Squaring 2 Digit numbers in your head fast is easy and fun once you know these three simple steps that break it into three simple calculations. There is also two special cases where we can take a shortcut on the general method and get the result even faster and easier.
The two special cases are.
- Numbers Ending in 5
- Numbers Starting with 5
You can watch the video to see several examples for both the special cases and the general method for squaring any 2 digit numbers or continue reading if you prefer not to watch.
2 Digit Numbers Ending in 5
For these numbers because they end in 5 the last two digits of the result are always going to be 25. Knowing this means the only thing we need to work out is what the leading digits are going to be.
The formula to calculate the leading digits is:
Multiply the tens digit by the next larger digit.
What is “the next larger digit”? It is the digit you get when you add 1 to the digit you are looking at.
As it ends in _5 we know the answer ends in _25 so we just need to calculate the leading digits.
The tens digit is 7, so we multiply it by the next larger digit, which is 8.
- And there we have our leading digits, so the answer to our problem is:
The tens digit is 2 so the next larger digit is 3 and we multiply them together.
- So our leading digit is 6.
We know it ends in _25 so the answer to our problem is:
- Now you know this little trick calculating the square of two-digit numbers ending in 5 should be easy to do in your head from now on.
2 Digit Numbers Starting with 5
The answer for two digit numbers starting with 5 can be calculated in two easy steps:
- Square the units digit.
- Add 25 to the units digit.
Step 1 – Square the units digit.
This step should always give us the last two digits of the result, so if we get only a single digit we put a leading zero to make up the two digits.
Step 2 – Add 25 to the units digit.
To get the leading digits of the result we simply add 25 to the units digit. Why 25, well that is the result of 5 squared.
- Putting it all together:
- Now we will move on to the General method for calculating the squares of any two digit number.
Squaring Any 2 Digit Numbers
This general method of squaring two digit numbers will work for any of the two digit numbers, including those special cases mentioned above.
There are three steps to the method:
- Square the units digit.
- Multiply the two digits together then double the result.
- Square the tens digit
The easiest way to explain these steps is with an example, say 47 squared.
Step 1 – Square the units digit.
The units digit is 7.
- So we would write 9, and we carry the 4
Step 2 – Multiply the two digits together then double the result.
We take the 4 and the 7 and simply multiply these two digits together and double the result.
- We add the 4 we carried from step 1 to the 56.
- We then write the 0 and carry the 6.
Step 3 – Square the tens digit.
The tens digit is 4, and we square this.
- We add the 6 we carried from step 2 to the 16.
- Write down the 22, and we have our result.
Once you know the rules then you should also be able to reverse direction and work the solution from left to right. This is better especially if you are doing the whole calculation mentally rather than writing it down as you go.
Lets reverse the order of the rules and work from left to right for .
Reverse Step 1 – Square the tens digit.
As the tens unit is 3 this means we are really looking at 30 so we will use 30 as we calculate as it will help to prevent us loosing our place in the answer.
Reverse Step 2 – Multiply the two digits together then double the result.
Again the 3 is 30 so we continue to use this as we work.
- Adding the results of each step we get:
Reverse Step 3 – Square the units digit.
The units digit is 6 so we square this.
- We add this to our running total.
- So our answer is: .
There you have it a very easy way to calculate the square of a two digit number mentally.
So don't even think about reaching for that calculator the next time you need to find the square of a two digit number you should be able to mentally solve the problem before you even hit the calculator's “on” button.
For those of you who have read through the How to multiply double digit numbers may have felt a sense of déjà vu because it is the SAME METHOD! It is just described a little differently. What is squaring a 2 digit number? It is just multiplying a 2 digit number by itself.
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Mental Math—BE THERE OR B2: SQUARING TWO-DIGIT NUMBERS
—Paul Ramirez, co-editor
Squaring numbers in your head (multiplying a number by itself) is one of the easiest yet most impressive feats of mental calculation you can do. I can still recall where I was when I discovered how to do it.
I was 13, sitting on a bus on the way to visit my father at work in downtown Cleveland. It was a trip I made often, so my mind began to wander.
I’m not sure why, but I began thinking about the numbers that add up to 20, and I wondered, how large could the product of two such numbers get?
I started in the middle with 10 × 10 (or 102), the product of which is 100. Next, I multiplied 9 × 11 = 99, 8 × 12 = 96, 7 × 13 = 91, 6 × 14 = 84, 5 × 15 = 75, 4 × 16 = 64, and so on. I noticed that the products were getting smaller, and their difference from 100 was 1, 4, 9, 16, 25, 36, … —or 12, 22, 32, 42, 52, 62, … ( see table below).
I found this pattern astonishing. Next I tried numbers that add to 26 and got similar results. First I worked out 132 = 169, then computed 12 × 14 = 168, 11 × 15 = 165, 10 × 16 = 160, 9 × 17 = 153, and so on. Just as before, the distances these products were from 169 were 12, 22, 32, 42, and so on (see table below).
There is actually a simple algebraic explanation for this phenomenon (see Why These Tricks Work). At the time, I didn’t know my algebra well enough to prove that this pattern would always occur, but I experimented with enough examples to become convinced of it.
Then I realized that this pattern could help me square numbers more easily. Suppose I wanted to square the number 13. Instead of multiplying 13 × 13,
Why not get an approximate answer by using two numbers that are easier to multiply but also add to 26? I chose 10 × 16 = 160. To get the final answer, I just added 32 = 9 (since 10 and 16 are each 3 away from 13). Thus, 132 = 160 + 9 = 169. Neat!
This method is diagrammed as follows:
Now let’s see how this works for another square:
To square 41, subtract 1 to obtain 40 and add 1 to obtain 42. Next multiply 40 × 42. Don’t panic! This is simply a 2-by-1 multiplication problem (specifically, 4 × 42) in disguise. Since 4 × 42 = 168, 40 × 42 = 1680. Almost done! All you have to add is the square of 1 (the number by which you went up and down from 41), giving you 1680 + 1 = 1681.
Can squaring a two-digit number be this easy? Yes, with this method and a little practice, it can. And it works whether you initially round down or round up. For example, let’s examine 772, working it out both by rounding up and by rounding down:
In this instance the advantage of rounding up is that you are virtually done as soon as you have completed the multiplication problem because it is simple to add 9 to a number ending in 0!
In fact, for all two-digit squares, I always round up or down to the nearest multiple of 10. So if the number to be squared ends in 6, 7, 8 or 9, round up, and if the number to be squared ends in 1, 2, 3 or 4, round down. (If the number ends in 5, you do both!) With this strategy you will add only the numbers 1, 4, 9, 16 or 25 to your first calculation.
Let’s try another problem. Calculate 562 in your head before looking at how we did it, below:
Squaring numbers that end in 5 is even easier. Since you will always round up and down by 5, the numbers to be multiplied will both be multiples of 10. Hence, the multiplication and the addition are especially simple. We have worked out 852 and 352, below:
As you saw in Chapter 0, when you are squaring a number that ends in 5, rounding up and down allows you to blurt out the first part of the answer immediately and then finish it with 25. For example, if you want to compute 752, rounding up to 80 and down to 70 will give you “Fifty-six hundred and … twenty-five!”
For numbers ending in 5, you should have no trouble beating someone with a calculator, and with a little practice with the other squares, it won’t be long before you can beat the calculator with any two-digit square number.
Even large numbers are not to be feared. You can ask someone to give you a really big two-digit number, something in the high 90s, and it will sound as though you’ve chosen an impossible problem to compute.
But, in fact, these are even easier because they allow you to round up to 100.
Let’s say your audience gives you 962. Try it yourself, and then check how we did it.
Wasn’t that easy? You should have rounded up by 4 to 100 and down by 4 to 92, and then multiplied 100 × 92 to get 9200. At this point you can say out loud, “Ninety-two hundred,” and then finish up with “sixteen” and enjoy the applause!
Why These Tricks Work
This section is presented for teachers, students, math buffs and anyone curious as to why our methods work. Some people may find the theory as interesting as the application.
Fortunately, you need not understand why our methods work in order to understand how to apply them. All magic tricks have a rational explanation behind them, and mathemagical tricks are no different.
It is here that the mathemagician reveals his deepest secrets!
- In this chapter on multiplication problems, the distributive law is what allows us to break down problems into their component parts. The distributive law states that for any numbers a, b and c:
- (b + c) × a = (b × a) + (c × a)
- That is, the outside term, a, is distributed, or separately applied, to each of the inside terms, b and c. For example, in our first mental multiplication problem of 42 × 7, we arrived at the answer by treating 42 as 40 + 2, then distributing the 7 as follows:
- 42 × 7 = (40 + 2) × 7 = (40 × 7) + (2 × 7) = 280 + 14 = 294
You may wonder why the distributive law works in the first place. To understand it intuitively, imagine having 7 bags, each containing 42 coins, 40 of which are gold and 2 of which are silver. How many coins do you have altogether? There are two ways to arrive at the answer.
In the first place, by the very definition of multiplication, there are 42 × 7 coins. On the other hand, there are 40 × 7 gold coins and 2 × 7 silver coins. Hence, we have (40 × 7) + (2 × 7) coins altogether. By answering our question two ways, we have 42 × 7 = (40 × 7) + (2 × 7).
Notice that the numbers 7, 40 and 2 could be replaced by any numbers (a, b or c) and the same logic would apply. That’s why the distributive law works!
Using similar reasoning with gold, silver and copper coins we can derive:
(b + c + d) × a = (b × a) + (c × a) + (d × a)
Hence, to do the problem 326 × 7, we break up 326 as 300 + 20 + 6, then distribute the 7, as follows: 326 × 7 = (300 + 20 + 6) × 7 = (300 × 7) + (20 × 7) + (6 × 7), which we then add up to get our answer. As for squaring, the following algebra justifies my method. For any numbers A and d
A2 = (A + d) × (A – d) + d2
Here, A is the number being squared; d can be any number, but I choose it to be the distance from A to the nearest multiple of 10. Hence, for 772, I set d = 3 and our formula tells us that 772 = (77 + 3) × (77 − 3) + 32 = (80 × 74) + 9 = 5929. The following algebraic relationship also works to explain my squaring method:
- (z + d)2 = z2 + 2zd + d2 = z( z + 2d) + d2
- Hence, to square 41, we set z = 40 and d = 1 to get:
- 41² = (40 + 1) 2 = 40 × (40 + 2) + 12 = 1681
- Similarly, (z − d)2 = z( z − 2d) + d2
- To find 772 when z = 80 and d = 3,
- 77² = (80 − 3)2 = 80 × (80 − 6) + 32 = 80 × 74 + 9 = 5929
Squaring 2-Digit Numbers
Let's see an example different from the ones at the index page. Say, find 32².
First add the last digit (2) to the number itself: 32 + 2 = 34. Multiply the sum by the first digit: 34 × 3 = 102. Square the last digit: 2² = 4. Append that square to the product just computed: 1024. If the square is a 2-digit number, append its last digit and carry the first digit to the last digit of the product.
Why does this work?
Let the number be N = 10a + b.
|(10a + b)²||= 100a² + 20ab + b²|
|= 10a(10a + 2b) + b²|
|= 10a(10a + b + b) + b²|
|= 10a(N + b) + b².|
So, to compute the square of N = 10a + b, first find N + b. Then multiply that by the first digit a to get a(N + b). Square the second digit to get b². “Appending b²” mean multiplying a(N + b) by 10 and adding b².
In fact the method is not restricted to 2-digit numbers. a may have 2 or more digits as well. The calculations become more complex of course.
Find 215². 215 + 5 = 220. 220 × 21 = 4400 + 220 = 4620. 5² = 25. 4620·10 + 25 = 46225..
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Square Tricks: Two, Three & Four digit numbers – JustQuant.com
The square trick explained in this section will help you find the square of any number between 1 and 10000. How? Using a Vedic Math Technique, we will first learn a shortcut to find the square of any two digit number. We will then apply this trick to calculate the squares of three and four digit numbers.
The Vedic Math Technique that we are going to use here is known as Duplex.
- According to duplex terminology,
- Duplex of single digit number, say ‘a’ is given byD(a) = $a^2$
- Duplex of double digit number, say ‘ab’ is given byD(ab) = 2ab
- Given below is the table containing the duplex of some numbers
NumberDuplex of the number 3 Duplex of 3 => D(3) = $3^2$ = 9 9 Duplex of 9 => D(9) = $9^2$ = 81 12 Duplex of 12 => D(12) = 2 x 1 x 2 = 4 43 Duplex of 43 => D(43) = 2 x 4 x 3 = 24
- Now with this knowledge of Duplexes, we will see how we can find square of two digit numbers easily.
Square of Two Digit Numbers
Using the Vedic math technique explained lets calculate the square of two digit numbers. At the end of this section you will be to square any number up to 100 in your head within seconds.
Once you understand this shortcut you can easily extend it to find the squares of three and four digit numbers as explained in the subsequent sections below.
Now, lets get started. You can check out the video below, where we tried to demonstrate this trick in an intuitive way that may find it easier to understand than through a normal learning through reading experience.
Math Trick to find Square of Two Digit Numbers:
- Consider a general two digit number, say, ‘ab’.
- The square of ‘ab’ will have three parts.
- ab2 = left most part| middle part | right most part
- During calculations, we shall pass from the leftmost duplex to the leftmost duplex.
- The left most part will be duplex of ‘a’, the middle part will be duplex of ‘ab’, finally the right most part will be duplex of ‘b’.
e ab2 = D(a)| D(ab) | D(b)
- = a2 | 2ab | b2
- 12$^2$ = D(1)| D(12) | D(2)
- = 1$^2$ | 2x1x2 | 2$^2$
- = 1 | 4 | 4
- Hence, 12 = 144 using the duplex methodology
- 23$^2$ = D(2)| D(23) | D(3)
- = 2$^2$ | 2x2x3 | 3$^2$
- = 4 | 12 | 9
Here, the middle portion has more than two digits, please note that only the leftmost part can have more than one digit. For the rest of the parts, we need to carry over the number preceding the units digit, to the immediate left part , and add it there respectively.
- Hence for 12 which has ‘1’ as the non units digit, we need to carry over ‘1’ to the immediate left which turns 4 to 5. Therefore,
- 23$^2$ = 4+1 | 2 | 9
- Hence, 23$^2$ = 529 using the duplex methodology
To practice two digit squares, check out our exercises in Two Digit Squares
Square of Three Digit Numbers
Now lets learn the trick to find square of 3 digit number using Vedic Math. At the end of this section you will be able to square numbers from 1 to 1000.
The shortcut to find the square of 3 digit numbers involves using the duplex methodology as explained above. To apply this shortcut, we will also need to find the duplex of 3 digit numbers in addition to the duplex of single and double digit numbers as described above.
Duplex of 3 digit numbers
- For a number with three digits, say ‘abc’,
- Duplex of ‘abc’ is given by D(abc) = 2ac + $b^2$
- For example,
- Duplex of 125 => D(125) = 2x1x 5 + $2^2$ = 14
- Duplex of 756 => D(756) = 2x7x6 + $5^2$ = 109
Now with this knowledge of Duplexes, we will see the shortcut on how we can find the square of three digit numbers.
Math Trick to find Square of Three Digit Numbers:
- Consider a general three digit number, say, ‘abc’.
- The square of ‘abc’ will have five parts as shown below(each part numbered with a digit for our convenience)
- abc2 = 5 | 4 | 3 | 2 | 1
- During calculations, we shall pass from the rightmost duplex to the leftmost duplex.
- The rightmost part(1) will be duplex of ‘c’, the next part(2) will be duplex of bc, the middle part(3) will be duplex of ‘abc’, the next part(4) will be duplex of ab and finally the left most part(5) will be duplex of ‘a’.
- i.e abc2 = D(a) | D(ab) | D(abc)| D(bc) | D(c)
- 321$^2$ =D(3) | D(32) | D(321)| D(21) | D(1)
- = 3$^2$ | 2x3x2 | 2x3x1 + 2$^2$ | 2x2x1 | 1$^2$
- = 9 | 12 | 10 | 4 | 1
As mentioned above only the leftmost part can have more than one digit.
For the rest of the parts, we need to carry over the number preceding the units digit, to the immediate left part , and add it there respectively.
- Hence for 10 which has ‘1’ as the non units digit, we need to carry over ‘1’ to the immediate left which turns 12 to 13 and for 13 which has ‘1’ as the non units digit, we need to carry over ‘1’ to the immediate left which turns 9 to 10.
- 321$^2$ = 10 | 3 | 0 | 4 | 1
- = 103041
- 7912 =D(7) | D(79) | D(791)| D(91) | D(1)
- = 7$^2$ | 2x7x9 | 2x7x1 + 9$^2$ | 2x9x1 | 1$^2$
- = 49 | 126 | 95 | 18 | 1
- = 625681
To practice two digit squares, check out our exercises in Three Digit Squares
Square of Four Digit Numbers
- Proceeding on the same methodology as for 2 and 3 digit squares as mentioned above, to quickly square a 4 digit number, we must know duplex of 4 digit number as well.
- For a number with four digits, say ‘abcd’, Duplex of ‘abcd’ => D(abcd) = 2ad + 2bc
- The square of ‘abcd’ will have seven parts as shown below
- abcd2 = D(a) | D(ab) | D(abc) | D(abcd)| D(bcd) | D(cd) | D(d)
- = a2 | 2ab | 2ac + b2 | 2ad+2bc | 2bd + c2 | 2cd | d2
- 1221$^2$= D(1) | D(12) | D(122) | D(1221)| D(221) | D(21) | D(1)
- = 1$^2$ | 2x1x2 | 2x1x2 + 2$^2$ | 2x1x1+2x2x2 | 2x2x1 + 2$^2$ | 2x2x1 | 1$^2$
- = 1 | 4 | 8 | 10 | 8 | 4 | 1
As mentioned above only the leftmost part can have more than one digit. For the rest of the parts, we need to carry over the number preceding the units digit, to the immediate left part , and add it there respectively.
- Hence for 10 which has ‘1’ as the non units digit, we need to carry over ‘1’ to the immediate left which turns 8 to 9
- 1221$^2$ = 1 | 4 | 9 | 0 | 8 | 4 | 1
- = 1490841
- 9654$^2$ = D(9) | D(96) | D(965) | D(9654)| D(654) | D(54) | D(4)
- = 9$^2$ | 2x9x6 | 2x9x5 + 6$^2$ | 2x9x4+2x6x5 | 2x6x4 + 5$^2$ | 2x5x4 | 4$^2$
- = 81 | 108 | 126 | 132 | 73 | 40 | 16
- = 93199716
Arithmetical calculations using only the human brain
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Mental calculation comprises arithmetical calculations using only the human brain, with no help from any supplies (such as pencil and paper) or devices such as a calculator.
People use mental calculation when computing tools are not available, when it is faster than other means of calculation (such as conventional educational institution methods), or even in a competitive context. Mental calculation often involves the use of specific techniques devised for specific types of problems . People with unusually high ability to perform mental calculations are called mental calculators or lightning calculators.
Many of these techniques take advantage of or rely on the decimal numeral system. Usually, the choice of radix is what determines which method or methods to use.
Methods and techniques
Casting out nines
Main article: Casting out nines
After applying an arithmetic operation to two operands and getting a result, the following procedure can be used to improve confidence in the correctness of result:
- Sum the digits of the first operand; any 9s (or sets of digits that add to 9) can be counted as 0.
- If the resulting sum has two or more digits, sum those digits as in step one; repeat this step until the resulting sum has only one digit.
- Repeat steps one and two with the second operand. There are two single-digit numbers, one condensed from the first operand and the other condensed from the second operand. (These single-digit numbers are also the remainders one would end up with if one divided the original operands by 9; mathematically speaking, they are the original operands modulo 9.)
- Apply the originally specified operation to the two condensed operands, and then apply the summing-of-digits procedure to the result of the operation.
- Sum the digits of the result that were originally obtained for the original calculation.
- If the result of step 4 does not equal the result of step 5, then the original answer is wrong. If the two results match, then the original answer may be right, though it is not guaranteed to be.
- Say that calculation results that 6338 × 79 equals 500702
- Sum the digits of 6338: (6 + 3 = 9, so count that as 0) + 3 + 8 = 11
- Iterate as needed: 1 + 1 = 2
- Sum the digits of 79: 7 + (9 counted as 0) = 7
- Perform the original operation on the condensed operands, and sum digits: 2 × 7 = 14; 1 + 4 = 5
- Sum the digits of 500702: 5 + 0 + 0 + (7 + 0 + 2 = 9, which counts as 0) = 5
- 5 = 5, so there is a good chance that the prediction that 6338 × 79 equals 500702 is right.
The same procedure can be used with multiple operations, repeating steps 1 and 2 for each operation.
While checking the mental calculation, it is useful to think of it in terms of scaling. For example, when dealing with large numbers, say 1531 × 19625, estimation instructs one to be aware of the number of digits expected for the final value.
A useful way of checking is to estimate. 1531 is around 1500, and 19625 is around 20000, so a result of around 20000 × 1500 (30000000) would be a good estimate for the actual answer (30045875). So if the answer has too many digits, a mistake has been made.