The task of Alexander Ivanovich Koreyko

Alexander Ivanovich Koreiko, one of the smallest servants of Hercules, was a man in the last attack of youth, he was 38 years old. On the red wax face were yellow wheat eyebrows and white eyes. English antennae color even looked like ripe grass. His face would have seemed quite young, if not for the gross corporal folds that crossed his cheeks and neck. At the service of Alexander Ivanovich, he behaved like a long-term soldier: he did not reason, he was executive, hardworking, searching and stupid.

“He’s some timid one,” said the chief of the financial account, “some too humble, a devotee too. Only announce a loan subscription, as he already climbs with his monthly salary. The first is signed. And the whole salary is 46 rubles. I would like to know how it exists with this money.

Alexander Ivanovich was an amazing feature. He instantly multiplied and divided large three-digit and four-digit numbers in his mind. But this did not free Alexander Ivanovich from the reputation of a stupid guy.
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- Listen, Alexander Ivanovich, - the neighbor asked, - how many will be 836 to 423?
(The Golden Calf, Ilya Ilf, Yevgeny Petrov)

Rereading the novel The Golden Calf, I decided to repeat the calculations of the underground millionaire. I do not have the ability to multiply large numbers in my mind and I do not set myself the task of such an ability to develop. My goal is more modest - to find a way to quickly calculate the above example. As a tool, I decided to use a collection of beautiful works that I collected on the way to work and stored in a cell phone. Here are some of them:

• 13 * 23 = 299
• 17 * 47 = 799
• 7 * 11 * 13 = 1001

The first thing I did was find the factorization:

• 423 = 3 * 3 * 47
• 836 = 2 * 2 * 11 * 19

Regarding the factorization. How to do it quickly. We take the first number 423, the sum of the digits is divided by 9, which means the number is also divided by 9. To divide use the fast algorithm, which is given below. 423/9 = 46, (9) = 47.
The second number is 836. The first divisor is almost obvious - it is 4. 836 = 4 * 209. 209 is divided by 11, according to the appropriate sign of divisibility. The method of dividing by 11 almost repeats, the method we used to divide by 9,209 / 11 = 2 (-2), 11 (-11) = 18.9999 .. = 19.

Now we will rewrite the original example as follows:

423 * 836 = 4 * 99 * 47 * (17 + 2) = 4 * 99 * 799 + 4 * 99 * 94, here I used the beautiful piece 17 * 47 = 799.

4 * 99 * 799 = 4 * (100-1) * (800-1) = 4 * (80000-900 + 1) = 320 000-3 600 + 4

4 * 99 * 94 = 376 * (100-1) = 37 600-376

It remains to add the resulting expressions:

320 000 + (37 600-3 600) + (4-376) = 320 000 + 34 000 - 372 = 354 000 - 372 = 353 628

- 353.628, - answered Koreiko, after a moment's pause. And the neighbor did not check the result of multiplication, for he knew that the blunt Koreiko never makes mistakes.

Our result coincides with Koreiko's answer. Of course, in the mind, the above calculations are most likely impossible to perform. I was helped by a pencil and paper. The question of how to calculate an example orally remains open.

Waiting for advice.

After reading the comments received, I must admit that I did not find what I was looking for. Apparently, misrepresented his thought. In the beautiful book “You are, of course, joking, Mr. Feynman!” There is a chapter called “Happy Numbers”. In it, Richard Feynman talks about his collection of methods and numbers for quick calculations.

I remembered the meanings of several logarithms and began to notice what was going on. For example, if someone asks: “What is 28 squared?”, You notice that the square root of two is 1, 4, and 28 is 20 times 1, 4, so 28 in the square should be approximately equal to 400, multiplied by 2, or 800.
If someone asks how much it will turn out, if you divide 1 by 1, 73, then you can immediately answer that 0, 577, because you know that 1, 73 is almost the square root of 3, therefore 1/1, 73 is one third of the square root of 3. And if you want to determine the ratio 1/1, 75, it is equal to the value of the reciprocal fraction 7/4, and you remember that if the denominator is 7, then the decimal digits are repeated: 0, 571428 ...

I was hoping that someone has his collection of lucky numbers, from which the Koreiko number is derived.
By the way, lucky numbers are found in the most unexpected places. In the same chapter, Feynman has this episode.

This very worried the Japanese, because he clearly knew how to perform arithmetic operations with the help of an account, and here he was almost defeated by a restaurant visitor.
“Raios cubicos!” He says vengefully. Cubic roots! He wants to take cubic roots with the help of arithmetic! It is difficult to find a more complex fundamental problem in arithmetic. It must have been his trick in billing exercises.
He writes a number on the paper — any large number — I still remember him: 1729, 03. He starts working with this number and at the same time mumbles and grumbles: “Bu-bu-bu-hm-hm-bu- boo, ”he works like a demon! It just sinks into this cube root!
In the meantime, I'm just sitting in my place.
One of the waiters says: “What are you doing?”
I point to the head. “I think!” I say. Then I write on paper 12. Even after a while - 12, 002.
A man with bills wipes sweat from his forehead and says: "Twelve!"
"Oh no! - I object. - More numbers! More numbers! ”I know that when using arithmetic you take a cube root, then each subsequent figure requires more work than the previous one. This is not an easy job.
He again goes to work and at the same time mumbles: "Uv-fyr-hm-uh-hm-hm ...". I add two more digits. Finally, he raises his head and says: "12, 0!"
The waiters just glow with happiness. They say to the Japanese: “Look! He does it in his mind, and you need bills! And he has more numbers! ”
He was completely exhausted and left, defeated and humiliated. The waiters congratulated each other.
How did the visitor beat the scores? The number was 1729, 03. I accidentally knew that in a cubic foot of 1728 cubic inches, so it was clear that the answer is a little over 12. The excess, equal to 1, 03, is just one part of almost 2000, and during of the calculus course, I remembered that for small fractions the surplus of a cubic root is one third of the surplus of a number. So I only had to find the fraction 1/1728, then multiply the result by 4 (divide by 3 and multiply by 12). That's how I managed to get a whole bunch of numbers.

But the number 1729 has an interesting property. Once Godfrey Hardy arrived to visit Srinivas Ramanujan’s hospital. Hardy arrived by taxi with the number 1729 and during the conversation with Ramanujan noticed that, they say, a surprisingly boring number came to him as a number. To this the Indian answered that this is not true - 1729 is the minimum natural number, representable as a sum of cubes of two natural numbers in two different ways. Really,

1729 = 1 ^ 3 + 12 ^ 3 = 9 ^ 3 + 10 ^ 3.

Isn't it an amazing coincidence?

However, I did not give up trying to find a quick solution for the example of Koreiko. I liked this method, although it is not perfect.

836 * 423 = (840-4) * (420 + 2) +836

The product (840-4) * (420 + 2) is analogous to the known formula (ab) * (a + b). In our case, the cross-pieces of 840 * 2 and 420 * 4 are equal and have opposite signs, so they are reduced. At the output we get the following:

840 * 420-8 + 836 = 2 * 42 * 42 * 100 + 828 = 8 * 21 * 21 * 100 + 828

21 can be squared in different ways. Probably the simplest, through the expression 21 ^ 2-20 ^ 2 = 41. Hence, 21 ^ 2 = 400 + 41 = 441;
441 * 8 = 3200 + 320 + 8 = 3528.

And finally, 352800 + 828 = 353 628.

Looking at the solution, I saw another simplification.

8 * 21 * 21 * 100 + 828 = 800 * 21 * 21 + 800 + 28 = 800 * (21 * 21 + 1) +28

And the last expression is already really considered in the mind.

800 * 442 + 28 = 353600 + 28 = 353 628.

Returning after some time to this example, I tried again to quickly calculate 840 * 420. The previous version with the multiplication of 8 * 21 * 21 did not quite suit me. We will try to calculate 2 * 42 * 42. The square of numbers in the interval 41 ... 49 can be obtained as follows.

(40 + i) ^ 2 = (15 + i) * 100 + (10-i) ^ 2. In our case, 42 ^ 2 = (15 + 2) * 100 + 64 = 1764. And the required value is 2 * 1764 = 3528

The final result, of course, has not changed and is 352800 + 828 = 353 628.

Well, the last remark about the example of Alexander Ivanovich. Let us return to the representation of the desired number through the factors 4 * 9 * 11 * 19 * 47. Having grouped decomposition coefficients, we will receive (400-4) * (900-7). In this form, the problem is solved very quickly and at this point I decided for myself that the question is closed.

It should be noted that, apart from Koreiko, other literary characters also thought well in their minds. For example, Voland.

“Nine months,” thought Woland thoughtfully, “two hundred and forty-nine thousand ... Is this a round of twenty-seven thousand a month?” Not much, but with a modest life is enough. Yes, dozens more.

“Dozens of projects cannot be realized,” the same voice got involved, chilling the heart of the barmaid, “after the death of Andrei Fokich, the house will be immediately broken and dozens will be sent to the state bank.

Reflections on the methods of rapid division by 9 led me to the following observations. Since we can replace division by multiplication by 0.111 ..., in some cases the following algorithm helps out. First, the period of the final fraction is searched for - this is the remainder of dividing the sum of the digits of the original number by 9, in our case it is (2 + 4 + 9)% 9 = 6. Then we start moving from left to right and record the sum of digits.

(2) (2 + 4 = 6) (2 + 4 + 9 = 15) (15) (15) ...

Tens by the rules of addition to the column is transferred to the left digit. The answer is 27,66666 ...

Another example is 734/9.

Period - (7 + 3 + 4)% 9 = 5

(7) (7 + 3 = 10) (7 + 3 + 4 = 14) (14) ...

The answer is 81.5

And here is another literary hero who was distinguished by the speed of his oral account.

- Well, you see, I suddenly comprehended your character. So, why not give me five hundred rubles per soul, but ... there is no state; five kopecks, if you please, is ready to add, so that each soul cost, thus, thirty kopecks.

- Well, sir, your will, at least two kopeks fasten:

- Two penny clasp, if you please. How many have you got? You seem to be talking seventy?

- Not. A total of seventy eight.

- Seventy eight, seventy eight, thirty kopecks per soul, it will be ... - our hero is here one second, no more, he thought and said suddenly: - it will be twenty four rubles ninety six kopecks! - he was strong in arithmetic.

The Chichikov example, of course, is simpler than the Koreiko problem, but there are different approaches here as well. I propose a solution from a programmer, that is, a person who remembers the powers of two.

78 * 32 = 39 * 64 = (40-1) * 64 = 2560-64 = 2496

But Chichikov was not a programmer, Pavel Ivanovich earned a living by other labor. And therefore, could not be considered the proposed method. He had a different approach, perhaps such.

78 = 80-2
32 = 30 + 2

(80-2) * (30 + 2) = 2,400 + 2 * (80-30) -4 = 2,500-4 = 2,496

And maybe this. Recall the following trick. If you need to multiply something by 33, then we can divide this something by 3, multiply the result by 100 and subtract the first quotient. In this way,
78 * 33 = 26 * 100-26, and 78 * 32 = 26 * 100-26-78 = 25 * 100-4 = 2 496

I tend to go back to old tasks, at least in order to recall the condition. This time I discovered that Chichikov’s example can be solved using the algorithm for multiplying numbers, the last digits of which give a total of 10. This method was discovered by one of the students of Rachinsky S.A. Here is how it is done:

32 * 78 = 3 * (7 + 1) * 100 + 2 * (78-30) = 2496

Recently discovered another literary example on multiplication.

But one member asked permission to ask another question:
- How much will it be if you multiply twelve thousand eight hundred ninety seven by thirteen thousand eight hundred sixty three?
“Seven hundred and twenty-nine,” answered Schweik without batting an eye.

Note that Schweik considers quickly but incorrectly. However, I admit that the answer was given to them in the numeral system, which is different from decimal.