Today we will learn how to quickly square large expressions without a calculator. By large, I mean numbers ranging from ten to one hundred. Big Expressions They are extremely rare in real problems, and you already know how to count values less than ten, because this is a regular multiplication table. The material in today's lesson will be useful to fairly experienced students, because beginner students simply will not appreciate the speed and effectiveness of this technique.
First, let's figure out what we're talking about we're talking about. I propose, as an example, to construct an arbitrary numerical expression, as we usually do. Let's say 34. We raise it by multiplying it 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 points:
1) it requires written documentation;
2) it is very easy to make a mistake during the calculation process.
Today we will learn how to quickly multiply without a calculator, orally and with virtually no mistakes.
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 fact is that any value in the range from 10 to 100 can be represented as the number $a$, which is divisible by 10, and the number $b$, which is the remainder of division by 10.
For example, 28 can be represented as follows:
\[\begin(align)& ((28)^(2)) \\& 20+8 \\& 30-2 \\\end(align)\]
We present the remaining examples in the same way:
\[\begin(align)& ((51)^(2)) \\& 50+1 \\& 60-9 \\\end(align)\]
\[\begin(align)& ((42)^(2)) \\& 40+2 \\& 50-8 \\\end(align)\]
\[\begin(align)& ((77)^(2)) \\& 70+7 \\& 80-3 \\\end(align)\]
\[\begin(align)& ((21)^(2)) \\& 20+1 \\& 30-9 \\\end(align)\]
\[\begin(align)& ((26)^(2)) \\& 20+6 \\& 30-4 \\\end(align)\]
\[\begin(align)& ((39)^(2)) \\& 30+9 \\& 40-1 \\\end(align)\]
\[\begin(align)& ((81)^(2)) \\& 80+1 \\& 90-9 \\\end(align)\]
What does this idea tell us? The fact is that with a sum or a difference, we can apply the calculations described above. Of course, to shorten the calculations, for each element you should choose the expression with the smallest second term. For example, from the options $20+8$ and $30-2$, you should choose the option $30-2$.
We similarly select options for the remaining 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 we strive to reduce the second term when fast multiplication? It's all about the initial calculations of the square of the sum and the difference. The fact is that the term $2ab$ with a plus or a minus is the most difficult to calculate when solving real problems. And if the factor $a$, a multiple of 10, is always multiplied easily, then with the factor $b$, which is a number ranging 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. That's 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 to six seconds to calculate any two-digit expression.
But that's not all. For those to whom the technique shown seems not fast enough and cool enough, I suggest even more quick 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 in literally a couple of lines. But, in fact, you can speed up, and this is done as follows. We write down the value that is a multiple of ten, which is closest to what we need. For example, let's take 51. Therefore, to begin with, let's build fifty:
\[{{50}^{2}}=2500\]
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 are counting, then we add numbers to the resulting square. If the desired number is smaller, as in the case of 39, then when performing the action, you need to subtract the value 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 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 multiply 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's not all! Using these expressions, you can instantly square numbers “adjacent” to the reference ones. For example, we know 152 ( reference value), but we need to find 142 (an adjacent number that is one less than the reference number). Let's write it down:
\[\begin(align)& ((14)^(2))=((15)^(2))-14-15= \\& =225-29=196. \\\end(align)\]
Please note: no mysticism! Squares of numbers that differ by 1 are actually obtained by multiplying by themselves reference numbers, if you subtract or add 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 are calculated 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 your high-stakes math tests and exams. And that's all for me. See you!
If you multiply number on itself, the result will be a construction in square. Even a first-grader knows that “twice two is four.” Three-digit, four-digit, etc. It’s better to multiply numbers in a column or on a calculator, but handle double-digit ones without an electronic assistant, multiplying in your head.
Instructions
1. Break down any two-digit number into components, highlighting the number of units. In the number 96, the number of units is 6. Consequently, we can write: 96 = 90 + 6.
2. Build in square the first of the numbers: 90 * 90 = 8100.
3. Do the same with the second one. number m: 6 * 6 = 36
4. Multiply the numbers together and double the total: 90 * 6 * 2 = 540 * 2 = 1080.
5. Add up the totals of the second, third and fourth steps: 8100 + 36 + 1080 = 9216. This is the result of construction in square number 96. After some training, you will be able to quickly take steps in your mind, amazing your parents and classmates. Until you get the hang of it, write down the results of the entire step so you don’t get confused.
6. To practice, raise to square number 74 and test yourself on the calculator. Sequence of actions: 74 = 70 + 4, 70 * 70 = 4900, 4 * 4 = 16, 70 * 4 * 2 = 560, 4900 + 16 + 560 = 5476.
7. Raise to the second power number 81. Your actions: 81 = 80 + 1, 80 * 80 = 6400, 1 * 1 = 1, 80 * 1 * 2 = 160, 6400 + 1 + 160 = 6561.
8. Remember non-standard method construction in square two-digit numbers that end with the number 5. Select the number of tens: in the number 75 there are 7 of them.
9. Multiply the tens number by the next digit in number in the first row: 7 * 8 = 56.
10. Write on the right number 25: 5625 - the result of construction in square number 75.
11. For practice, raise to the second power number 95. It ends with the number 5, so the sequence of actions is: 9 * 10 = 90, 9025 is the total.
12. Learn to build in square negative numbers: -95 in square e is equal to 9025, as in the eleventh step. Similar to -74 in square e is equal to 5476, as in the sixth step. This is due to the fact that when multiplying 2 negative numbers invariably the correct result is obtained. number: -95 * -95 = 9025. Consequently, when erected in square you can easily ignore the minus sign.
Raising a number to a power is one of the simplest algebraic operations. IN everyday life construction is rarely used, but in production when performing calculations - virtually everywhere, so it’s useful to remember how this is done.
Instructions
1. Let's imagine that we have some number a, the power of which is the number n. To construct a number to a power means that you need to multiply the number a by itself n times.
2. Let's look at a few examples. In order to construct the number 2 to the second power, you need to perform the action: 2x2 = 4
3. In order to construct the number 3 to the fifth power, you need to perform the action: 3x3x3x3x3 = 243
4. There is a generally accepted notation for 2nd and 3rd degree numbers. The phrase “second degree” is usually replaced by the word “square”, and instead of the phrase “third degree” they traditionally say “cube”.
5. As can be seen from the above examples, the duration and complexity of calculations depends on the value of the exponent of the number. Squaring or cube is enough simple task; Raising a number to the fifth or highest power requires a lot of time and accuracy in calculations. To speed up this process To avoid errors, you can use special mathematical tables or an engineering calculator.
To briefly write the product of the same number and itself, mathematicians came up with the power representation. Consequently, the expression 16*16*16*16*16 can be written more short method. It will look like 16^5. The expression will read as the number 16 to the fifth power.
You will need
- Paper, pen.
Instructions
1. In general degree written as a^n. This notation means that the number a is multiplied by itself n times. The expression a^n is called degree yu,a is the number degree base,n is a number, an exponent. Let's say a = 4, n = 5, Then we write 4^5 = 4*4*4*4*4 = 1,024
2. The power n can be a negative numbern = -1, -2, -3, etc. To calculate the negative degree number, it must be omitted into the denominator.a^(-n) = (1/a)^n = 1/a*1/a*1/a* … *1/a = 1/(a^n) Let's look at example2 ^(-3) = (1/2)^3 = 1/2*1/2*1/2 = 1/(2^3) = 1/8 = 0.125
3. As can be seen from the example, -3 degree the number 2 can be calculated using various methods.1) First, calculate the fraction 1/2 = 0.5; and then build in degree 3, i.e. 0.5^3 = 0.5*0.5*0.5 = 0.1252) First, construct the denominator in degree 2^3 = 2*2*2 = 8, and then calculate the fraction 1/8 = 0.125.
4. Now let's calculate -1 degree for a number, i.e. n = -1. The rules discussed above are suitable for this case.a^(-1) = (1/a)^1 = 1/(a^1) = 1/aFor example, construct the number 5 in -1 degree 5^(-1) = (1/5)^1 = 1/(5^1) = 1/5 = 0,2.
5. The example clearly shows that a number to the -1 power is reciprocal fraction from the number. Let us assume the number 5 in the form of a fraction 5/1, then 5^(-1) can not be counted arithmetically, but immediately write the fraction inverse of 5/1, this is 1/5. So, 15^(-1) = 1 /15.6^(-1) = 1/6.25^(-1) = 1/25
Pay attention!
When raising a number to a negative power, remember that the number cannot be equal to zero. According to the rule, we must omit the number into the denominator. And zero cannot be in the denominator, since it is impossible to divide by zero.
Useful advice
Occasionally when working with powers to facilitate calculations fractional number deliberately replaced by an integer to the -1 power1/6 = 6^(-1)1/52 = 52^(-1).
When solving arithmetic and algebraic problems occasionally required to build fraction V square. It's easier for everyone to do this when fraction decimal is a fairly ordinary calculator. However, if fraction ordinary or mixed, then when such a number is raised to square Some difficulties may arise.
You will need
- calculator, computer, Excel application.
Instructions
1. To construct a decimal fraction V square, take engineering calculator, type on it the one being built in square fraction and press the raise to the second power key. On most calculators this button is labeled "x?". On a standard Windows calculator, the function of raising to square looks like "x^2". Let's say square the decimal fraction 3.14 will be equal to: 3.14? = 9.8596.
2. In order to build in square decimal fraction on an ordinary (accounting) calculator, multiply this number by itself. By the way, some models of calculators provide the possibility of raising a number to square even in the absence of a special button. Therefore, please read the instructions for specific calculator. Occasionally, examples of “tricky” exponentiation are given on the back cover or on the box of the calculator. For example, on many calculators to raise a number to square Just press the “x” and “=” buttons.
3. For construction in square common fraction(consisting of a numerator and a denominator), raise to square separately the numerator and denominator of this fraction. That is, use the further rule: (h/z)? = h? / z?, where h is the numerator of the fraction, z is the denominator of the fraction. Example: (3/4)? = 3?/4? = 9/16.
4. If being built in square fraction– mixed (consists of an integer part and an ordinary fraction), then reduce it in advance to normal appearance. That is, apply the following formula:(ts h/z)? = ((c*z+h) / z)? = (ts*z+h)? / з?, where ц – whole part mixed fraction. Example: (3 2/5)? = ((3*5+2) / 5)? = (3*5+2)? / 5? = 17? / 5? = 289/25 = 11 14/25.
5. If built in square regular (not decimal) fractions are added continuously, then use MS Excel. To do this, enter the following formula into one of the cells of the table: = DEGREE(A2;2) where A2 is the address of the cell into which the raised value will be entered square fraction.To inform the program that the entered number should be treated as an ordinary fraction yu (i.e. do not convert it to decimal form), type before fraction I have the number “0” and the sign “space”. That is, to enter, say, the fraction 2/3, you need to enter: “0 2/3” (and press Enter). In this case, the decimal representation of the entered fraction will be displayed in the input line. The meaning and representation of the fraction at ease in the cell will be preserved in initial form. In addition, when using mathematical functions, whose arguments are ordinary fractions, the result will also be presented in the form of an ordinary fraction. Consequently square the fraction 2/3 will be represented as 4/9.
The method of squaring a binomial is used to simplify massive expressions, as well as to solve quadratic equations. In practice, it is traditionally combined with other techniques, including factorization, grouping, etc.
Instructions
1. The method for isolating the complete square of a binomial is based on the use of 2 formulas for abbreviated multiplication of polynomials. These formulas are special cases of Newton’s Binomial for the 2nd degree and allow us to simplify the desired expression so that it is possible to carry out further reduction or factorization: (m + n)² = m² + 2 m n + n²;(m – n)² = m² – 2·m·n + n².
2. According to this method, from the initial polynomial it is necessary to extract the squares of 2 monomials and the sum/difference of their double product. The use of this method makes sense if the highest degree of the terms is not less than 2. Imagine, you are given the task of factoring the following expression into factors with decreasing degree: 4 y^4 + z^4
3. To solve the problem, you need to use the method of selecting a complete square. It turns out that the expression consists of 2 monomials with variables of even degree. Consequently, it is possible to denote each of them by m and n: m = 2·y²; n = z².
4. Now we need to bring initial expression to the form (m + n)². It already contains the squares of these terms, but lacks the double product. It is necessary to add it unnaturally and then subtract it: (2 y²)² + 2 2 y² z² + (z²)² – 2 2 y² z² = (2 y² + z²)² – 4 y² z².
5. In the resulting expression you can see the formula for the difference of squares: (2 y² + z²)² – (2 y z)² = (2 y² + z² – 2 y z) (2 y² + z² + 2 y z).
6. It turns out that the method consists of 2 stages: isolating the monomials of a perfect square m and n, adding and subtracting their double product. The method of isolating the complete square of a binomial can be used not only independently, but also in combination with other methods: removing the universal factor from brackets, replacing a variable, grouping terms, etc.
7. Example 2: Select perfect square in the expression: 4 y² + 2 y z + z². Solution. 4 y² + 2 y z + z² = = (2 y)² + 2 2 y z + (z) ² – 2 y z = (2 y + z)² – 2 y z.
8. The method is used when finding roots quadratic equation. Left side equation is a trinomial of the form a·y? + b·y + c, where a, b and c are some numbers, and a ? 0. a·y? + b y + c = a (y? + (b/a) y) + c = a (y? + 2 (b/(2 a)) y) + c = a ( y? + 2 (b/(2 a)) y + b?/(4 a?)) + c – b?/(4 a) = a (y + b/(2 a )) ? – (b? – 4 a c)/(4 a).
9. These calculations lead to the representation of the discriminant, which is equal to (b? – 4·a·c)/(4·a), and the roots of the equation are equal to: y_1,2 = ±(b/(2 a)) ± ? ((b? – 4 a c)/(4 a)).
The erection operation degree is “binary”, that is, it has two indispensable input parameters and one output parameter. One of the initial parameters is called the exponent and specifies the number of times the multiplication operation must be applied to the second parameter, the radix. The reason can be either correct or negative number .
Instructions
1. When raising a negative number to a power, use the usual rules for this operation. As for positive numbers, raising to a power means multiplying the initial value by itself a number of times, one less than the exponent. Let's say, in order to construct the number -2 to the fourth power, you need to multiply it by itself three times: -2?=-2*(-2)*(-2)*(-2)=16.
2. Multiplying 2 negative numbers invariably gives a positive value, and the result of this operation for quantities with various signs will be a negative number. From this we can conclude that when negative values are raised to a power with an even exponent, the result should invariably be a positive number, and with odd exponents the result will invariably be less than zero. Use this quality to check your calculations. Let's say -2 to the fifth power should be a negative number -2?=-2*(-2)*(-2)*(-2)*(-2)=-32, and -2 to the sixth power should be positive -2 ?=-2*(-2)*(-2)*(-2)*(-2)*(-2)=64.
3. When raising a negative number to a power, the exponent can be given in the format of a common fraction - say, -64 to the power?. This indicator means that the initial value should be built to a power equal to the numerator of the fraction, and the root of the power should be extracted from it, equal to the denominator. One part of this operation was discussed in the previous steps, but here you should pay attention to another.
4. Root extraction – odd function, that is, for negative real numbers it can only be used with an odd exponent. If even, this function has no meaning. Consequently, if the conditions of the problem require constructing a negative number in fractional power with an even denominator, then the problem has no solution. In other cases, first perform the operations from the first 2 steps, using the numerator of the fraction as an exponent, and then extract the root with the degree of the denominator.
The power format of writing a number is a shortened form of writing the operation of multiplying a base by itself. With a number presented in this form, you can carry out the same operations as with any other numbers, including raising them to degree. Let's say it is allowed to build in an arbitrary degree square numbers and obtaining the result at the modern level of technology development will not be any difficulty.
You will need
- Internet access or Windows calculator.
Instructions
1. For construction square and in degree use the general rule of construction in degree numbers that already have power exponent. With this operation, the indicators are multiplied, but the base remains the same. If the base is designated as x, and the initial and additional exponents are designated as a and b, this rule can be written in general form as follows: (x?)?=x??.
2. For utilitarian calculations, it is easier for everyone to use search engine Google – it has a very easy to use calculator built into it. Let's say, if you need to build in the fifth degree square number 6, go to the main page of the search engine and enter the appropriate query. It can be formulated as follows: (6^2)^5 – here the ^ symbol means degree. You can independently calculate the resulting exponent in accordance with the formula from the previous step and formulate the query as follows: 6^10. Or trust Google to do this by entering the following query: 6^(2*5). For each of these options, the search engine’s calculator will return an identical result: 60,466,176.
3. If there is no access to the Internet, the Google calculator can be replaced, say, with a built-in Windows calculator. If you are using the Seven or Vista versions of this OS, open the main menu of the system and type two letters for each: “ka”. The system will display in the main menu all programs and files that it associates with this combination. In the first line there will be a link “Calculator” - click on it with the mouse and the application will be launched.
4. Press the key combination Alt + 2 so that a button with the function of raising to arbitrary appears in the application interface degree. After that, enter the base - in the example from the second step, this is the number 6 - and first click on the x? button, and then on the x? button. Enter the exponent to which you want to construct square– in the example used, this number is 5. Press the Enter button, and the calculator will display the final result of the operation.
Video on the topic
Useful advice
To make the workout less boring, call a friend to help. Let him write a two-digit number, and you write the conclusion of squaring this number. After that, change places.
As you know, the area of a rectangle is calculated by multiplying the lengths of its two various sides. A square has all sides equal, so you need to multiply the side by itself. This is where the expression “squaring” came from. Perhaps the easiest way to square any number is to take a regular calculator and multiply the desired number by itself. If you don’t have a calculator at hand, you can use the built-in calculator in mobile phone. For more advanced users, we can recommend using the application Office Microsoft Excel, especially if such calculations need to be carried out quite often. To do this, you need to select an arbitrary cell, for example G7, and enter the formula =F7*F7 into it. Next, enter any number in cell F7, and get the result in cell G7.
How to square a number last digit which is 5. To square this number, you need to discard the last digit of the number. The resulting number must be multiplied with a larger number by 1. Then you need to add the number 25 to the right after the result. Example. Let's say you want to get the square of the number 35. After the last digit 5 is discarded, the number 3 remains. Add 1 and you get the number 4.3x4=12. Add 25 and the result is 1225. 35x35=3*4 add 25=1225.
How to square a number whose last digit is 6. This algorithm is suitable for those who have figured out the question of how to square a number ending in 5. As is known from mathematics, the square of a binomial can be calculated using the formula (A + B) x (A+B) =AxA+2xAxB + BxB. In the case of squaring a number A, the last digit of which is 6, this number can be represented as A=B+1, where B is the number that is 1 less number And, therefore, its last digit is 5. In this case, the formula can be represented in more in simple form(B+1) x(B+1) =BxB+2xBx1+1x1=BxB + 2xB+1. For example, let this number be 16. Solution 16 x16=15 x15+2x15 x1+1x1=225+30+1=256 Oral rule: in order to find the square of a number ending in 6: you need to square the previous number, add two times the previous number and add 1.
How to square numbers from 11 to 29. To square numbers from 11 to 19, you need to add the number of ones to the original number, multiply the resulting result by 10 and add the squared number of ones to the right. Example. Square 13. The number of units in this number is 3. Next, you need to calculate the intermediate number 13+3=16. Then multiply it by 10. It turns out 160. The square of the number of units is 3x3=9. The final result is 169. For numbers in the third ten, a similar algorithm is used, only you need to multiply by 20 and add the square of the units rather than add them. Example. Calculate the square of the number 24. The number of ones is found – 4. The intermediate number is calculated – 24+4=28. After multiplying by 20, the result is 560. The square of the number of ones is 4x4=16. The final result is 560+16=576.
How to square numbers from 40 to 60. The algorithm is quite simple. First you need to find how much given number more or less than the middle of the range of the number 50. Add to the resulting result (if the number is greater than 50) or subtract (if the number is less than 50) 25. Multiply the resulting sum (or difference) by 100. To the resulting result add the square of the difference between the number whose square you need find, and the number 50. Example: you need to find the square of the number 46. The difference is 50-46=4.5-4=1.1x100=0.4x4=6.0+16=2116. Result: 46x46=2116.
Another trick is how to square numbers from 40 to 60. In order to calculate the square of a number from 40 to 49, you need to increase the number of units by 15, multiply the resulting result by 100, and to the right of it add the square of the difference between the last digit given number and 10. Example. Calculate the square of the number 42. The number of units of this number is 2. Add 15: 2+15=17. The difference between the same number of units and 10 is found. It is equal to 8. Squared: 8x8 = 64. The number 64 is added to the right to the previous result 17. The final number is 1764. If the number is in the range from 51 to 59, then the same algorithm is used to square it, only 25 must be added to the number of units.
How to square any two-digit number in your head. If a person knows how to square single digit numbers, in other words, if he knows the multiplication table, he will not have problems calculating the squares of two-digit numbers. Example. You need to square the two-digit number 36. This number is multiplied by the number of its tens. 36x3=8. Next you need to find the product of the digits of the number: 3x6=18. Then add both results. 108+18=126. The next step: you need to square the units of the original number: 6x6=36. In the resulting product, the number of tens is determined - 3 and added to the previous result: 126 + 3 = 129. And the last step. To the right of the obtained result is assigned the number of units of the original number, in in this example - 6. End result– number 1296.
There are many ways to square different numbers. Some of the given algorithms are quite simple, others are quite cumbersome and incomprehensible at first glance. People have been using many of them for centuries. Each person can develop their own more understandable and interesting algorithms. But if there are problems with verbal counting or other difficulties arise, you will have to attract technical means.
The ability to count squares of numbers in your head can be useful in different life situations, for example, for quick assessment of investment transactions, for calculating areas and volumes, as well as in many other cases. In addition, the ability to count squares in your head can serve as a demonstration of your intellectual abilities. This article discusses methods and algorithms that allow you to learn this skill.
Squared sum and squared difference
One of the simplest ways to square two-digit numbers is a technique based on the use of the squared sum and squared difference formulas:
To use this method, you need to decompose a two-digit number into the sum of a multiple of 10 and a number less than 10. For example:
- 37 2 = (30+7) 2 = 30 2 + 2*30*7 + 7 2 = 900+420+49 = 1 369
- 94 2 = (90+4) 2 = 90 2 + 2*90*4 + 4 2 = 8100+720+16 = 8 836
Almost all squaring techniques (which are described below) are based on the squared sum and squared difference formulas. These formulas made it possible to identify a number of algorithms that simplify squaring in some special cases.
A square close to a known square
If the number being squared is close to a number whose square we know, we can use one of four techniques for simplified mental arithmetic:
1 more:
Methodology: to the square of a number one less we add the number itself and the number one less.
- 31 2 = 30 2 + 31 + 30 = 961
- 16 2 = 15 2 + 15 + 16 = 225 + 31 = 256
1 less:
Methodology: From the square of a number that is one more, we subtract the number itself and the number that is one more.
- 19 2 = 20 2 - 19 - 20 = 400 - 39 = 361
- 24 2 = 25 2 - 24 - 25 = 625 - 25 - 24 = 576
2 more
Methodology: to the square of the number 2 less we add twice the sum of the number itself and the number 2 less.
- 22 2 = 20 2 + 2*(20+22) = 400 + 84 = 484
- 27 2 = 25 2 + 2*(25+27) = 625 + 104 = 729
2 less
Methodology: From the square of a number 2 more, subtract twice the sum of the number itself and the number 2 more.
- 48 2 = 50 2 - 2*(50+48) = 2500 - 196 = 2 304
- 98 2 = 100 2 - 2*(100+98) = 10 000 - 396 = 9 604
All these techniques can be easily proven by deriving algorithms from the squared sum and squared difference formulas (mentioned above).
Square of numbers ending in 5
To square numbers ending in 5. The algorithm is simple. The number up to the last five, multiply by the same number plus one. We add 25 to the remaining number.
- 15 2 = (1*(1+1)) 25 = 225
- 25 2 = (2*(2+1)) 25 = 625
- 85 2 = (8*(8+1)) 25 = 7 225
This is also true for more complex examples:
- 155 2 = (15*(15+1)) 25 = (15*16)25 = 24 025
Square of numbers close to 50
Count the square of the numbers that are in range from 40 to 60, you can very in a simple way. The algorithm is as follows: to 25 we add (or subtract) as much as the number is greater (or less) than 50. We multiply this sum (or difference) by 100. To this product we add the square of the difference between the number being squared and fifty. See the algorithm in action using examples:
- 44 2 = (25-6)*100 + 6 2 = 1900 + 36 = 1936
- 53 2 = (25+3)*100 + 3 2 = 2800 + 9 = 2809
Square of three-digit numbers
Squaring three-digit numbers can be done using one of the abbreviated multiplication formulas:
It cannot be said that this method is convenient for oral counting, but especially difficult cases you can use it:
436 2 = (400+30+6) 2 = 400 2 + 30 2 + 6 2 + 2*400*30 + 2*400*6 + 2*30*6 = 160 000 + 900 + 36 + 24 000 + 4 800 + 360 = 190 096
Training
If you want to improve your skills on the topic this lesson, you can use the following game. The points you receive are affected by the correctness of your answers and the time spent on completion. Please note that the numbers are different each time.
The book “The Magic of Numbers” talks about dozens of tricks that simplify the usual mathematical operations. It turned out that multiplication and long division are last century, but there is much more effective ways divisions in the mind.
Here are 10 of the most interesting and useful tricks.
Multiplying "3 by 1" in your head
Multiplying three-digit numbers by single-digit numbers is very simple operation. All you have to do is break big task for several small ones.
Example: 320×7
- Divide the number 320 into two more prime numbers: 300 and 20.
- We multiply 300 by 7 and 20 by 7 separately (2,100 and 140).
- Add up the resulting numbers (2,240).
Squaring two-digit numbers
Square double figures not much more difficult. You need to divide 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 advice (40 × 42 = 1,680).
- We add the square of the number by the amount by which we decreased and increased 41 (1,680 + 1^2 = 1,681).
The key rule here is to turn the number you are looking for into a couple of other numbers, which 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.
Instantly square 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 higher and add 25 to the end of the number.
Example: 75^2
Division by a single digit number
Mental division is a fairly useful skill. Think about how often we divide numbers every day. For example, a bill at a restaurant.
Example: 675: 8
- Let's find approximate answers by multiplying 8 by convenient numbers that give extreme results (8 × 80 = 640, 8 × 90 = 720). Our answer is more than 80.
- 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 to get 15%
To quickly find out 15% of any number, you first need to count 10% of it (moving the decimal place one place to the left), then divide the resulting number by 2 and add it to 10%.
Example: 15% of 650
- We find 10% - 65.
- We find half of 65 - this is 32.5.
- Add 32.5 to 65 and get 97.5.
Trivial trick
We've probably all come across 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 wish for, you will still get a 6. 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 elementary rules algebra. Therefore, if you ever hear that someone is riddled with it, put on your most arrogant 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 to reverse order and subtract it from the original number. Add the same answer to the answer you receive, only in reverse order.
Whatever number you choose, the result will be 1,089.
Quick 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, find cube root from any number it will be elementary simple.
Example: cube root of 19,683
- We take the value of thousands (19) and look between which numbers it is located (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 once in the table 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 for your money to double, you 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 to triple your money, you divide 110 by the annual interest rate.
Example: The number of years it takes to triple the money at an annual interest rate of 12%.
110: 12 = 9 years
Mathematics is a magical science. Even if such simple tricks surprise, what other tricks can you come up with?
