N satisfies: n has exactly 144 different divisors; among all divisors of N, there are 10 continuous non-zero natural numbers Explanation! Thank you very much

This problem should be based on this idea. There are 10 divisors of continuous non-zero continuous natural numbers, so this number must contain at least 5 2, at least 3 3, at least 2 5 and at least 1 7 after factoring the prime factor
So let's first express it as 2 ^ 5 * 3 ^ 3 * 5 ^ 2 * 7 * n, and the following n represents other items
Now let's not consider N and see how many divisors are contained in 2 ^ 5 * 3 ^ 3 * 5 ^ 2 * 7. First, there are five divisors only containing factor 2, namely 2,4,8,16,32, and so on. There are three divisors only containing 3, two divisors only containing 5, and one divisor only containing 7, with a total of 11 divisors. Second, there are two divisors in 2,3,5,7, five divisors with 2,3, five divisors with 2,5, two divisors with 2, There are 5 * 1 divisors of 7, 3 * 2 divisors of 3,5, 3 * 1 divisors of 3,7, 2 * 1 divisors of 5,7, a total of 41 divisors; thirdly, there are 5 * 3 * 2 divisors of 2,3,5, 5 divisors of 2,3,5, 5 * 3 * 1 divisors of 2,3,7, 5 * 2 * 1 divisors of 2,5,7, 3 * 2 * 1 divisors of 3,5,7, a total of 61 divisors; finally, there are 4 divisors of 2,3,5,7, There are 5 * 3 * 2 * 1 = 30. Then all these divisors add up to 143, and this number must have divisor 1, so the total number is 144, which is consistent with the meaning of the title, so this number is 2 ^ 5 * 3 * 5 ^ 2 * 7 = 151200, and its 10 adjacent divisors are 1,2,3,4,5,6,7,8,9,10. We can find that 11 is not its divisor
thank you.

What is the smallest of nine natural numbers with different divisors?
When 9 = (8 + 1) × 1, the eighth power of 2 = 256; when 9 = 3 × 3 = (2 + 1) × (2 + 1), the square of 2 × 3 = 36. Therefore, among the nine natural numbers with different divisors, 36 is the smallest (36 < 256),

The total number of a divisor is the exponent of different prime factors after it is decomposed into prime factors and multiplied by 1
(for example, 24 = 3 * 2 ^ 3, then the total number of divisors is (1 + 1) (3 + 1) = 8 divisors
The number 9 = (8 + 1) × 1 only contains 8 identical prime factors, so it has 9 divisors. Taking 2 as the minimum, we can get the 8th power of 2 = 256
9 = 3 × 3 = (2 + 1) × (2 + 1) means that the number contains two different prime factors, both of which are 2. The minimum natural values are 2 and 3
That is, 2 * 2 * 3 * 3 = 36
By the way, we don't talk about divisors now. They are called factors in the book

There is a natural number whose sum of the two largest divisors is 9

1+8=9
2+7=9
3+6=9
4+5=9
For positive integers, there are four combinations of which the result of adding two numbers is 9, and the maximum divisor is itself

It is known that the coefficient with X term in the expansion of F (x) = (1 + 2x) m + (1 + 4x) n (m, n ∈ n *) is 36. The minimum coefficient with x 2 term in the expansion and the value of M, n are obtained

∵ f (x) = (1 + 2x) m + (1 + 4x) n the coefficients of x-term in the expansions of ∵ f (x) = (1 + 2x) m + (1 + 4x) n (m, n ∈ n *) are 36, ∵ m + 2n = 18, ∵ f (x) = (1 + 2x) m + (1 + 4x) n the coefficients of x-term in the expansions of ? f (x) = (2m + 4N) x, ? m + 2n = 18, ∵ f (x) = (1 + 2x) m + (1 + 4x) n the coefficients of x-term in the expansions of ? f (x) = (1 + 2x) m + (1 + 4x) m are t = c2m · 22 + C2n · 42 = 2m2-2m + 8n2- -2 (18-2n) + 8n2-8n = 16n2-148n + 612 = 16 (n2-374n + 1534), when n = 378, t is the minimum, but when n ∈ n *, ｜ n = 5, t is the minimum, that is, the coefficient of x2 term is the minimum, and the minimum is 272, then n = 5, M = 8

It is known that the sum of binomial coefficients in the expansion of (2x + 1) ^ n is equal to 2 ^ 20

The sum of binomial coefficients in the expansion of (2x + 1) ^ n is equal to 2 ^ 20
The binomial coefficient is 2 ^ n
therefore
2^n=2^20
We get n = 20

[it is known that x, y, Z A is a natural number, and X

Because x, y, Z, a are natural numbers, so 1 / x + 1 / y + 1 / Z

1 / x + 1 / y + 1 / z = 1

x=y=z=3
x=2,y=4,z=4
x=4,y=2,z=4
x=4,y=4,z=2

If the natural number x
A is an integer! X, y, Z are natural numbers

There must be few conditions in this problem, such as the limitation of A
Don't tell me to use a for XYZ
A is a constant, this constant can take any value, take it as a complex number, then XYZ has no solution
A is an integer, a = 1
x. Y, Z are 2,3,6 respectively

It is known that n =. 13xy45z can be divided by 792, then X= ，y= ，z=__ ．

792 = 8 × 9 × 11, 13xy45z can be divisible by 8, 9, 11: ① when it can be divisible by 8: 13xy45z = (13xy4 × 100 + 5Z) mod8 = 8, ∧ 5Z can be divisible by 8, ∧ z = 6, (13xy456), ② when it can be divisible by 9: 1 + 3 + 4 + 5 + 6 + X + y can be divisible by 9, ∧ x + y = 8 or x + y = 17, ③ when it can be divisible by 11: 1 + X + 4 + 6 = 3 + y + 5 or 1 + X + 4 + 6 = 3 + y + 5 (± 11), ∧ by X, y If x = y = 8, x = y + 8: x = 8, y = 0, (x + y = 17, y = x + 3: x = 7, y = 10 does not conform, give up), then 13xy45z = 1380456 = 792 × 1743

A three digit natural number ABC subtracts the sum of its digits to get a number of 74, where a = 74_ ，b= _ ．

From the question meaning: 100A + 10B + c-a-b-c, = (100-1) a + (10-1) B, = 99A + 9b, = 9 × (11a + b), so 100A + 10B + c-a-b-c can be divided by 9, let □ = x, because 100A + 10B + c-a-b-c = x + 7 + 4, so x + 7 + 4 is a multiple of 9, so x = 7; so 9 (11a + b) = 774, that is 11a + B = 86

N satisfies: n has exactly 144 different divisors; among all divisors of N, there are 10 continuous non-zero natural numbers Explanation! Thank you very much