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In honor of the anniversary of this column, the Internet kindly provided me with a Singaporean mathematical task that had become viral . In mid-May, the web was agitated by a task that Singapore’s first-graders supposedly give to solve, and these are children aged 5 to 7 years old, and which turned out to be so complex that no one can solve it.

But our story is really about how phrases like “math problem shook the Internet” became boring and predictable attempts to attract visitors to the page. Since even a brief look at this question, which first appeared on the Singapore Technical Forum , allows us to say that this photo is an obvious fake. The photo looks edited, but there is no explanation for the problem.

Apparently, the task was taken (and modified) from the site devoted to mathematical puzzles, which is led by Gordon Burgin, an American retired teacher. And in the version from the site in the lower left quarter the figure is 20. In Singapore photo 0 is smeared. No wonder there is no obvious solution!

“I am amazed at this fake and I don’t know what they were trying to achieve with this,” Burgin says. “If their goal was a heated discussion and subsequent despair, then they achieved their goal!”

Next comes the correct version of the puzzle.

This riddle reminded me of a puzzle from one of the most interesting books on the history of riddles: " Wakoku chiekurabe ", the oldest Japanese book of riddles, published in 1727. This is a wonderful mystery - and, at least, she was born in the East!

These are four sums, each of which consists of five items, and all sums are the same. Here is a picture from the book itself, which shows a similar task.

If we call the positions for the numbers North, East, West, and South, then the candidates for the West and the South will be 8 and 9, since Yu + W + 3 = 20, or Yu + W = 17. But we know that Yu + B + 3 = 12, or Yu + B = 9. But Yu cannot be 9, because then B = 0, and this is prohibited. Therefore, ω = 8, 3 = 9, C = 6, and B = 1.

2. I liked most of all the assignment sent by a reader named Tom Flannery, because it turned out to be very simple.

Fill the circles with integers so that the sum of the sectors in each of the semicircles equals the sum of the numbers in the circles.

3. And the last task, from the Japanese book of riddles of 1727.

There may be many solutions. Here is my solution:

The path to the solution helps to make a guess - in the center there can be only an odd number. Having chosen such a number, it is necessary to divide the remaining digits into pairs, which together give the same number, and arrange them in opposite circles. I chose 1 for the center, respectively, there are pairs 2 and 9, 3 and 8, 4 and 7, 5 and 6 - they all add up to 11.

Source: https://habr.com/ru/post/404559/