I have a formula to do the calculation from 7 set of numbers to 4 digits, for example:
04, 05, 19, 21, 22, 31, 13 ====> 4382
These 7 sets of numbers are ranging from 01 to 45, each set will only exists once.
Calculation are as follow:
-
Sum up 1st to 6th number, and times with 2
(4 + 5 + 19 + 21 + 22 + 31) * 2 = 204
-
Take the result of step 1, minus 1st number and 6th number
204 – 4 – 31 = 169
-
Take the result of step 2, plus with the 7th number
169 + 13 = 182
- From here, take the last 2 digits of the result from step 3
82
- From here, take the last 2 digits of the result from step 3
-
Sum up the 4th and 5th number
21 + 22 = 43
- From here, take the last digit of the result from step 4
3
- From here, take the last digit of the result from step 4
-
Sum up the 2nd and 3rd number
5 + 19 =24
- From here, take the last digit of the result from step 5
4
- From here, take the last digit of the result from step 5
There you get the final result of 4382
If any result from any step above contains 0 (zero), take that as a value in one of the digit into the final result.
So anyone has any idea if I have "4382" with me, can you get back the 7 sets of numbers?
Not necessary must be exactly the same as I've written on top, as long as the 7 sets of numbers can produce the final result into "4382".
Really need a master to gimme some clues. Thanks.
1st to 6th number must be in a sequence from smallest to largest number in the range of 01 to 45. It's a game that comes out with this calculation, and I need to find out the way to reverse it.
Best Answer
Your problem sets out to solve the following system of modular equations:
The first equivalence comes from steps 1-3 above, the second from step 4 and the third from step 5. The basic math that you need to solve a system of equivalences is the Chinese Remainder Theorem.
This system will not have a unique solution. To see how redundant it is, count the possibilities for the sequence
(x1,x2,x3,x4,x5,x6,x7)
and(C,B,A)
separately. For the former,x1,x6,x7
are interchangeable as arex4,x5
andx2,x3
, so we count them separately. There are(45+3-1 choose 3)
ways to choosex1,x6,x7
each between01
and45
(remember to choose with repetition), and(45+2-1 choose 2)
ways to choose each of the other pairs. This gives a total ofways of choosing the
x
s. For the 4-digit sequence, there are10 000
possibilities. The number above is far larger, so certainly there is a lot of repetition.If you want to pin down just one
x
sequence that yields a givenCBA
sequence, then you can simplify the problem by choosing arbitrary values forx6,x7,x5,x3
and then using the CRT to solve forx1,x4,x2
(which may or may not exist).For example, suppose you take
Then you solve for the remaining
x
s by solving this systemThe first two equations are easy, though the solutions are not unique
Adding in the final equation to solve for
x1
and pin downx2,x4
giveswhich has many solutions, among them
x1 = 4, x2 = 5, x4 = 21
, but alsox1 = 44, x2 = 5, x4 = 1
as well as many others.