Integral division of mathematical problems There are five numbers of 0, 1, 4, 7 and 9 to form different four digit numbers. If you arrange the four digit numbers that can be divided by 3 from small to large, what is the fifth one

Integral division of mathematical problems There are five numbers of 0, 1, 4, 7 and 9 to form different four digit numbers. If you arrange the four digit numbers that can be divided by 3 from small to large, what is the fifth one

1047 1074 1407 1470 1479 1497 1704 1740 1749 1794
The fifth is 1479

Characteristics of numbers divisible by 13

(1) The characteristics of 1 and 0 are as follows
1 is the divisor of any integer, that is, for any integer a, there is always 1|a
If 0 is a multiple of any non-zero integer, a ≠ 0 and a is an integer, then a | 0
(2) If the last bit of an integer is 0, 2, 4, 6 or 8, then the number can be divided by 2
(3) If the sum of the numbers of an integer can be divided by 3, then the integer can be divided by 3
(4) If the last two digits of an integer can be divided by 4, then the number can be divided by 4
(5) If the last bit of an integer is 0 or 5, the number can be divided by 5
(6) If an integer can be divided by 2 and 3, then the number can be divided by 6
(7) If the number of digits of an integer is truncated, and then 2 times of the number of digits is subtracted from the remaining number, if the difference is a multiple of 7, then the original number can be divided by 7. If the difference is too large or it is difficult to see whether it is a multiple of 7 by mental arithmetic, the above process of "truncation, multiplication, subtraction and difference checking" needs to be continued until it can be clearly judged, The process of judging whether 133 is a multiple of 7 is as follows: 13-3 × 2 = 7, so 133 is a multiple of 7; for example, the process of judging whether 6139 is a multiple of 7 is as follows: 613-9 × 2 = 595, 59-5 × 2 = 49, so 6139 is a multiple of 7, and so on
(8) If the last three digits of an integer can be divided by 8, then the number can be divided by 8
(9) If the number sum of an integer can be divided by 9, then the integer can be divided by 9
(10) If the last bit of an integer is 0, the number can be divided by 10
(11) If the difference between the sum of odd digits and the sum of even digits of an integer can be divided by 11, then the number can be divided by 11. The multiple test method of 11 can also be processed by the "tail cutting method" of check 7. The only difference in the process is that the multiple is not 2, but 1!
(12) If an integer can be divided by 3 and 4, then the number can be divided by 12
(13) If the number of one digit of an integer is truncated, and then four times of the number of one digit is added to the remaining number, if the difference is a multiple of 13, then the original number can be divided by 13. If the difference is too large or it is difficult to see whether it is a multiple of 13 by mental arithmetic, we need to continue the above process of "truncation, multiplication, addition and difference checking" until we can make a clear judgment
(14) If the number of one digit of an integer is truncated, and then five times of the number of one digit is subtracted from the remaining number, if the difference is a multiple of 17, then the original number can be divided by 17. If the difference is too large or it is difficult to see whether it is a multiple of 17 by mental arithmetic, we need to continue the above process of "truncation, multiplication, subtraction and difference checking" until we can make a clear judgment
(15) If the number of one digit of an integer is truncated, and then two times of the number of one digit is added to the remaining number, if the difference is a multiple of 19, then the original number can be divided by 19. If the difference is too large or it is difficult to see whether it is a multiple of 19 by mental arithmetic, we need to continue the above process of "truncation, multiplication, addition and difference checking" until we can make a clear judgment
(16) If the difference between the last three digits of an integer and the preceding number of 3 times can be divided by 17, then the number can be divided by 17
(17) If the difference between the last three digits of an integer and the preceding number of 7 times can be divided by 19, then the number can be divided by 19
(18) If the difference between the last four digits of an integer and the first five times of the number separated can be divided by 23 (or 29), then the number can be divided by 23

Characteristics of numbers divisible by 13

For reference:
(1) The characteristics of 1 and 0 are as follows
1 is the divisor of any integer, that is, for any integer a, there is always 1|a
If 0 is a multiple of any non-zero integer, a ≠ 0 and a is an integer, then a | 0
(2) If the last bit of an integer is 0, 2, 4, 6 or 8, then the number can be divided by 2
(3) If the sum of the numbers of an integer can be divided by 3, then the integer can be divided by 3
(4) If the last two digits of an integer can be divided by 4, then the number can be divided by 4
(5) If the last bit of an integer is 0 or 5, the number can be divided by 5
(6) If an integer can be divided by 2 and 3, then the number can be divided by 6
(7) If the number of digits of an integer is truncated, and then 2 times of the number of digits is subtracted from the remaining number, if the difference is a multiple of 7, then the original number can be divided by 7. If the difference is too large or it is difficult to see whether it is a multiple of 7 by mental arithmetic, the above process of "truncation, multiplication, subtraction and difference checking" needs to be continued until it can be clearly judged, The process of judging whether 133 is a multiple of 7 is as follows: 13-3 × 2 = 7, so 133 is a multiple of 7; for example, the process of judging whether 6139 is a multiple of 7 is as follows: 613-9 × 2 = 595, 59-5 × 2 = 49, so 6139 is a multiple of 7, and so on
(8) If the last three digits of an integer can be divided by 8, then the number can be divided by 8
(9) If the number sum of an integer can be divided by 9, then the integer can be divided by 9
(10) If the last bit of an integer is 0, the number can be divided by 10
(11) If the difference between the sum of odd digits and the sum of even digits of an integer can be divided by 11, then the number can be divided by 11. The multiple test method of 11 can also be processed by the "tail cutting method" of check 7. The only difference in the process is that the multiple is not 2, but 1!
(12) If an integer can be divided by 3 and 4, then the number can be divided by 12
(13) If the number of one digit of an integer is truncated, and then four times of the number of one digit is added to the remaining number, if the difference is a multiple of 13, then the original number can be divided by 13. If the difference is too large or it is difficult to see whether it is a multiple of 13 by mental arithmetic, we need to continue the above process of "truncation, multiplication, addition and difference checking" until we can make a clear judgment
(14) If the number of one digit of an integer is truncated, and then five times of the number of one digit is subtracted from the remaining number, if the difference is a multiple of 17, then the original number can be divided by 17. If the difference is too large or it is difficult to see whether it is a multiple of 17 by mental arithmetic, we need to continue the above process of "truncation, multiplication, subtraction and difference checking" until we can make a clear judgment
(15) If the number of one digit of an integer is truncated, and then two times of the number of one digit is added to the remaining number, if the difference is a multiple of 19, then the original number can be divided by 19. If the difference is too large or it is difficult to see whether it is a multiple of 19 by mental arithmetic, we need to continue the above process of "truncation, multiplication, addition and difference checking" until we can make a clear judgment
(16) If the difference between the last three digits of an integer and the preceding number of 3 times can be divided by 17, then the number can be divided by 17
(17) If the difference between the last three digits of an integer and the preceding number of 7 times can be divided by 19, then the number can be divided by 19
(18) If the difference between the last four digits of an integer and the first five times of the number separated can be divided by 23 (or 29), then the number can be divided by 23

In the gas mixture of 32 g O2 and O3, why does the content of oxygen atoms have to be 2 moles? If it is 32 g O2 and N2, can it be calculated?

Both oxygen and ozone are made up of oxygen atoms. There are only oxygen atoms and no other atoms. The relative atomic mass of oxygen atom is 18
32 / 18 = 2mol, but oxygen and nitrogen can't be calculated in this way. If we don't know the contents of two kinds of atoms, we can't calculate the number of moles of oxygen atoms

Given the vector M1 = (0, x) N1 = (1,1) m2 = (x, 0) N2 = (y ^ 2,1) (where x and y are real numbers), let m = M1 + radical 2n2 n = M2 - radical 2n1 and m parallel n
The locus of point P (x, y) is curve C
Let the intersection of the positive half axis of the Y-axis and the curve C be m, and make a straight line L through the point m to intersect the curve C at another point n. when |||||||= (4 times root 2) / 3, the equation of the straight line L is obtained

Vector M = (square of radical 2 & nbsp; y, radical + x) vector n = (x-radical 2 & nbsp;, - radical 2 & nbsp;) & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (x1, Y1) & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; vector & nbsp; & nbsp; X2, Y2) vector M / / vector n & nbsp; (vector n, vector M's x1y2 = x2y1) so the square of X + 2Y = 2. Set the point oblique equation m (0,1) l: Y-1 = KX, and combine it with the curve C. use the Veda theorem to find x1x2 and X1 + X2, then use the chord length formula to find Mn, and then Mn = | = (4 times root 2) / 3 & nbsp; to get the value of K

Given the direction vector of incident light (M1, N1, L1) and the coordinates of the intersection point with the reflecting surface (x, y, z), how to find the direction vector of reflected light (m2, N2, L2)?
The reflector is a paraboloid x ^ 2 + y ^ 2 = 4 * f * Z. M1, N1, L1, x, y, Z can be considered as known. It's better to get the expressions of M2, N2, L2, or can we list the equations?

x^2+y^2=4*f*z,
Let z = Z (x, y) be the parabolic equation,
p=dz/dx,q=dz/dy,
The tangent plane equation at points (x0, Y0, Z0) is z-z0 = P0 (x-x0) + Q0 (y-y0),
The normal vector is (P0, Q0, - 1)
The normal equation is (x-x0) / P0 = (y-y0) / Q0 = (z-z0) / (- 1)
The equation of incident light is (x-x) / M1 = (Y-Y) / N1 = (Z-Z) / L1
Among them, P0 = DZ / DX | (x0, Y0), Q0 = DZ / dy | (x0, Y0),
Take a point (x1, Y1, z1) on the incident light, and let its symmetry point about the normal be (X2, Y2, Z2)
The midpoint of the connecting line between point (x1, Y1, z1) and point (X2, Y2, Z2) ((x1 + x2) / 2, (Y1 + Y2) / 2, (z1 + Z2) / 2) is on the normal,
arcsinx-x
[(x1+x2)/2-x0]/p0=[(y1+y2)/2-y0]/q0=[(z1+z2)/2-z0]/(-1)(=t) .(1)
The line between point (x1, Y1, z1) and point (X2, Y2, Z2) is perpendicular to normal. The product of vector (x1-x2, y1-y2, z1-z2) and normal vector (P0, Q0, - 1) is 0,
(x1-x2)p0+(y1-y2)q0+(z1-z2)(-1)=0.(2).
The point coordinates (X2, Y2, Z2) are solved by (1) (2),
The direction vector of reflected light is (m2, N2, L2) = (x-x2, Y-Y2, z-z2)

There are two vectors: N1 (a, B, c) and N2 (C, D, e), then | N1 &; N2 | =?; | N1 | & _; | N2 | =?, use the given letter to indicate the answer?

ac+bd+ce

(m ^ 2 + n ^ 2) ^ 2 - (m ^ 2 + n ^ 2) - 6 = 0, find the value of M2 + N2
There needs to be a calculation process

A=m^2+n^2
A^2-A-6=(A-3)(A+2)=0
A=m^2+n^2=3

If / M + 1 / + (n-2) 2 = 0, then the root of the original equation (m2-3m-4) X2 - (N2 + 4) X-1 = 0 is

∵｜m+1｜≥0,(n-2)^2≥0
∴m+1=0,n-2=0
∴m=1,n=2
∴（1-3-4）x^2-(4+4）x-1=0
-6x^2-8x-1=0
∴x1=(-4-√10）/3,x2=(-4+√10）/3

When m is an integer, the equation (M2-1) x2-6 (3m-1) x + 72 = 0 has two unequal positive integer roots

∵ M2-1 ≠ 0 ∵ m ≠ ± 1 ∵ (?) = 36 (M-3) 2 ＞ 0 ∵ m ≠ 3 using the root formula, we can get: X1 = 6m − 1, X2 = 12m + 1 ∵ x1, X2 is a positive integer ∵ M-1 = 1, 2, 3, 6, m + 1 = 1, 2, 3, 4, 6, 12, the solution is m = 2. Then X1 = 6, X2 = 4