# The Criss-Cross Method: An Alternative Form of Multiplication

Traditionally, multiplication of multiple digit numbers is done as a series of multiplications that are eventually added together to form a final answer. The criss-cross method is a variation on this technique that allows for much quicker processing of the problem without the need for a calculator or extensive use of paper space. There are many situations, such as trips to the grocery store, where you will find a need to perform multiplication of odd numbers in order to stay within a budget as you shop.

This system of multiplication is adopted from Vedic Mathematics’ *URDHVA-TIRYAK SUTRA, *which means vertically and cross-wise.

To start with, we will look at a simple example just to get a grasp on the steps involved in the method. Later we will apply it to a slightly more advanced problem to show how to handle carrying numbers from one digit to the next. For now, we will multiply 111 by 111.

111

x 111

First, you will take the right-hand digits and multiply them together. This will give you the one’s digit of the answer, as shown below with the digits used encased in brackets.

11[1]

x11[1]

____

* – — 1

Next, multiply the one’s digit of the top number by the ten’s digit of the bottom number, and the one’s digit of the bottom number by the ten’s digit of the top number. Once you have those values, add them together, and you will have the ten’s digit of the answer. The digits you multiply together are enclosed in the same type of bracket. This gives (1*1)+(1*1), so the ten’s digit is equal to two.

1{1}[1]

x 1[1]{1}

______

* 321

For the next step, all digits of the number will be involved in order to find the middle of the answer. Multiply the one’s digit of one to the hundreds digit of the other, and then multiply the ten’s digit of both together, then finally add them all together. This will give you (1*1)+(1*1)+(1*1) for a value of 3. As above, the digits paired together are enclosed in the same type of bracket.

{1}[1]

x [1]{1}

_______

* 321

The fourth step is similar to the second step, just moved one place to the left. You multiply the ten’s digit of one number by the hundred’s digit of the other number. Again you will get (1*1) + (1*1), showing the thousand’s digit of the answer is equal to 2.

{1}[1]1

x {1}[1]1

______

* 2321

For the final step, simply multiply the left-hand number of both numbers together to get a value of 1 for the ten-thousand’s place.

{1}11

x {1}11

____

12321

If you are a teacher you can go ahead and teach this wonderful teaching technique to your folks. If not, teaching certification programs like the St Joseph University online program offer aspiring math teachers a way to earn a degree at their own pace.

To give a real-world example, consider that you want to buy 15 of some product for $1.25 each. You can consider 15 to be a three digit number, where the third digit is equal to 0.

$1.25

x 015

_____

* – — –

You will notice that immediately you will have to deal with carrying over value from one digit to the other. This works very similar to regular multiplication methods. You simply take any value in the tens digit of that step as an addition to the next step. When you multiply 5 by 5 in this example to get 25, you would place the 5 as the one’s digit of the answer, and add the 2 to the next step to find the ten’s digit. This makes the second step equal to (5*2) + (1*5) +2, for a total of 17. Use the 7 as the value for the ten’s digit, and carry the one over to the next step. Here you would end up with (5*1) + (1*2) + (0*5) + 1, coming to a value of 8. This time there is no number to carry over, so proceed through the rest of the problem as normal. The thousands place is 1 via (0*2)+(1*1), and the ten-thousands place is equal to 0 via 0*1. This comes to 01875, then drop the 0 from the end.

As in regular multiplication, you count the total number of places behind the decimal point, and add the same number to the answer. This means the items would cost $18.75 to purchase.

Justin McGenity is a freelance writer, science geek, and self-proclaimed philosopher. His hobbies include studying, video games, and socializing. You can find him writing for such sites as Degree Jungle a resource for university students.

thanks vineet its indeed a very useful technique , but when the number of digits for multiplication increases it not that easy , but definitely better than what taught in schools

The first comment i guess is not correct relating to finding the angles of the triangle given the sides of the triangle. the example given is also absurd where 2 6 10 will not form a triangle as(2+6<10 which cant be a triangle) also one can cross check for 3 4 5 sides of a right triangle where the given formula does not give right angle to any angle of the triangle. Correct me if i am wrong.

sorry i was referring to the flamefletcher method to find angles given the sides of a triangle.

yeah surya you are right.

I found out now that my theory is only true for Acute angled triangles.

Not for right angled traingles and for obtuse triangles.

The example i took below is an obtuse angled triangle.

You can test it with any acute angled triangle and tell me the result and if you can …. prove thats also wrong.

To find out the angles of an Obtuse angled triangle and other side.

Use the Parallelogram law of vectors in Physics.

Google it and you will find the formula.

And for a right angled triangle.

This is the formula.

A,B,C are Vertices

AC is hypotenuse.

90/(AB+BC)=P

And then multiply P with other sides except the hypotenuse

Ok well let me submit something.

I hope you will like it vineet.

In a triangle.

we know the relation ship between sides and angles with trignometry.

but how can we find out angles if we know the sides in a simple method?

use ‘&’ as the proportionality symbol.

in a triangle ABC the side opposite to angleA is BC , side opposite to angleC is AB etc.

so as the side AB increases then angleC definitely increases. right?

so

AngleA & BC (& —> means directly proportional to)

AngleB & AC

AngleC & AB

Let ‘P’ be a constant.

AngleA = P(BC) —–> 1

AngleB = P(AC) ——>2

AngleC = P(AB) ——>3

Adding 1,2,3 we get……

AngleA+AngleB+AngleC = P(BC+AC+AB)

But AngleA+AngleB+AngleC = 180

therefore, 180= P(BC+AC+AB)

=> P= 180/(BC+AC+AB)

from formula if we know the 3 sides of a triangle we can find the constant ‘P’.

and to get an angle.

AngleA = P(BC)

so multiply P with a side opposite to the angle to get that angle.

here is an Example sum.

2,6,10 are the 3 sides of a triangle. Find out all their angles.

From my formula above, P = 180/(AB+BC+CA)

AB+BC+CA = 2+6+10 = 18

=> P=180/18

=10

therefore P=10 degrees per unit distance.

that means if there is an increase of 1 unit distance then the angle opposite to that side increases by 10 degrees.

AngleA = 10(BC) = 10(2) = 20 degrees

AngleB = 10(CA) = 10(6) = 60 degrees

AngleC = 10(AB) = 10(10)=100 Degrees

Therefore the angles are found.

Note: This formula can be used for any type of triangle and this formula is 100% usable and proved.

Note:The value of P is different for different triangles. So we always have to find P before finding angles.

Feel Free to use it and please publish this on this site.

Hello , thanks admin for writing so many methods , tricks , techniques for solving Additions ,Subtractions , Multiplications & Divisions in Quicker math”s ! Interested students are requested to Learn one by one to know Basic Math’s & Vedic Maths !! I am also one of the Tutor to teach Basic Math’s & Vedic Maths as Home Tuition Classes & through Mail id for Outstation Students .To know more details please do write on my mail id is basic.maths483@gmail.com thanks once again Quicker Math’s * I have passed M.Sc. ( STAT ) in Ist class from Nagpur University in 1981 .

Good working. But its some what confuse & difficult to understand at a glance.