 5-2=3, 12-5=7, 23-12=11, 38-23=15, 57-38=19. The next difference should be 23 (19 + 4), so the next number is 57 + 23 = 80. Therefore, the correct answer is C (80).

Let the four consecutive two-digit odd numbers be x, x+2, x+4, and x+6. Their sum is 4x + 12. When this sum is divided by 10, the quotient must be a perfect square. We want (4x + 12)/10 = k, where k is a perfect square integer. Thus, 4x + 12 = 10k. Since x is odd and two-digit, testing perfect squares for k shows that k can be 16 or 36. For k=16, sum = 160, so 4x + 12 = 160 => 4x = 148 => x = 37, giving numbers 37, 39, 41, 43. For k=36, sum = 360, so 4x + 12 = 360 => 4x = 348 => x = 87, giving numbers 87, 89, 91, 93. Among the options given (21, 25, 41, 67), only 41 appears in the valid set (37, 39, 41, 43). Therefore, option C (41) is the correct answer.

The least common multiple (LCM) of 70 and 80 is the smallest number divisible by both. The prime factors of 70 are 2, 5, and 7; for 80, they are 2^4 and 5. Taking the highest powers, LCM = 2^4 × 5 × 7 = 16 × 5 × 7 = 560.

The number 1 is the only number whose square (1² = 1) and cube (1³ = 1) are equal. Other numbers have different values for their square and cube.

Scientific notation expresses numbers as a product of a number between 1 and 10 and a power of ten, making option A correct.

Prime numbers are natural numbers greater than 1 that have no divisors other than 1 and themselves. This means they can only be divided evenly by 1 and the number itself, which matches option C.

Let the two consecutive positive integers be n and n+1. The sum of their squares is n^2 + (n+1)^2, and their product is n(n+1). According to the problem, n^2 + (n+1)^2 = n(n+1) + 91. Simplifying, we get 2n^2 + 2n + 1 = n^2 + n + 91, which reduces to n^2 + n - 90 = 0. Solving this quadratic equation gives n = 9 or n = -10 (discard negative). Thus, the integers are 9 and 10. Checking the original condition: 9^2 + 10^2 = 81 + 100 = 181; their product is 90; 181 - 90 = 91, which matches the problem statement. Therefore, the correct answer is A (9, 10).

Let the three consecutive even natural numbers be x, x+2, and x+4. Their squares sum to x^2 + (x+2)^2 + (x+4)^2 = 1460. Simplifying leads to 3x^2 + 12x + 20 = 1460, or 3x^2 + 12x - 1440 = 0. Dividing by 3 gives x^2 + 4x - 480 = 0. Solving this quadratic yields x = 20 (positive root). Thus, the numbers are 20, 22, and 24, which matches option B.

The prime factorization of 252 is 2 × 2 × 3 × 3 × 7, which can also be written as 2² × 3² × 7. This matches option A. The previously given answer B (2 × 2 × 2 × 3 × 7) is incorrect because it represents 168, not 252.

The greatest common divisor (GCD) of 80, 144, and 368 is 16. This means each can should contain 16 liters of a single drink to maximize the volume per can without mixing drinks. The number of cans required is calculated by dividing each drink quantity by 16 and summing the results: (80/16) + (144/16) + (368/16) = 5 + 9 + 23 = 37 cans. Therefore, the least number of cans required is 37, which corresponds to option B.

To find the G.C.D of decimal numbers, convert them to integers by multiplying by a power of 10 that removes the decimals. Here, multiply each by 100: 1.08 = 108, 0.36 = 36, 0.9 = 90. The G.C.D of 108, 36, and 90 is 18. Dividing back by 100 gives 0.18. Therefore, the G.C.D of 1.08, 0.36, and 0.9 is 0.18, which corresponds to option C.

The highest common factor (HCF) of 36 and 84 is 12, as 12 is the largest number that divides both 36 and 84 without leaving a remainder.

The question asks for the least number which, when doubled, is exactly divisible by 12, 18, 21, and 30. This means that twice the number must be a common multiple of these numbers. The least common multiple (LCM) of 12, 18, 21, and 30 is 1260. Therefore, 2 × number = 1260, so number = 1260 ÷ 2 = 630. Among the options given, 630 corresponds to option B, making it the correct answer.

Let the three co-prime numbers be a, b, and c. Given a*b = 551 and b*c = 1073. Since the numbers are co-prime in pairs, b must be 1 (the only common factor). Then a = 551 and c = 1073. The sum is 551 + 1 + 1073 = 1625, which is not among the options, so this suggests b is not 1. Instead, factorize 551 = 19 * 29 and 1073 = 29 * 37. Since the numbers are co-prime in pairs, the common factor between the two products is 29, so b = 29, a = 19, and c = 37. Sum = 19 + 29 + 37 = 85, which corresponds to option C.

Let the three numbers be 3x, 4x, and 5x. Since 3, 4, and 5 have no common factors other than 1, the HCF of the three numbers is x. The LCM of 3, 4, and 5 is 60, so the LCM of the three numbers is 60x. Given that the LCM is 2400, we have 60x = 2400, which gives x = 40. Therefore, the HCF of the three numbers is 40.

The number of plants leaves a remainder of 5 when divided by 6, 8, 10, and 12, but is divisible by 13. The least common multiple (LCM) of 6, 8, 10, and 12 is 120. The number of plants is therefore 120k + 5. Checking multiples of 120 plus 5 for divisibility by 13, 845 (120*7 + 5) is divisible by 13, so the number of plants is 845.

Given the LCM (495) and HCF (5) of two numbers and their sum (100), we can find the numbers using the relationships: product = HCF × LCM = 5 × 495 = 2475, and sum = 100. Let the numbers be a and b. Then a + b = 100 and ab = 2475. Using the formula for difference: (a - b)^2 = (a + b)^2 - 4ab = 100^2 - 4 × 2475 = 10000 - 9900 = 100, so a - b = 10. Therefore, the difference between the two numbers is 10, which corresponds to option A.

The numbers are in the ratio 1:2:3, so let the numbers be 12x, 24x, and 36x. Since their H.C.F is 12, the common factor x must be 1. Therefore, the numbers are 12, 24, and 36.

For two co-prime numbers, their H.C.F is 1. The product of two numbers equals the product of their H.C.F and L.C.M. Since H.C.F is 1, L.C.M equals the product, which is 117.