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[The Problem of the Week] Long sequence of addition and subtraction to solve Kakuro Tutorial

A new problem, quite simple if we approach it properly.

In a book has appeared rather long operation:

999 to 998 + 997-996 + 995-994 + ... + 5 to 4 + 3 - 2 + 1

words, this is going adding and subtracting numbers in decreasing sequence, from the 999 to 1, the odd numbers are added, the pairs are subtracted. Can you calculate this operation?

The solution, as always, below, so do not expect to see our deductions.

[This illustration matenavegábamos found while looking for images of sums, has been extracted from this page . As you can see, there is a table that presents all sums of two squares less than or equal to 100. What's so funny? Fermat realized that these sums could give composite numbers and prime numbers, but the premiums collected were those of the form 4n +1, that is, that, when divided by 4 gives the remainder 1 (with the exception of 2, that is prime, is the sum of two squares and does not give remainder 1 when divided by 4). In the table are appearing all the cousins \u200b\u200band none of 4n +1 primes 4n +3. This led to enunciate the so-called Fermat Theorem Christmas, a name sometimes given because it is in a letter to Marin Mersenne dated December 25, 1640]

Solution:

may seem silly, we will add to the sum the term zero, which does not alter the result, and wrote:

999 to 998 + 997-996 + 995-994 + ... + 5 - 4 + 3 - 2 + 1 - 0

This way we can match the numbers:

(999-998) + (997-996) + (995-994) + ... + (5 - 4) + (3 - 2) + (1 - 0)

Each pair gives the same result, 1, then we have a large sum of ones. How many around there? Given that the numbers ranging from 999 to 0, there are exactly 1000 numbers, and therefore there are 500 pairs. (The zero we have added to match pairs and counting easier). There are 500 pairs, each one gives 1, then the sum is 500 :

1 + 1 + 1 + ... [500 times] ... + 1 + 1 + 1 = 500.

Note: this problem, slightly modified, has been taken from the textbook 4 º ESO publisher Anaya.

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present this tutorial, which is a English translation of that comes as a flash animation page of the Japanese magazine Nikoli .
The numeric Kakuro is a hobby, family, sudoku. Kakuro

The numbers must be partitioned into sums of smaller numbers to be placed in the appropriate cells.

white cells must be filled with numbers from 1 to 9. For example, in the cells listed below, the numbers must add up to 5, and in principle can come in any order (could be, for example, 1, 4, 4 and 1, 2 and 3, 3 and 2).

listed below in the cells, the numbers must add up to 14.

numbers can not be repeated in consecutive cells. The following example may be correct:

But the following example it is not, it should not repeat numbers in the sum:

The following figure There are two numbers 1, but it is correct, because they are not in consecutive cells and not in the same amount:

Let's start solving the kakuro. Look at the sum 4 below on the right. To get 4 can only be done by adding 1 and 3, but do not know in what order:

But if you look at the 3 which is on the right, only can be obtained by adding 1 and 2, and numbers can be placed in two possible orders:

The common number is the sum of 4 and of 3 is 1, then 1 in the cell must be common to both:

Placing on 1 then you can fill cells missing:

continue with the further sum of 3 that is in the center. There are two possibilities:

But the two possibilities represented, is valid only on the left, because the right of the 2 would be repeated in the same row.

We can complete the sum 10. Just keep in mind that four boxes totaling 10 only allowed the numbers 1, 2, 3 and 4. As already in place on 1 and 2, just fill with 3 and 4 properly so there is no repetition in the columns.

Now let's look at another type of reasoning. 6 Look at the sum of two squares, center left, and the sum 14 in column to the left. Both sums have a common box.

6 The sum of two squares can be expressed in several ways: 1 and 5, 2 and 4. The same goes with 14, which can be broken down into 5 and 9, or at 6 and 8. But if the marked box is a number equal to or greater than 6, it would not be compatible with the sum 6, and if the box indicated the number was equal to or less than 4, then to complete the $ 14 would be 10 or older. Then the following two options are wrong:

The number of box must be marked, therefore, a 5, so that is compatible with the two sums.

Following this type of logical reasoning, you can complete the kakuro in the only way possible.

To solve the Kakuro is very useful to know the list of unique sums . For example, two cells or boxes, 3 can only be obtained with 1 and 2, and 4 with 1 and 3, in addition to 17 can only be obtained with 8 and 9, and 16 with 7 and 9. With three cells, 6 can only be obtained with 1, 2 and 3, 7 to 1, 2 and 4, in addition to 24 can only be obtained with 7, 8 and 9, and 23 with 6, 8 and 9. A complete list of all amounts only by the number of boxes is, for example in this direction .

This tutorial can be expanded more, but each person, with practice and developing their own logic resources to be able to go face Kakuro increasingly advanced level. One page recommended to play online kakuro is www.kakuro.com .

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[The Problem of the Week ] Counting decimal

The new boat tour of our school, the first issue of the week is as follows:

Some numbers may have many decimal places, even infinite. Consider the decimal number

0'012345670123456701234567 ...

If you look you will see that decimals are repeated in a succession very easy. Note that the first decimal place is 0, the second a 1, the third a 2, etc., And if you keep counting, the figure is, for example, in tenth place is a 1, and the figure is twentieth place is a 3.

Would you know how to tell which decimal is in the thousandths place?

We put the image shown below, and then put the solution.

[In the picture we see the Eye of Horus, divided into parts, each corresponding to a fraction. The Egyptians did not use our Hindu-Arabic positional system, and smaller amounts represent the unit used unit fractions, ie with a numerator equal to one. They were able to express any fractional amount adding unit fractions. The image and more information on Egyptian fractions, can be found on page corresponding wikipedia]

Solution:

It is clear that the recurring decimal eight eight, and also that the 7 is in all positions multiples of eight: the eighth decimal place is 7, the sixteenth decimal is 7, the twenty-fourth again becomes 7, etc..

Coincidentally, a thousand is a multiple of eight, then the decimal is in the thousandths place is 7 .

Note: the number of which is our problem is a pure periodic infinite decimal. Matenavegante as any apprentice should know, this number is an expression as a fraction. In this case the fraction is

This fraction is irreducible, because numerator and denominator have no common prime factors. Specifically, 1234567 = 127 ° 9721, and 99,999,999 = 3 • 3 • 11 · 73 ° 101 ° 137. Factorizations we have done with this calculator page. However, on that same page is another calculator passing periodic infinite decimal fraction , but Wrong.

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[The Problem of the Week] Comparing rectangles

resumed the pleasant task of proposing the problem of the week. The one we have today is actually the last to put the past year apprentices:

look at the two rectangles in Fig. Which of the two takes up more surface area, the ABCD or the BEFD? Explain your answer.

Of course, do not wait for our beloved public, the answer to the problem is below the illustration.

[This photograph mathematics is taken from the web's will . For now, I do not know where the photo was taken, but I would like to find out, is a good collection of cuboids in it, with their corresponding rectangles that the prospect has become simple parallelograms]

Solution:

First, if you look carefully at the figure, one might argue that ABCD is a rectangle, but BEFD not, because their angles are not straight. The chart did with the Windows Paint program or similar program, and in fact, I went to BEFD angles as straight as intended. But this detail has no bearing on the problem. Or we can draw correctly for BEFD is a rectangle, or we can assume that is a rectangle, or we can completely ignore the graphics, in any case the solution will be the same: surface ABCD is the same as BEFD .

The reasoning is simple: Both boxes have a triangle in common, the BCD. Then we need only prove that the remains of each area are equal, ie the area ABD is equal to the sum of the areas BEC + DCF.

This is easy, just draw a segment parallel to BE at point C to cut the segment BD at point G, as shown in the figure below:

BECG You can see that is a parallelogram, and therefore, because of parallel sides, the triangle equals the triangle BEC BCG. Similarly, GCFD is another parallelogram, and the triangles are equal GCD and DCF. So DCF = BEC + BCG + GCD, but the latter sum is equal to the triangle BCD.

Since ABCD is a rectangle, triangles BCD and ABD are equal, and hence we conclude that we needed to show: ABD = BEC + DCF.

Notes:

1) Geometrical reasoning that we used is a logical process that is being used in the geometry from the time of Euclid. Is independent of the measures of the sides, not connected with any algebraic operation, uses only geometric properties such as parallelism and similarity of triangles. Today is not usually teach this kind of reasoning to apprentices, and it is very difficult for them, motu proprio , what used to solve the problem. In fact, apprentices solved the problem by measuring the sides of the rectangles on the sheet where they were printed. Since the measures are never going to be accurate by the imperfections of the design and lack of precision of the rules used to measure the areas of rectangles leaving them different, and that's the answer I got. Not one of the apprentices found the correct answer.

2) I had a language problem in saying area, because I had entered the doubt that has been accepted recently that the area as area is a feminine noun. Thanks to the Royal English Academy I resolved: it has to say area, but with the indefinite article a , a normal ie area, but is not incorrect to say an area .

3) As has been several months since I proposed this problem and have been transferred to another vessel School, can not remember where I got it, but I think it was a textbook publisher Anaya.