Number puzzles by Michael Pousen in Hörzu - Guide to the solution
Written by Ingo in IQ , school and learning , tags: number puzzle
I love number puzzles by Michael Pousen. These appear in many years in the TV magazine Hörzu. This link takes you to his homepage, where each time you view / Reload the page a new version of this puzzle is generated at. The solution is also displayed there.
As many have certainly come at the sight of these puzzles in a panic, I would just like to explain this puzzle for one example of how to solve it. This example puzzle is indeed a very easy copy, but the principle is certainly clear. Here comes the example puzzle:
So, from the structure they are always the same: a total of nine two-digit numbers are up on mathematical operations linked horizontally and vertically. It is a symbol for each digit, ie there are 10 icons instead of the 10 digits 0 to 9 Task is, of course, find out what the symbol stands for the digit.
First, I look 'views on the mystery about it "and look for abnormalities. Such abnormalities are produced for example by the numbers 0 and 1. The zero can be found, for example, often by scanning the last digits of the additions or subtraction: Here you can see, for example, that the white circle (red) plus the thick rimmed circle (green) again gives the white circle (ie a + b = a). It is known that the thick rimmed circle must be zero.
So we replace the thick circle around the zero:
Let us continue to the last digits of each number (in this case the computational laws are directly applicable, as no transfers can be added for the following posts). The last line is that blue-green times is zero, as is green pink zero times.
In order for a Malaufgabe in the last place one is 0, there are blue and green, only the two possibilities (a) is an even number of times the 5 - or (b) any number times zero. Possibility (b leaves), since in this case there would have the symbol for the zero must stand, but this is not the case. So the only (a), where it ("just five" or "five times even") is still unclear. But as green pink times resulting zero, must be zero, the green, green, otherwise the figure would be just, but which both blue and pink that would be five, and both will now not work. So, green is the fifth If we replace this symbol everywhere by 5:
Let us now consider the quartered circle (green), who is somewhat common. As is: number in the top left of the middle number + top = number in the top right, which in turn + = middle right number five thousand-something, it follows from this that the green circle can only be a 1 or 2. If he namely 3 or greater, could not be below the 5xx0, for two times in the series which would be adding up three already six. So 1 or 2
Looking at the number below to be in the middle and on the second Middle row, we see that 2 is not possible, because 25 times 25 times would be about 6250 ("by" because I like here in Geteiltaufgaben always "turned around" look), which would be much too large, it might not be right below the 5xx0 stand. So, no two, say it is the first
ABB is now cracked the riddle almost 15 times 15 times results in a number of 2250-2385, as here (top center) the first two digits are different symbols, it must be just 2310:
First line: 1xxx plus 2310 with what must be three or four thousand (if his transfer), 3 is already gone, there is thus only 4 This is right above the white circle is 6, because 6 + 4 = 10 results (zero rear). The rest is easy, and it gives the following solution:
This example was pretty simple, as many numbers occurred in "verrätersichen" positions, and also lacked the 7 (there are usually almost always represent all digits). As I said, try the homepage of Michael Pousen there is always such a new puzzles and costs you nothing. In the beginning it will probably take a little longer, go with a little practice it becomes clear quickly. I need a few minutes up to fifteen minutes. Have fun tinkering!
Yes, I like this puzzle quite as happy with Gehinrjogging experience guarantee success (as opposed to patience as the number puzzle is always on).
I do find it (as a visual person) always nice to get a 10 × 10-field matrix on paper and rauszustreichen fall out possibilities. Often sinds only 9 × 9 squares, when the 0 at the first "scan" seem to recognize.
But I have no fixed schedule approach (otherwise it would only be stupid, too routine). I also tend to search for simple anomalies, and the rest will be gradually narrowed after the knockout principle.
You could also think of a computer program for the solution. However, for this is probably a simple try all possible solutions as quickly enough, so you would need no experts to implement strategies.