Quick squaring of numbers. The beauty of numbers
Today we will learn how to quickly square large expressions without a calculator. By large I mean numbers between ten and one hundred. Large expressions are extremely rare in real problems, and you already know how to count values \u200b\u200bless than ten, because this is a regular multiplication table. The material of today's lesson will be useful for fairly experienced students, because novice students simply will not appreciate the speed and effectiveness of this technique.
To begin with, let's understand in general what we are talking about. For example, I propose to do the construction of an arbitrary numeric expression, as we usually do. Let's say 34. We raise it by multiplying by itself with a column:
\[((34)^(2))=\times \frac(34)(\frac(34)(+\frac(136)(\frac(102)(1156))))\]
1156 is the square 34.
problem this method can be described in two ways:
1) it requires written registration;
2) it is very easy to make a mistake in the process of calculation.
Today we will learn how to quickly multiply without a calculator, verbally and practically without errors.
So let's get started. To work, we need the formula for the square of the sum and difference. Let's write them down:
\[(((a+b))^(2))=((a)^(2))+2ab+((b)^(2))\]
\[(((a-b))^(2))=((a)^(2))-2ab+((b)^(2))\]
What does this give us? The point is that any value between 10 and 100 can be represented as $a$, which is divisible by 10, and $b$, which is the remainder of division by 10.
For example, 28 can be represented as follows:
\[\begin(align)& ((28)^(2)) \\& 20+8 \\& 30-2 \\\end(align)\]
Similarly, we present the remaining examples:
\[\begin(align)& ((51)^(2)) \\& 50+1 \\& 60-9 \\\end(align)\]
\[\begin(align)& ((42)^(2)) \\& 40+2 \\& 50-8 \\\end(align)\]
\[\begin(align)& ((77)^(2)) \\& 70+7 \\& 80-3 \\\end(align)\]
\[\begin(align)& ((21)^(2)) \\& 20+1 \\& 30-9 \\\end(align)\]
\[\begin(align)& ((26)^(2)) \\& 20+6 \\& 30-4 \\\end(align)\]
\[\begin(align)& ((39)^(2)) \\& 30+9 \\& 40-1 \\\end(align)\]
\[\begin(align)& ((81)^(2)) \\& 80+1 \\& 90-9 \\\end(align)\]
What gives us such an idea? The fact is that with the sum or difference, we can apply the above calculations. Of course, in order to shorten the calculations, for each of the elements one should choose an expression with the smallest second term. For example, from the $20+8$ and $30-2$ options, you should choose the $30-2$ option.
Similarly, we choose options for other examples:
\[\begin(align)& ((28)^(2)) \\& 30-2 \\\end(align)\]
\[\begin(align)& ((51)^(2)) \\& 50+1 \\\end(align)\]
\[\begin(align)& ((42)^(2)) \\& 40+2 \\\end(align)\]
\[\begin(align)& ((77)^(2)) \\& 80-3 \\\end(align)\]
\[\begin(align)& ((21)^(2)) \\& 20+1 \\\end(align)\]
\[\begin(align)& ((26)^(2)) \\& 30-4 \\\end(align)\]
\[\begin(align)& ((39)^(2)) \\& 40-1 \\\end(align)\]
\[\begin(align)& ((81)^(2)) \\& 80+1 \\\end(align)\]
Why should one strive to reduce the second term in fast multiplication? It's all about the initial calculations of the square of the sum and difference. The fact is that the plus or minus term $2ab$ is the most difficult to calculate when solving real problems. And if the factor $a$, a multiple of 10, is always easily multiplied, then with the factor $b$, which is a number in the range from one to ten, many students regularly have difficulties.
\[{{28}^{2}}={{(30-2)}^{2}}=200-120+4=784\]
\[{{51}^{2}}={{(50+1)}^{2}}=2500+100+1=2601\]
\[{{42}^{2}}={{(40+2)}^{2}}=1600+160+4=1764\]
\[{{77}^{2}}={{(80-3)}^{2}}=6400-480+9=5929\]
\[{{21}^{2}}={{(20+1)}^{2}}=400+40+1=441\]
\[{{26}^{2}}={{(30-4)}^{2}}=900-240+16=676\]
\[{{39}^{2}}={{(40-1)}^{2}}=1600-80+1=1521\]
\[{{81}^{2}}={{(80+1)}^{2}}=6400+160+1=6561\]
So in three minutes we did the multiplication of eight examples. This is less than 25 seconds per expression. In reality, after a little practice, you will count even faster. It will take you no more than five or six seconds to calculate any two-digit expression.
But that's not all. For those to whom the shown technique seems not fast enough and not cool enough, I suggest even more fast way multiplication, which, however, does not work for all tasks, but only for those that differ by one from multiples of 10. In our lesson, there are four such values: 51, 21, 81 and 39.
It would seem much faster, we already count them literally in a couple of lines. But, in fact, it is possible to accelerate, and this is being done in the following way. We write down the value, a multiple of ten, which is closest to the desired one. For example, let's take 51. Therefore, to begin with, we will raise fifty:
\[{{50}^{2}}=2500\]
Values that are multiples of ten are much easier to square. And now we simply add fifty and 51 to the original expression. The answer will be the same:
\[{{51}^{2}}=2500+50+51=2601\]
And so with all numbers that differ by one.
If the value we are looking for is greater than the one we think, then we add numbers to the resulting square. If the desired number is less, as in the case of 39, then when performing the action, the value must be subtracted from the square. Let's practice without using a calculator:
\[{{21}^{2}}=400+20+21=441\]
\[{{39}^{2}}=1600-40-39=1521\]
\[{{81}^{2}}=6400+80+81=6561\]
As you can see, in all cases the answers are the same. Moreover, this technique is applicable to any adjacent values. For example:
\[\begin(align)& ((26)^(2))=625+25+26=676 \\& 26=25+1 \\\end(align)\]
At the same time, we do not need to remember the calculations of the squares of the sum and difference at all and use a calculator. The speed of work is beyond praise. Therefore, remember, practice and use in practice.
Key points
With this technique, you can easily do the multiplication of any natural numbers ranging from 10 to 100. Moreover, all calculations are performed orally, without a calculator and even without paper!
First, remember the squares of values that are multiples of 10:
\[\begin(align)& ((10)^(2))=100,((20)^(2))=400,((30)^(2))=900,..., \\ & ((80)^(2))=6400,((90)^(2))=8100. \\\end(align)\]
\[\begin(align)& ((34)^(2))=(((30+4))^(2))=((30)^(2))+2\cdot 30\cdot 4+ ((4)^(2))= \\& =900+240+16=1156; \\\end(align)\]
\[\begin(align)& ((27)^(2))=(((30-3))^(2))=((30)^(2))-2\cdot 30\cdot 3+ ((3)^(2))= \\& =900-180+9=729. \\\end(align)\]
How to count even faster
But that is not all! Using these expressions, you can instantly do the squaring of numbers that are “adjacent” to the reference ones. For example, we know 152 (the reference value), but we need to find 142 (an adjacent number that is one less than the reference). Let's write:
\[\begin(align)& ((14)^(2))=((15)^(2))-14-15= \\& =225-29=196. \\\end(align)\]
Please note: no mysticism! The squares of numbers that differ by 1 are indeed obtained by multiplying the reference numbers by themselves by subtracting or adding two values:
\[\begin(align)& ((31)^(2))=((30)^(2))+30+31= \\& =900+61=961. \\\end(align)\]
Why is this happening? Let's write down the formula for the square of the sum (and difference). Let $n$ be our reference value. Then they count like this:
\[\begin(align)& (((n-1))^(2))=(n-1)(n-1)= \\& =(n-1)\cdot n-(n-1 )= \\& ==((n)^(2))-n-(n-1) \\\end(align)\]
- this is the formula.
\[\begin(align)& (((n+1))^(2))=(n+1)(n+1)= \\& =(n+1)\cdot n+(n+1) = \\& =((n)^(2))+n+(n+1) \\\end(align)\]
- a similar formula for numbers greater than 1.
I hope this technique will save you time on all the important tests and exams in mathematics. And that's all for me. See you!
The book "Magic Numbers" tells about dozens of tricks that simplify the usual mathematical operations. It turned out that multiplication and division in a column is last century, but there are much more effective ways divisions in the mind.
Here are 10 of the most interesting and useful tricks.
Multiplication "3 by 1" in the mind
Multiplying three-digit numbers by single-digit numbers is a very simple operation. All you have to do is break the big task down into smaller ones.
Example: 320×7
- We break the number 320 into two simpler numbers: 300 and 20.
- We multiply 300 by 7 and 20 by 7 separately (2100 and 140).
- Add up the resulting numbers (2240).
Squaring two-digit numbers
Squaring two-digit numbers is not much more difficult. You need to split the number by two and get an approximate answer.
Example: 41^2
- Subtract 1 from 41 to get 40 and add 1 to 41 to get 42.
- We multiply the two resulting numbers using the previous tip (40 × 42 = 1680).
- We add the square of the number by which we decreased and increased 41 (1680 + 1^2 = 1681).
The key rule here is to turn the desired number into a pair of other numbers that are much easier to multiply. For example, for the number 41 these are the numbers 42 and 40, for the number 77 - 84 and 70. That is, we subtract and add the same number.
Instant squaring of a number ending in 5
With squares of numbers ending in 5, there is no need to strain at all. All you have to do is multiply the first digit by the number that is one more and add 25 to the end of the number.
Example: 75^2
Division by a single number
Division in the mind is quite a useful skill. Think about how often we divide numbers each day. For example, a bill in a restaurant.
Example: 675: 8
- Find approximate answers by multiplying 8 by convenient numbers that give extreme results (8 × 80 = 640, 8 × 90 = 720). Our answer is 80 plus.
- Subtract 640 from 675. Having received the number 35, you need to divide it by 8 and get 4 with a remainder of 3.
- Our final answer is 84.3.
We do not get the most accurate answer (the correct answer is 84.375), but you will agree that even such an answer will be more than enough.
Easy get 15%
To quickly find out 15% of any number, you must first calculate 10% of it (moving the comma one character to the left), then divide the resulting number by 2 and add it to 10%.
Example: 15% off 650
- We find 10% - 65.
- We find half of 65 - this is 32.5.
- We add 32.5 to 65 and get 97.5.
Banal trick
Perhaps we all stumbled upon this trick:
Think of any number. Multiply it by 2. Add 12. Divide the sum by 2. Subtract the original number from it.
You got 6, right? No matter what you think, you will still get 6. And here's why:
- 2x (double the number).
- 2x + 12 (add 12).
- (2x + 12) : 2 = x + 6 (divide by 2).
- x + 6 − x (subtract the original number).
This trick is based on the elementary rules of algebra. Therefore, if you ever hear that someone is guessing it, put on your most haughty grin, make a contemptuous look and tell everyone the solution. 🙂
The magic of the number 1089
This trick has been around for centuries.
Write down any three-digit number, whose digits are in decreasing order (for example, 765 or 974). Now write it in reverse order and subtract it from the original number. Add it to the answer you received, only in reverse order.
Whatever number you choose, the result will be 1,089.
Fast cube roots
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 1 | 8 | 27 | 64 | 125 | 216 | 343 | 512 | 729 | 1 000 |
Once you remember these values, finding the cube root of any number will be elementary easy.
Example: cube root of 19,683
- We take the value of thousands (19) and see what numbers it is between (8 and 27). Accordingly, the first digit in the answer will be 2, and the answer lies in the range of 20+.
- Each digit from 0 to 9 appears in the table once as the last digit of the cube.
- Since the last digit in the problem is 3 (19,683), this corresponds to 343 = 7^3. Therefore, the last digit of the answer is 7.
- The answer is 27.
Note: The trick only works when the original number is the cube of an integer.
Rule 70
To find the number of years it takes to double your money, divide 70 by the annual interest rate.
Example: the number of years it takes for money to double at an annual interest rate of 20%.
70:20 = 3.5 years
Rule 110
To find the number of years it takes for money to triple, divide 110 by the annual interest rate.
Example: the number of years it takes to triple money at an annual interest rate of 12%.
110:12 = 9 years old
Mathematics is a magical science. If even such simple tricks are surprising, then what other tricks can you think of?
One of the most frequent mathematical operations, used in engineering and other calculations, is the raising of a number to the second power, which is otherwise called the square. For example, this method calculates the area of an object or figure. Unfortunately, there is no separate tool in Excel that would square a given number. However, this operation can be performed using the same tools that are used for raising to any other power. Let's find out how they should be used to calculate the square of a given number.
As you know, the square of a number is calculated by multiplying it by itself. These principles, of course, underlie the calculation of this indicator in Excel. In this program, there are two ways to square a number: using the exponentiation sign for formulas «^» and applying the function DEGREE. Consider the algorithm for applying these options in practice in order to evaluate which one is better.
Method 1: construction using a formula
First of all, let's look at the simplest and most commonly used method of raising to the second power in Excel, which involves using a formula with the symbol «^» . In this case, as an object that will be squared, you can use a number or a reference to the cell where this numeric value is located.
The general form of the formula for squaring is as follows:
In it instead "n" you need to substitute a specific number that should be squared.
Let's see how it works with specific examples. Let's start by squaring the number that will be integral part formulas.
Now let's see how to square a value that is located in another cell.
Method 2: Use the POWER function
You can also use the built-in Excel function to square a number. DEGREE. This operator is included in the category of mathematical functions and its task is to raise a certain numerical value to a specified power. The function syntax is as follows:
POWER(number, degree)
Argument "Number" can be a specific number or a reference to the sheet element where it is located.
Argument "Degree" indicates the power to which the number is to be raised. Since we are faced with the question of squaring, in our case this argument will be equal to 2 .
Now let's look at specific example, how squaring is done using the operator DEGREE.
Also, to solve the problem, instead of a number as an argument, you can use a reference to the cell in which it is located.
* squares up to hundreds
In order not to mindlessly square all numbers according to the formula, you need to simplify your task as much as possible with the following rules.
Rule 1 (cuts off 10 numbers)
For numbers ending in 0.If a number ends in 0, multiplying it is no more difficult than single digit. All you have to do is add a couple of zeros.
70 * 70 = 4900.
The table is marked in red.
Rule 2 (cuts off 10 numbers)
For numbers ending in 5.To square two-digit number ending in 5, you need to multiply the first digit (x) by (x + 1) and add “25” to the result.
75 * 75 = 7 * 8 = 56 … 25 = 5625.
The table is marked in green.
Rule 3 (cuts off 8 numbers)
For numbers from 40 to 50.XX * XX = 1500 + 100 * second digit + (10 - second digit)^2
Hard enough, right? Let's take an example:
43 * 43 = 1500 + 100 * 3 + (10 - 3)^2 = 1500 + 300 + 49 = 1849.
The table is marked in light orange.
Rule 4 (cuts off 8 numbers)
For numbers from 50 to 60.XX * XX = 2500 + 100 * second digit + (second digit)^2
It's also quite difficult to understand. Let's take an example:
53 * 53 = 2500 + 100 * 3 + 3^2 = 2500 + 300 + 9 = 2809.
The table is marked in dark orange.
Rule 5 (cuts off 8 numbers)
For numbers from 90 to 100.XX * XX = 8000+ 200 * second digit + (10 - second digit)^2
Similar to rule 3, but with different coefficients. Let's take an example:
93 * 93 = 8000 + 200 * 3 + (10 - 3)^2 = 8000 + 600 + 49 = 8649.
The table is marked in dark dark orange.
Rule #6 (cuts off 32 numbers)
It is necessary to memorize the squares of numbers up to 40. It sounds crazy and difficult, but in fact, up to 20, most people know the squares. 25, 30, 35 and 40 lend themselves to formulas. And only 16 pairs of numbers remain. They can already be memorized using mnemonics (which I also want to talk about later) or by any other means. Like a multiplication table :)The table is marked in blue.
You can remember all the rules, or you can remember selectively, in any case, all numbers from 1 to 100 obey two formulas. The rules will help, without using these formulas, to quickly calculate more than 70% of the options. Here are the two formulas:
Formulas (24 numbers left)
For numbers from 25 to 50XX * XX = 100(XX - 25) + (50 - XX)^2
For example:
37 * 37 = 100(37 - 25) + (50 - 37)^2 = 1200 + 169 = 1369
For numbers from 50 to 100
XX * XX = 200(XX - 50) + (100 - XX)^2
For example:
67 * 67 = 200(67 - 50) + (100 - 67)^2 = 3400 + 1089 = 4489
Of course, do not forget about the usual formula for expanding the square of the sum ( special case binomial Newton):
(a+b)^2 = a^2 + 2ab + b^2. 56^2 = 50^2 + 2*50*6 + 6*2 = 2500 + 600 + 36 = 3136.
UPDATE
Products of numbers close to 100, and, in particular, their squares, can also be calculated according to the principle of "shortcomings up to 100":
In words: from the first number we subtract the “flaw” of the second to a hundred and attribute the two-digit product of the “flaws”.
For squares, respectively, even easier.
92*92 = (92-8)*100+8*8 = 8464
(by sielover)
Squaring may not be the most useful thing in the household. You will not immediately remember the case when you may need the square of a number. But the ability to quickly operate with numbers, apply the appropriate rules for each of the numbers, perfectly develops the memory and “computing abilities” of your brain.
By the way, I think all Habra readers know that 64^2 = 4096, and 32^2 = 1024.
Many squares of numbers are remembered at the associative level. For example, I easily memorized 88^2 = 7744 because of the same numbers. Everyone will surely have their own characteristics.
Two unique formulas I first found in the book "13 steps to mentalism", which has little to do with mathematics. The fact is that earlier (perhaps even now) unique computing abilities were one of the numbers in stage magic: a magician told a bike about how he got superpowers and, as proof of this, instantly squares numbers up to a hundred. The book also shows how to cube, how to subtract roots and cube roots.
If the topic of quick counting is interesting, I will write more.
Please write comments about errors and corrections in PM, thanks in advance.