Ehrlich

1. The matrix is not a square matrix, its length is 10¹² and its width is 12.

2. In my solution i HAD to count those zeroes (in the second one. In the first one i just used them to show we can look at it as a set of combinations)

3. You keep claimimg that the zeroes i added are exactly one half of the digits. Your calculation is wrong, because you are assuming A TON of things you think are right. I calculated it, and the number of added zeroes is less than 1/100 of all of the digits (after the zeroes were added)

What do you NOT understand from my solution? tell me and ill explain it further

We know that cos(A)=sin(90-A)

We also know that if sin(A)=sin(B) then B=A+360K or B=180-A+360K (think about it)

sin(x+37)=cos(2x+8)=sin(90-8-2x)=sin(82-2x)

FIRST OPTION:

x+37=82-2x+360K   / add 2x, and subtract 37

3x=45+360K    / divide by 3

x=15+120K

SECOND OPTION:

x+37=180-(82-2x)+360K=98+2x+360K   / subtract (37+2x)

-x=61+360K   / divide by -1

x=-360K-61

Half zeroes? could you prove that? And How did you infer from that that there are 6*10¹¹ threes? it doesnt make sense. I know alot of the digits are zeroes, but it doesnt change anything.

Please post your proof.

Here's a nice trick:

There's a way to find every single rational root for some of the polynomials:

step one:

we have the polynomial a₀+a₁*x+a₂*x²+a₃*x³+........+a_n*xⁿ=0. We need to multiply the polynomial by a number in a way that all coefficients will be whole numbers. Sometimes we cant do it. sometimes we can.

step two:

find all of the whole divisors of a_n and a₀ (including the negative divisors). lets call the set of the divisors of a₀ A and the set of the divisors of a_n B. there are k divisors in A and m divisors in B.

Every rational root of the polynomial=A_i/B_j, 0 i is a member of A and B _j is a member of B, meaning every rational root can be expressed using one divisor from A divided by one divisor from B).

step three:

Check for every two divisors (one from A and one from B) if A_i/B_j is a root. after a finite amount of time (unless your calculator is broken) you'll have every rational root of the polynomial.

ok. Lets count together:

1

2

3 NUMBER OF 3's=1

4

5

.

.

.

13 NUMBER OF 3's=1

.

.

23 NUMBER OF 3's=1

.

.

.

30 NUMBER OF 3's=1

31 NUMBER OF 3's=1

32 NUMBER OF 3's=1

33 NUMBER OF 3's=2

34 NUMBER OF 3's=1

35 NUMBER OF 3's=1

36 NUMBER OF 3's=1

37 NUMBER OF 3's=1

38 NUMBER OF 3's=1

39 NUMBER OF 3's=1

.

.

.

43 NUMBER OF 3's=1

.

.

.

53 NUMBER OF 3's=1

.

.

.

63 NUMBER OF 3's=1

.

.

.

73 NUMBER OF 3's=1

.

.

.

83 NUMBER OF 3's=1

.

.

.

93  NUMBER OF 3's=1

1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2=20

But by that he meant he wants to know the number of times the digit "3" appears. Lets test MY theory, but except for the number of 3's between 1 and 10¹², the number of 3's between 1 and 100. when it was between 1 and 10¹² the answer was 10¹¹*12, and between 1 and 10² the answer will be 10¹*2=20. You can count the 3's, the anwer will be 20 (The number of numbers containing 3 as a digit will be 10²-9²=19).

Heres another way to think about it: Every number is supposed to appear an equal number of times (if we count the zeroes after that number, the way i suggested writing the numbers in my first comment) . imagine we're counting the number of times ANY digit appears. every number has 12 digits on total, meaning our answer for the number of digits will be 12*10¹². Every digit appears an equal amount of times, meaning we have to divide by 10 (the number of digits) to get 12*10¹¹, the number of times ANY of the digits appear.

no no no. You found the number of numbers between 1 and 10¹² that contain 1 or more 3's. he asked to find the number of 3's.

For my solution, 1=000000000001, 2=000000000002 and so on (we will write every number's digits up to the 12^nd digit even if that digit is 0)

the numbers between 1 and 10¹² are all the combination of 12 digits (10 posibilities for every digit because there are 10 digits) except for the combination 000000000000 and including the 13 digits combination 1000000000000. But both of these combinations dont contain any 3, so for us they are the same. so the question is "how many 3's are there between 0 and 10¹²-1.

We have exactly 10¹² combinations. Lets look at the digit in the n^th place (the units digit is in the first place, the tens digit is in the second place and so on). there are exactly 1/10*(the number of combinations=10¹²) where that digit is 3 (because there are 10¹¹ combinations for the other 11 digits, and if we want it to have a 3 as its digit in the n place we need to put 3 as the digit in that place) meaning there are 10¹¹ numbers between 0 and 10¹²-1 that contain 3 as their digit in the n place. there are 12 places, so we repeat this 12 times (we count the times 3 appears as the first digit, then count the times 3 appears as the second digit.... and so on until the 12^nd digit). So, there are 12*10¹¹ 3's between 0 and 10¹²-1, meaning there are 12*10¹¹ between 1 and 10¹².

You forgot a solution. If cos(x)=1/2 then sin(x)=(3/4)^1/2 OR sin(x)=-(3/4)^1/2

That means tan(x)=-(3^1/2) OR tan(x)=3^1/2