# The Prince of Mathematicians It is said that when Carl Friedrich Gauss was nine years old, in math class the teacher collectively punished the students to add the numbers from 1 to 100, surely in the hope of having them entertained for a good while. But little Carl found the result in a matter of seconds: he realized that the 2 and 99, the 3 and the 98, the 4 and the 97... they added the same as 1 and 100, so, matching the hundred numbers in this way, to find the total sum you only had to multiply 101 x 50 x 5,050. And generalizing this method is easy to find the formula of the sum of the terms of an arithmetic progression (can you deduce it?).

To add up the 11 consecutive numbers of last week's problem, we can use the Gaussian method: (1985 + 1995) x 11/2, and we don't even have to do the operation to see that the result will be a multiple of 11, which makes it considerably easier to solve the problem, because it is easy to see that all the 44-digit numbers resulting from the reordering of the numbers from 1985 to 1995 will also be multiples of 11, so none of them can be prime.

We also talked last week about the divisibility criteria, generally easy to understand and apply. Since we can put any number, for example 324, in the form 324 x 300 + 24, and 100 is divisible by 4, it will suffice that the last two figures, 24 in this case, will be divisible by 4. Similarly, since 1,000 is divisible by 8, it is sufficient for the last three digits of a number to be divisible by 8 for the number to be

.

What condition must a number meet so that when reversing the order of its figures, the difference between the number and its invert is divisible by 9?

The criterion of divisibility by 3 is not so obvious, but it does not involve further difficulty. Following, by the example above:

324 x 3 x 100 + 2 x 10 + 4 x 3(99 + 1) + 2(9 + 1) + 4 x 3 x 99 + 2 x 9 + 3 + 2 + 4

And since 99 and 9 are divisible by 3, it will suffice that the sum 3 + 2 + 4 for 324 to be also. And since 99 and 9 are divisible by 9, the criterion is also applicable in this case. For a number to be divisible by 3 or 9, it is sufficient that the sum of its figures,

And curling the curl a little, what condition must a number meet so that, when reversing the order of its figures, the difference between the number and its invert is divisible by 9? (For example, 712 – 217 x 495 is divisible by 9).

### 'Princeps Mathematicorum'

Turning Gauss, his early discovery of arithmetic progressions was but the beginning of a dazzling career that earned him the nickname Prince of Mathematicians. To his fundamental contributions to number theory, it is worth adding his important achievements in the fields of algebra, analysis, differential geometry or statistics (just remember the famous Gaussian bell).

And speaking of Gaussians, and since in previous weeks we have talked about Ramsey's theory and the "happy ending problem", it is necessary to mention the excellent mathematical dissemination page Gaussians, managed by Miguel Angel Morales, where you will find, among many other things, a wide and enjoyable introduction to the aforementioned topics

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