Methods for quick squaring. Squaring a Number in Microsoft Excel

*squares up to hundreds

In order not to mindlessly square all the numbers using 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 a single-digit number. You just need to add a couple of zeros.
70 * 70 = 4900.
Marked in red in the table.

Rule 2 (cuts off 10 numbers)

For numbers ending in 5.
To square a 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.
Marked in green in the table.

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 look at an example:
43 * 43 = 1500 + 100 * 3 + (10 - 3)^2 = 1500 + 300 + 49 = 1849.
In the table they are 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 is also quite difficult to understand. Let's look at an example:
53 * 53 = 2500 + 100 * 3 + 3^2 = 2500 + 300 + 9 = 2809.
In the table they are 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 look at an example:
93 * 93 = 8000 + 200 * 3 + (10 - 3)^2 = 8000 + 600 + 49 = 8649.
In the table they are marked in dark dark orange.

Rule No. 6 (cuts off 32 numbers)

You need to memorize the squares of numbers up to 40. It sounds crazy and difficult, but in fact most people know the squares up to 20. 25, 30, 35 and 40 are amenable to formulas. And only 16 pairs of numbers remain. They can already be remembered using mnemonics (which I also want to talk about later) or by any other means. Like a multiplication table :)
Marked in blue in the table.

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 days left)

For numbers from 25 to 50
XX * 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 the expansion of the square of a sum (a special case of Newton’s binomial):
(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 using the principle of “disadvantages to 100”:

In words: from the first number we subtract the “disadvantage” of the second to a hundred and assign a two-digit product of “disadvantages”.

For squares, accordingly, it is even simpler.
92*92 = (92-8)*100+8*8 = 8464
(from sielover)

Squaring may not be the most useful thing on the farm. You won’t immediately remember a case when you might need to square a number. But the ability to quickly operate with numbers and apply appropriate rules for each number perfectly develops the memory and “computing abilities” of your brain.

By the way, I think all readers of Habra know that 64^2 = 4096, and 32^2 = 1024.
Many squares of numbers are memorized at the associative level. For example, I easily remembered 88^2 = 7744 because of the same numbers. Each one will probably have their own characteristics.

I first found two unique formulas in the book “13 steps to mentalism,” which has little to do with mathematics. The fact is that previously (perhaps even now) unique computing abilities were one of the numbers in stage magic: a magician would tell a story about how he received superpowers and, as proof of this, instantly squares numbers up to a hundred. The book also shows methods of cube construction, methods of subtracting 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.

Squaring three-digit numbers is an impressive feat of mental magic. Just as squaring a two-digit number involves rounding it up or down to get a multiple of 10, squaring a three-digit number requires rounding it up or down to get a multiple of 100. Let's square the number 193.

By rounding 193 to 200 (the second factor became 186), the 3 by 3 problem became a simpler 3 by 1, since 200 x 186 is just 2 x 186 = 372 with two zeros at the end . Almost done! Now all you have to do is add 7 2 = 49 and get the answer - 37,249.

Let's try squaring 706.

When rounding the number 706 to 700, you must also change the same number up by 6 to get 712.

Since 712 x 7 = 4984 (a simple 3 by 1 problem), 712 x 700 = 498,400. Adding 6 2 = 36 gives 498,436.

The last examples aren't that scary because they don't involve addition as such. In addition, you know by heart what 6 2 and 7 2 are equal to. It is much more difficult to square a number that is more than 10 units away from a multiple of 100. Try your hand at 314 2.

In this example, 314 is decreased by 14 to round to 300 and increased by 14 to 328. Multiply 328 x 3 = 984 and add two zeros at the end to get 98,400. Then add the square of 14. If that immediately comes to mind (thanks to memory or quick calculations) that 14 2 = 196, then you're in good shape. Next, simply add 98,400 + 196 to get the final answer of 98,596.

If you need time to count 14 2, repeat "98,400" several times before continuing. Otherwise, you can calculate 14 2 = 196 and forget which number you need to add the product to.

If you have an audience you'd like to impress, you might say "279,000" out loud before you find 292. But that won't work for every problem you solve.

For example, try squaring 636.

Now your brain is really working, isn't it?

Remember to repeat “403,200” to yourself several times as you square 36 in the usual way to get 1296. The hardest part is adding 1296 + 403,200. Do this one digit at a time, from left to right, and you get the answer 404,496 I promise that once you become more familiar with squaring two-digit numbers, problems with three-digit numbers will become much easier.

Here's an even more complex example: 863 2 .

The first problem is to decide which numbers to multiply. Undoubtedly, one of them will be 900, and the other will be more than 800. But which one? This can be calculated in two ways.

1. The hard way: the difference between 863 and 900 is 37 (63's complement), subtract 37 from 863 and get 826.

2. Easy way: double the number 63, we get 126, now we add the last two digits of this number to the number 800, which ultimately gives 826.

Here's how the easy way works. Since both numbers have the same difference with the number 863, their sum must be equal to twice the number 863, that is, 1726. One of the numbers is 900, which means the other will be equal to 826.

Then we carry out the following calculations.

If you're having trouble remembering the number 743,400 after squaring the number 37, don't worry. In the following chapters you will learn the mnemonic system and learn how to remember such numbers.

Try your hand at the most difficult task so far - squaring the number 359.

To get 318, either subtract 41 (59's complement) from 359, or multiply 2 x 59 = 118 and use the last two digits. Next, multiply 400 x 318 = 127,200. Adding 412 = 1681 to this number gives a total of 128,881. That's it! If you did everything right the first time, you're great!

Let's finish this section with a big but easy task: calculating 987 2 .

EXERCISE: SQUARING THREE-DIGIT NUMBERS

1. 409 2 2. 805 2 3. 217 2 4. 896 2

5. 345 2 6. 346 2 6. 276 2 8. 682 2

9. 413 2 10. 781 2 11. 975 2

What's behind door number 1?

A mathematical platitude that stumped everyone in 1991 was an article by Marilyn Savant - the woman with the world's highest IQ (as registered in the Guinness Book of Records) - in Parade magazine. This paradox has become known as the Monty Hall problem, and it goes as follows.

You're on Monty Hall's show Let's Make a Deal. The host gives you the opportunity to choose one of three doors, behind one of which is a big prize, behind the other two are goats. Let's say you choose door number 2. But before showing what is hidden behind this door, Monty opens door number 3. There is a goat. Now, in his teasing way, Monty asks you: do you want to open door #2 or risk seeing what's behind door #1? What should you do? Assuming that Monty is going to tell you where the main prize is not, he will always open one of the "consolation" doors. This leaves you with a choice: one door with a big prize, and the other with a consolation prize. Now your chances are 50/50, right?

But no! The chance that you chose correctly the first time is still 1 in 3. The chance that the big prize will be behind the other door increases to 2/3, because the probabilities must add up to 1.

Thus, by changing your choice, you will double your chances of winning! (The problem assumes that Monty will always give the player a new choice by showing a "non-winning" door, and, when your first choice is correct, will open the "non-winning" door at random.) Think about a game with ten doors. After your first choice, let the host open eight “non-winning” doors. This is where your instincts will most likely be to change the door. People usually make the mistake of thinking that if Monty Hall doesn't know where the main prize is and opens door number 3, which turns out to be a goat (even though there could be a prize), then door number 1 has a 50 percent chance of being the right one. Such reasoning defies common sense, yet Marilyn Savant received piles of letters (many from scientists, even mathematicians) telling her that she should not have written about mathematics. Of course, all these people were wrong.

Let us now consider the squaring of a binomial and, applying an arithmetic point of view, we will speak of the square of the sum, i.e. (a + b)², and the square of the difference of two numbers, i.e. (a – b)².

Since (a + b)² = (a + b) ∙ (a + b),

then we find: (a + b) ∙ (a + b) = a² + ab + ab + b² = a² + 2ab + b², i.e.

(a + b)² = a² + 2ab + b²

It is useful to remember this result both in the form of the above-described equality and in words: the square of the sum of two numbers is equal to the square of the first number plus the product of two by the first number and the second number, plus the square of the second number.

Knowing this result, we can immediately write, for example:

(x + y)² = x² + 2xy + y²
(3ab + 1)² = 9a² b² + 6ab + 1

(x n + 4x)² = x 2n + 8x n+1 + 16x 2

Let's look at the second of these examples. We need to square the sum of two numbers: the first number is 3ab, the second 1. The result should be: 1) the square of the first number, i.e. (3ab)², which is equal to 9a²b²; 2) the product of two by the first number and the second, i.e. 2 ∙ 3ab ∙ 1 = 6ab; 3) the square of the 2nd number, i.e. 1² = 1 - all these three terms must be added together.

We also obtain a formula for squaring the difference of two numbers, i.e. for (a – b)²:

(a – b)² = (a – b) (a – b) = a² – ab – ab + b² = a² – 2ab + b².

(a – b)² = a² – 2ab + b²,

i.e. the square of the difference of two numbers is equal to the square of the first number, minus the product of two by the first number and the second, plus the square of the second number.

Knowing this result, we can immediately perform the squaring of binomials, which, from an arithmetic point of view, represent the difference of two numbers.

(m – n)² = m² – 2mn + n²
(5ab 3 – 3a 2 b) 2 = 25a 2 b 6 – 30a 3 b 4 + 9a 4 b 2

(a n-1 – a) 2 = a 2n-2 – 2a n + a 2, etc.

Let's explain the 2nd example. Here we have in brackets the difference of two numbers: the first number is 5ab 3 and the second number is 3a 2 b. The result should be: 1) the square of the first number, i.e. (5ab 3) 2 = 25a 2 b 6, 2) the product of two by the 1st and the 2nd number, i.e. 2 ∙ 5ab 3 ∙ 3a 2 b = 30a 3 b 4 and 3) the square of the second number, i.e. (3a 2 b) 2 = 9a 4 b 2 ; The first and third terms must be taken with a plus, and the 2nd with a minus, we get 25a 2 b 6 – 30a 3 b 4 + 9a 4 b 2. To explain the 4th example, we only note that 1) (a n-1)2 = a 2n-2 ... the exponent must be multiplied by 2 and 2) the product of two by the 1st number and by the 2nd = 2 ∙ a n-1 ∙ a = 2a n .

If we take the point of view of algebra, then both equalities: 1) (a + b)² = a² + 2ab + b² and 2) (a – b)² = a² – 2ab + b² express the same thing, namely: the square of the binomial is equal to the square of the first term, plus the product of the number (+2) by the first term and the second, plus the square of the second term. This is clear because our equalities can be rewritten as:

1) (a + b)² = (+a)² + (+2) ∙ (+a) (+b) + (+b)²
2) (a – b)² = (+a)² + (+2) ∙ (+a) (–b) + (–b)²

In some cases, it is convenient to interpret the resulting equalities in this way:

(–4a – 3b)² = (–4a)² + (+2) (–4a) (–3b) + (–3b)²

Here we square a binomial whose first term = –4a and second = –3b. Next we get (–4a)² = 16a², (+2) (–4a) (–3b) = +24ab, (–3b)² = 9b² and finally:

(–4a – 3b)² = 6a² + 24ab + 9b²

It would also be possible to obtain and remember the formula for squaring a trinomial, a quadrinomial, or any polynomial in general. However, we will not do this, because we rarely need to use these formulas, and if we need to square any polynomial (except a binomial), we will reduce the matter to multiplication. For example:

31. Let us apply the obtained 3 equalities, namely:

(a + b) (a – b) = a² – b²
(a + b)² = a² + 2ab + b²
(a – b)² = a² – 2ab + b²

to arithmetic.

Let it be 41 ∙ 39. Then we can represent this in the form (40 + 1) (40 – 1) and reduce the matter to the first equality - we get 40² – 1 or 1600 – 1 = 1599. Thanks to this, it is easy to perform multiplications like 21 ∙ 19; 22 ∙ 18; 31 ∙ 29; 32 ∙ 28; 71 ∙ 69, etc.

Let it be 41 ∙ 41; it’s the same as 41² or (40 + 1)² = 1600 + 80 + 1 = 1681. Also 35 ∙ 35 = 35² = (30 + 5)² = 900 + 300 + 25 = 1225. If you need 37 ∙ 37, then this is equal to (40 – 3)² = 1600 – 240 + 9 = 1369. Such multiplications (or squaring two-digit numbers) are easy to perform, with some skill, in your head.

*squares up to hundreds

In order not to mindlessly square all the numbers using 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 a single-digit number. You just need to add a couple of zeros.
70 * 70 = 4900.
Marked in red in the table.

Rule 2 (cuts off 10 numbers)

Rule 3 (cuts off 8 numbers)

Rule 4 (cuts off 8 numbers)

Rule 5 (cuts off 8 numbers)

Rule No. 6 (cuts off 32 numbers)

Formulas (24 digits left)

For numbers from 25 to 50
XX * 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 - 25) + (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 the expansion of the square of a sum (a special case of Newton’s binomial):
(a+b)^2 = a^2 + 2ab + b^2.
56^2 = 50^2 + 2*50*6 + 6*2 = 2500 + 600 + 36 = 3136.

If the topic of quick counting is interesting, I will write more.
Please write comments about errors and corrections in PM, thanks in advance.

Today we will learn how to quickly square large expressions without a calculator. By large, I mean numbers ranging from ten to one hundred. Large expressions 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 in general. As an example, I propose 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.

The problem with 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 multiplying quickly? 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 does not seem fast enough and cool enough, I propose an even faster method of 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, it is possible to 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

Using 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 is 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 value). 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 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 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!