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#91
Britton, the ! in math is a factorial.
For example:
4! = fact(4) = 1*2*3*4=24
7! = fact(7) = 1*2*3*4*5*6*7=5040
Funny enough, don't ask me why, the 0! = fact(0) = 1. Trying to find an explanation I just Googled "Math 0!" and got this page: Math Forum: Ask Dr. Math FAQ: 0! = 1
You mean this:
You count first the formula within the first (blue) parenthesis 2 + 0 which gives 2, then using that sum you solve the formula within next (green) parenthesis using the sum from first parenthesis as a factor, 2 * 1 = 2.
Without (double) parenthesis it would be 2 + 0 x 1 + 3 = 5, so although the parenthesis are correct in this formula from Colin, they are this time unnecessary.
An example where using parenthesis would give a different outcome:
2 * 0 + 1 + 3 = 4
2 * (0 + 1) + 3 = 5
No special skills needed, just addition:
A three by three table:
And 9 prime numbers: 7, 37, 43, 67, 73, 79, 103, 109, 139
Place the given prime numbers so that each horizontal, vertical and diagonal line gives the exactly same sum. For instance if you think that the first row should be 73 + 103 + 7 = 183, then the two other horizontal lines, all three vertical lines and both diagonal lines should also give 183.
If this is too difficult, I will tell what the sum of all 8 lines should be :).
solution:
How did I arrive at the solution?Code:103 7 109 79 73 67 37 139 43
I don't know if this works in every case, but I used some logic as follows.
- given: 3,37,43,67,73,79,103,109,139
- the sum of the numbers is 657
- for the sum of each row or column to be equal, the sum has to be 657/3=219
- because there are 9 squares, I decided to divide the total by 9 and see what comes up: 657/9=73
- looking at the sequence, I found 73 to be the median --> why not try that in the center square?
- so 219-73=146 --> is there a pattern?
- the sums of the outer pair, second from outer pair, etc. all are 146
- this provides the following pairs: 7,139; 37,109; 43,103; 67,79
- because a center out perspective is working so far, let's examine using the two center pairs above (37,109; 43,103) for the diagonals
- placing the pairs in diagonal opposite corners provides the same sums as calculatd earlier for the rows and columns
- remaining squares are just a matter of simple math to provide a column or row sum of 219
- magically, all of these solutions used one of the remaining numbers given for the puzzle
I was posting my solution while you were providing the hint. I did not peak.