Quicker Maths

Quickly Multiply by 21

Posted on September 2, 2011

Rule for multiplying any number by 21

Start from left. Double the first digit and add it to left side neighboring digit. Repeat the steps for subsequent digits. The last number will be same as the last number of the multiplied number.

 

This rule is very much like the shortcut for multiplying by 11. Since 21 is sum of 11 and 10, it does belong to the same family of short cuts.

Let’s understand the whole concept with an example. Let’s multiply 5392 by 21.

The first digit of the answer will be equal to twice the first digit of 5392. To make the rule consistent assume there is a zero before the number.

Shortcut for Addition of Consecutive Numbers

Posted on August 26, 2011

In this post I’ll share with you a useful shortcut maths trick for “finding out the sum of consecutive numbers”. For example, this trick I am talking about can help you in finding the sum of all the numbers from 23 to 31 or any other set of numbers.

Shortcut Addition Trick

Add the smallest number to the largest number of the given set of consecutive numbers. Then multiply the result by the number of numbers in the set. Finally divide the result by two.

Solving the above example, let’s find: 23+24+25+26+27+28+29+30+31

Polish Hand Magic

Posted on July 26, 2011

This is a guest post by Danielle

I learned this problem from The Puzzler’s Elusion (flipkart link) by Dr. Dennis E. Shasha. It’s called Polish Hand Magic. It’s not a method of counting faster, but it is a fun little trick to show young kids (and adults) who know their multiplication tables.

In this Magical Polish tradition, a closed fist equals 5. Let’s say you want to multiply 7 x 8.

7 is represented by …||, or three fingers down and two up. 8 is represented by ..|||, or two fingers down and three up. Find the sum of the fingers that are up, in this case, the amount of vertical lines. Then multiply the number of finger down. So:

Mentally Multiply by 5, 25, 50, 250

Posted on June 4, 2011

This is a simple quicker math trick but it can be very useful for young students to solve seemingly difficult calculations. I will be glad to get your feedback on this.

Mental multiplication by 5, 25, 50, 250, 500 and so on.

Any number can be expressed in different ways. For example, 5 can be expressed as 10x(1/2).

Trick: Multiplication by 5

Step 1: Multiply the number by 10, i.e. simply place a zero after the number.

Step 2: Halve the resultant number.

Example 1:

5 × 136 = ten times of 136 i.e. 1360 should be divided by 2 = 1360/2 = 680

Example 2:

5 × 343, half of 3430 is 1715

Also check out, how to mentally multiply by 111?

How to convert from decimal to other number systems

Posted on March 12, 2011

This post will be of special interest for people who are regularly in touch with mathematics. Students preparing for competitive examinations usually have Base System (Number Systems) in the list of their topics under quantitative aptitude.

Conversion from decimal to binary and other number bases

In order to convert a decimal number into its representation in a different number base, we have to be able to express the number in terms of powers of the other base. For example, if we wish to convert the decimal number 100 to base 4, we must figure out how to express 100 as the sum of powers of 4.

100 = (1 x 64) + (2 x 16) + (1 x 4) + (0 x 1)

= (1 x 4^3) + (2 x 4^2) + (1 x 4^1) + (0 x 4^0)

Then we use the coefficients of the powers of 4 to form the number as represented in base 4:

Squaring any 2-digit number

Posted on December 7, 2010

A guest post by Maria Rainier

Shortcut to Squaring Any 2-Digit Number

What do you do when your calculator has been confiscated and the world is depending on you to square a two-digit number within a minute? Don’t panic – just follow three simple steps that require basic addition and multiplication, and you’ll be able to solve the problem in no time. If you practice enough, you’ll even be able to complete each step mentally, rendering scratch paper unnecessary. This will save you time on drills and strengthen your skills so you can tackle other challenges. Eventually, you’ll be able to solve multi-step squaring problems without ever breaking a sweat – or a pencil.

Divisibility Rules for 7 , 11 and 13

Posted on October 20, 2010

A guest post by Dr. Cecily Zacharias from Oklahoma city, The United Sates of America. Currently she is an instructor at Oklahoma City Community College

Rules for divisibility of 7 , 11 and 13

It is equally good for 11 and 13.

Step 1    Divide into groups of three from the right.                                                      245782          245  782

______________________________________________________________-1           1
Step 2.   Write 1,-1,1,-1(alternate 1 and -1) in a row above the number          245       782
( start at the right end  and go left)

____________________________________________________________-1            1
Step 3.   Divide each Group by 7 (or 11 or 13, whatever the divisor is )     245         782
0             5

* You can avoid step 2, by simply subtracting first remainder from the second. In this case it will be simply, 5 – 0 = 5
Step 4.  Multiply the corresponding numbers in the top row and bottom row and add     0x -1  +  5x 1   = 5

** Step 4 can also be avoided.
Step 5. a.  If the sum obtained is zero, The number is divisible by 7 (or 11 or 13 )
b. If the sum is positive, then that is the remainder when we divide the number by 7 (  or 11 or 13 )
c. If the sum  is negative, then add 7 (or 11 or 13 )  to get the remainder.
The sum is always less than the divisor.

In the example given, the sum is 5 . Which can be verified.
When 245782 is divided by 7 by long division, the quotient is 35111 and remainder is 5.

If we test for divisor 11, the bottom row will be    3      1
The sum of products of the two rows is -2. since it is negative , add 11 .
So the remainder will be 9
Actual division gives the quotient to be 22343 and remainder 9.

It is the same  method for dividing by 13 too.

If you want you can simplify the  steps 2 ,to 4 as
Find the remainders in each group and alternately add and subtract the remainders starting from  the right. Then use step 5.

On behalf of all the QuickerMaths.com users, I  am highly grateful for her contribution.

Divisibility Rule of 7, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47

Posted on September 28, 2010

You might have seen divisibility rules of various numbers. But most of them very conveniently skip the ones which are very difficult and a divisibility rule for which is very much required.  This post includes the divisibility rule for some such numbers like 7, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47.

 

While reading this you have to be little patient. Read this carefully and try to apply it practically. If you master divisibility rules or tests explained below, I am sure these will come very handy in various examinations including competitive ones.

Quick Multiplication up to 20 x 20

Posted on September 22, 2010

“I’m having trouble above 10x10.”

This was a statement I heard many times while interacting with students preparing for competitive examinations including CAT. This was in response to my appeal to them to memorize tables up to 20x20.

Today I am posting here on QuickerMaths.com, the method which I recommend to my students too.

How to multiply up to 20x20 in your head?

Assumption: You know your multiplication table reasonably well up to 10×10.

I am trying to explain this with an example,

Quick method to evaluate polynomials – Horner’s method

Posted on July 20, 2010

This is a guest post by Nandeesh H.N. of Kolkata

How to find the value of a Polynomial Function?

Horner's method is commonly used to find the roots of a polynomial function. However it can also be used to evaluate the polynomial function for a given value of x.

Suppose, we want to evaluate the polynomial

p(x) = 4x^5 - 3x^4 + 7x^3 + 6x^2 + 3x + 9 at x = 2.41.

The usual method of evaluation is to evaluate each product (such as 4*2.41^5 or 7*2.41^3) separately and then add. The drawback is that to evaluate any power of x, we go through all of the previous powers.

A slightly better method is to make a table of powers of 2.41 and put them in the given polynomial.