The two numbers a and B only contain 2 and 3 prime factors. The number a has 21 divisors and the number B has 10 divisors. Their greatest common divisor is 18. How many are these two numbers

The two numbers a and B only contain 2 and 3 prime factors. The number a has 21 divisors and the number B has 10 divisors. Their greatest common divisor is 18. How many are these two numbers

18=2×3×3
Both a and B contain at least one factor 2 and two factor 3
21=3×7
10=2×5
A: 2 factor 3, 6 factor 2
It is: 2 × 2 × 2 × 2 × 2 × 3 × 3 = 576
B: 1 factor 2, 5 factors 3
It is: 2 × 3 × 3 × 3 × 3 = 486

In natural numbers, the smallest number with both divisor 2 and divisor 3 is___ The smallest number with both divisor 2 and divisor 5 is___ The smallest number with both divisor 3 and divisor 5 is___ ．

2 × 3 = 6, 2 × 5 = 10, 3 × 5 = 15

A number is not only a multiple of 15, but also a divisor of 15. This number is (). A, 1 B, 3
A number is not only a multiple of 15, but also a divisor of 15. This number is (). A, 1 B, 3 C, 30 d, 45

No, it's 15

A and B both contain prime factors 2 and 7, and their maximum divisor is 98. It is known that a has 12 divisors and B has 8 divisors. What are the two numbers?

The problem is not perfect, unless it is stated that both a and B contain and only contain prime factors 2 and 7
98 = 2×7^2
It is shown that the two numbers a and B, except 2 and 7, do not contain any other common prime factors
When both numbers a and B contain and only contain prime factors 2 and 7
According to the formula of divisor
8 = 2×4 = (1 + 1)×(3 + 1)
B = 2 × 7 ^ 3 = 686
Then the number of times of factor 7 in a must appear and only appear twice
12 = (2 + 1) × (3 + 1) the factor 2 appears three times
A = 2 ^ 3 × 7 ^ 2 = 392
The numbers of a and B are 392 and 686 respectively

The greatest common factor of a and B is 5, and the least common multiple is 150. If a is 25, B is 25______ If B is 15, then a is 15______ ．

If the number a is 25, then the number B is 30; if the number B is 15, then the number a is 50. So the answer is: 30, 50

The number a is 3x5x7xa and the number B is 2x5x7xa. (1) if the group common factor of a and B is 70, find a. (2) if the least common multiple of a and B is 630, find a

Because: a = 3 * 5 * 7 * a, B = 2 * 5 * 7 * a (the greatest common factor is the product of all the common prime factors)
The greatest common factor of a and B is: 5 * 7 * a = 70
So a = 70 / 35 = 2
The least common multiple of a and B is all common prime factors multiplied by their own unique prime factors
The least common multiple of a and B is 5 * 7 * a * 2 * 3 = 630
So a = 630 / 210 = 3

In the complex plane, the quadrant of the corresponding point of the conjugate complex z = (1 + I) I is ()
A. First quadrant B. second quadrant C. third quadrant D. fourth quadrant

∵ z = (1 + I) I = I + I2 = - 1 + I, z = (1 + I) I conjugate complex. Z = - 1-I, z = (1 + I) I conjugate complex corresponding point (- 1, - 1), z = (1 + I) I conjugate complex corresponding point in the third quadrant

In the complex plane, the point corresponding to the conjugate complex of complex z = 2i1 + I (I is the imaginary unit) is at the second point______ Quadrant

The complex z = 2i1 + I = 2I (1 − I) (1 + I) (1 − I) = 1 + I, the conjugate complex of the complex Z. the point (1, - 1) corresponding to Z = 1-I is in the fourth quadrant

On the equation AX ^ 2 - X - A - 1 = 0 of X, there is only one root of real number

My name is Chen Xiaojiu,
If a = 0, then the equation AX & # 178; - x-a-1 = 0 is changed to - X-1 = 0, and x = - 1 is obtained. In this case, a = 0
If a ≠ 0, then △ = (- 1) &# 178; - 4A × 1 × (- A-1) = 0, the solution is a = - 0.5
So a = 0 or a = - 0.5

If x 1, x 2 are two real roots of the equation x 2 - (2k + 1) x + K 2 + 1 = 0, and x 1, x 2 are greater than 1. (1) find the value range of real number k; (2) find the value of K if x 1, x 2 = 12

(1) ∵ the two roots of equation X2 - (2k + 1) x + K2 + 1 = 0 are greater than 1, Let f (x) = X2 - (2k + 1) x + K2 + 1 ∵ Δ = 4k-3 ≥ 0, 2K + 12 > 1F (1) > 0, 34 ≤ K < 1 (2) ∵ x1x2 = 12, ∵ 2x1 = X2, ① X1 + x2 = 2K + 1, ② x1 · x2 = K2 + 1 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ③ substitute ① into ② ③ to get 3x1 = 2K + 1, 2x12 = K2 + 1 to get k = 7 or K = 1 (rounding off); so k = 7