Note the set t = {0, 1, 2, 3, 4, 5, 6}, M = {A17 + A272 + a373 + a474 | AI ∈ T, I = 1, 2, 3, 4}, arrange the elements in m from large to small, then the number 2011 is () A. 17+172+073+174B. 17+072+673+574C. 17+172+073+074D. 17+072+673+674

Note the set t = {0, 1, 2, 3, 4, 5, 6}, M = {A17 + A272 + a373 + a474 | AI ∈ T, I = 1, 2, 3, 4}, arrange the elements in m from large to small, then the number 2011 is () A. 17+172+073+174B. 17+072+673+574C. 17+172+073+074D. 17+072+673+674

Because A17 + A272 + a373 + a474 = 174 (A1 × 73 + A2 × 72 + a3 × 7 + A4), the maximum value of the 7-ARY number in brackets is 6666, which is 2400 in decimal system. In decimal system, the number 2011 in descending order from 2400 is 2400-2010 = 390, which is 1065 in 7-ARY system. Divide this number by 74

Note the set t = {0,1,2,3,4,5,6}, M = {A1 ／ 7 + A2 ／ 7 ^ 2 + a3 ／ 7 ^ 3 + A4 ／ 7 ^ 4 ∣ AI ∈ T, I = 1,2,3,4}, arrange the elements in m from large to small, then the number 2005 is ()
A.5÷7+5÷7^2+6÷7^3+3÷7^4
B.5÷7+5÷7^2+6÷7^3+2÷7^4
C.1÷7+1÷7^2+0÷7^3+4÷7^4
D.1÷7+1÷7^2+0÷7^3+3÷7^4
In a simple way --

Option C
m1=a1+……
m2=a1‘+……
If A1 > A2, then M1 > m2; if A1 = A1 'and A2 > A2', then M1 > m2; and so on
Thus, if A1 = 1, then at least 5 * 7 ^ 3 numbers are greater than M1; when A1 = 1, A2 = 1, then at least 5 * 7 ^ 3 + 5 * 7 ^ 2 numbers are greater than M1; when A1 = 1, A2 = 1, A3 = 0, then at least 5 * 7 ^ 3 + 5 * 7 ^ 2 + 6 * 7 numbers are greater than M1; when A1 = 1, A2 = 1, A3 = 0, A4 = 4, then 5 * 7 ^ 3 + 5 * 7 ^ 2 + 6 * 7 + 2 numbers are greater than M1, that is, M1 is the number 5 * 7 ^ 3 + 5 * 7 ^ 2 + 6 * 7 + 3 = 2005
So A1 = 1, A2 = 1, A3 = 0, A4 = 4

The number of sets m satisfying m {A1, A2, A3, A4} and m ∩ {A1, A2, A3} = {A1 & # 8226; A2} is
（A）1 (B)2 (C)3 (D)4

Two m {A1, A2, A4} m {A1, A2}

A1 + A2 = 39, A3 + A4 = 33, then A5 + A6 =?

27

Let {A1, A2} &; a &; {A1, A2, A3, A4, A5} find the number of sets a satisfying the above conditions?
Why eight instead of nine. A = {A1, A2} doesn't count?

They are {A1, A2}, {A1, A2, A3}, {A1, A2, A4}, {A1, A2, A5}
{a1,a2,a3,a4},{a1,a2,a3,a5},{a1,a3,a4,a5}
{a1,a2,a3,a4,a5}

The number of all subsets of set {A1, A2, A3..., an}
Find the formula
Additional: the number of all subsets containing element A1 in this subset

1. The number of all subsets of the set {A1, A2, A3..., an} is 2 ^ n
The number of all subsets of this subset containing element A1 is equal to the number of subsets
The number of all subsets of set {A2, A3..., an} is 2 ^ (n-1)

The known set a = {A1, A2, A3 An} to find the sum of elements of all subsets of set a
I know a formula: (a1 + A2 + a3 +an)*[2^(n-1)]
How did you get here?

You can first analyze the situation of each element in yourself, taking A1 as an example
Its subset can be {A1} {A1, A2} {A1, A2 an｝
So A1 appears C (0) / (n-1) times in the subset of [1 element]
In the subset of [2 elements], C (1) / (n-1) times appear
……
In the subset of [n elements], C (n-1) / (n-1) times appear
So the sum of A1 is A1 [C (0) / (n-1) + C (1) / (n-1) + C(n-1)/(n-1)]
The same is true for other elements. For the sum of A2, A2 [C (0) / (n-1) + C (1) / (n-1) + C(n-1)/(n-1)]
……
On the sum of an [C (0) / (n-1) + C (1) / (n-1) +... " C(n-1)/(n-1)]
According to binomial theorem: [C (0) / (n-1) + C (1) / (n-1) + C(n-1)/(n-1)]=2^(n-1)
So add all the formulas together,
The sum of the elements of all subsets of set a
S=(a1+a2+…… an）×2^(n-1)

How many subsets, proper subsets, nonempty subsets and nonempty proper subsets are there in {A1, A2, A3, A4,. An}?

According to my understanding, subsets include empty sets, proper subsets and complete sets. Non empty proper subsets are even true subsets, while complete sets (including the set of all elements should not be classified as true subsets), so
N = 1, 2 subsets, 0 proper subsets, 1 nonempty subset and 0 nonempty proper subset
N = 2, 4 subsets, 2 proper subsets, 3 nonempty subsets and 2 nonempty proper subsets
N = 3, 8 subsets, 6 proper subsets, 7 nonempty subsets and 6 nonempty proper subsets
:
N = n, there are 2 ^ n subsets, 2 ^ n-2 proper subsets, 2 ^ n-1 nonempty subsets and 2 ^ n-2 nonempty proper subsets

Let s = {Ao, A1, A2, A3,), and define the operation on S@ ,Ai@Aj=Ak Where AK is the remainder of I + J divided by 4, I, j = O, 1,2,3, satisfying the relation（ x@X ）@The number of X (x belongs to s) of A2 = Ao is? The answer is 2. Please give the solution and what do I, j = O, 1,2,3 mean

If it is, please see the following: I, J are subscripts of AI, AJ, I, j = 0,1,2,3, which means that the values of I, J can be taken as 0,1,2,3 respectively, that is, I = 0 or 1 or 2 or 3; J = 0 or 1 or 2 or 3; solution ideas（ x@X...

Let s = {A0, A1, A2, A3}, define the operation on S
Is: AI AJ = AK, where k is the remainder of I + J divided by 4, I, j = 0,1,2,3, then the number of X (x ∈ s) satisfying the relation (x) A2 = A0 is

When x = A0, (x ♁ x) ♁ A2 = (A0 ♁ A0) ♁ A2 = A0 ♁ A2 = A2 ≠ A0
When x = A1, (x ♁ x) ♁ A2 = (A1 ♁ A1) ♁ A2 = A2 ♁ A2 = A4 = A0
When x = A2, (x ♁ x) ♁ A2 = (A2 ♁ A2) ♁ A2 = A0 ♁ A2 = A2 ≠ A0
When x = A3, (x ♁ x) ♁ A2 = (A3 ♁ A3) ♁ A2 = A2 ♁ A2 = A0 = A0
Then the number of X (x ∈ s) satisfying the relation (x ♁ x) ♁ A2 = A0 is 2