Remainder Theorem & its application

We have all learnt the Remainder Theorem in class 10 (now i am in 11) that when you divide a polynomial f(x) by x-c the remainder r will be f(c). Now let’s see how we can use this theorem in other situations.

Example #1

Let’s consider the following Product: 65 x 32.

We want to find out what is the remainder when it is divided by a number say 7.

To solve such questions we just need find the individual remainders when the numbers are divided by the divisor.

In this case 65 gives remainder 2 (65 -63) and 32 gives remainder 4 (32 – 28) when divide by 7. Multiplying the remainders we get 2*4=8

Since this number is greater than divisor, divide it again by the divisor again, i.e. 8/7 gives remainder 1.

Thus, when 65*32 is divided by 7 it gives remainder of 1. Isn’t it amazing! We save time and effort of multiplying large numbers and doing complex divisions.

Example #2

Let’s see another example to find the remainder when 1421 * 1423 * 1425 is divided by 12

By this method 1421 * 1423 * 1425

1st step remainders = 5 * 7* 9 = 35*9
2nd step remainders = 11*9
3rd step remainder = 99/12 = 3

So the monstrous product gives a remainder of 3 when divided by 12.

Example #3

Let’s suppose we want to find the last two digits of the product
22 * 31 * 44 * 27 * 37 * 43

For such problems we just need to find the remainder when it is divided by 100

(22 * 31) * (44 * 27) * (37 * 43)

1st step remainders =  82*88*91
2nd step remainder =  2 * 28

THATS IT!! The last two digits of the lengthy product is found within seconds and as you see it is 56

This is a guest post by one of the regular QuickerMaths.com follower Debasis Basak. On behalf of all the readers, I  thank him for his contribution.

The Mysterious Number 22

Number 22 Everywhere?

Numbers never fail to surprise us. This post talks about one such amazing property of number 22.

Select any three-digit number with all digits different from one another. Write all possible two-digit numbers that can be formed from the three-digits selected earlier. Then divide their sum by the sum of the digits in the original three-digit number.

You’ll always get the same answer, 22. Isn’t this wonderful!

For example, take the three-digit number 786. The 2 digit-numbers which can be made using the digits 7, 8 and 6 are 78, 87, 76, 67, 86, 68. Hence sum = 78 + 87 + 76 + 67 + 86 + 68 = 462. Sum of digits of 786 = 7+8+6 = 21. Then 462/21 = 22

This will be true for any three-digit number with all digits different.

Is it actually Mysterious?

Not really! If we go deeper and try to analyze this unusual result using, we’ll be able to appreciate the logic of it.

The general representation of any three digit number with all digits different will be 100x+10y+z.  Now to find the sum of all the two-digit numbers taken from the three digits

= (10x+y)+ (10y+x)+(10x+z)+(10z+x)+(10y+z)+(10z+y)
= 20(x+y+z) + 2(x+y+z)
=22(x+y+z)

This when divide by the sum of the digits, (x+y+z), is 22. This shows the importance of algebra in explaining such simple yet interesting mathematical phenomenon.

Do you know any such interesting property of any number?

Relationship between Miles and Kilometers

Mile is an English unit of measuring length or distance and it is equal to 5280 feet or 1760 yards  (for your information, 1 yard = 3 feet).
Whereas, Kilometer is the measure of length or distance in metric system, where it is equal to 1000 meters.

1 Mile = 1.609344 kilometers

Now if you have to convert few miles to kilometers, it will be a tedious task, isn’t it? Would you like to learn a method of doing this conversion mentally?

Here is simple yet effective method using which you will get an approximate value but in most cases that will suffice the purpose. So here you go.

You need to understand a very simple idea – the relationship between kilometers and miles can be closely represented by Fibonacci Series. Now what’s this Fibonacci Series?

For those of you who don’t know, we first need to understand the concept of Fibonacci sequence. First few numbers of the Fibonacci Series are the following. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144…

If you’ve noticed in the above series the numbers are the sum of the previous two numbers in the series, starting with 0 and 1. As you see 0+1=1; 1+1=2; 1+2=3;…….34+55=89 and so on. Hence a series of numbers in which each number is the sum of preceding tweo numbers is called Fibonacci series or sequence. So to understand the calculation of conversion from miles to kilometers or the other way round,  let us use Fibonacci series

Conversion of Miles to Kilometres

Relationship of succeeding numbers in Fibonacci series closely matches the relationship between miles and km.

• 3 miles = 5 km
• 5 miles = 8 km
• 8 miles = 13 km (12.8748 km to be exact)
• 13 miles = 21 km (20.9215 km to be exact)
• 21 miles = 34 km (33.796224 km to be exact)
• 89 miles = 144 km (143.232 km to be exact)

However, what if you have to convert number (miles) which is not in Fibonacci series into km? Not to worry. There is a way out. You need to break out that number in to Fibonacci numbers. Covert the numbers as per the above scheme and add the resultant numbers to get the answer. For example; 85 = 55+21+8+1 After the conversion: 89+34+13+1 = 137 km (136.794 km to be exact)

Fibonacci sequence has many other wonderful properties and wide occurrence in nature, music, design, etc. You’ve just seen another application of Fibonacci sequence. The answers you get are close enough.

Most Amusing Property of 1089

Select a three digit number (where the units and hundreds digits are not the same) and follow these instructions:

Step 1: Choose any three-digit number (where the units and hundreds digits are not the same).

Let us randomly select the number 469

Step 2: Reverse the digits of the number you have selected

So reverse of 469 is 964

Step 3: Subtract the smaller number from the bigger one

964 – 469 = 495

Step 4: Once again reverse the digits of this difference

Reverse of 495 is 594

Step 5: Add the last two numbers

594+495 = 1089

This result will be the same for any 3-digit number chosen in step 1. Isn’t it astonishing that regardless of which number you select at the beginning, you will get 1089 as the result. Check out similar amusing property of 6174, which is also called Kaprekar Constant, named after Indian recreational mathematician D.R.Kaprekar.

Another Interesting property of 1089

Let’s look at the first nine multiples of 1,089:

1089 x 1 = 1089

1089 x 2 = 2178

1089 x 3 = 3267

1089 x 4 = 4356

1089 x 5 = 5445

1089 x 6 = 6534

1089 x 7 = 7623

1089 x 8 = 8712

1089 x 9 = 9801

I am sure you notice a pattern in the products. Look at the first and ninth products. They are the reverses of one another. The second and the eighth are also reverses of one another. And so the pattern continues, until the fifth product is the reverse of itself, known as a palindromic number

One More Unique Property of 1089

33^2 = 1089 = 65^2 – 56^2

The above representation is also unique among two digit numbers.

Do you agree that there is a particular beauty in the number 1089?

Squaring Numbers in a Simple and Fun way

Squaring numbers is a tedious job if you do not know vedic maths shortcuts and other tricks and tips. Imagine having to waste your precious seconds in carrying out actual multiplications to find out squares during an exam. It can prove disastrous.

It is always advisable to learn various time saving tricks and alternate methods to find out percentages, fractions, squares, cubes, etc. because these will be useful in Quantitative Aptitude as well as Logical Reasoning sections of any competitive examination you take.

The idea of this post is contributed by Piyush Goel and it’s further edited by me to share with you a trend that he observed in squares of numbers.

If we talk about Square, Square is the result of multiplying a number by itself. So, square of 3 is 3×3=9. Similarly, square of 10 is 10×10=100. Since you know what is a square, observe this:

If we have to square 11, instead of 11×11=121,
we can simply put 1 (1*2)(1^2) and get 121.
Similarly, we can write the square of 12 as: 1(2*2)(2^2) and get 144. In the same fashion we can get the square of 13 as 1(3*2)(3^2)=169 and the square of 14 as 1(4*2)(4^2)=1|8|16 =196(after carrying over the 1 of 16 and adding it to 8).

Look at (2,4,6,8,10,12,14,16,18,20….difference of 2) & (1,4,9,16,25,36,49,64,81…….. difference is 3,5,7,9,11,13,15,17,19….) and difference of 3,5,7,9,11 is 2 so there is True Symmetry of 2.

11^2 = 1 2 1
12^2 = 1 4 4
13^2 = 1 6 9
14^2 = 1 8 16 = 100 + 80 + 16 = 196
15^2 = 110 25 = 100 + 100 + 25 =225
16^2 = 112 36 = 100 + 120 + 36= 256
17^2 = 114 49 = 100 + 140 + 49= 289
18^2 = 116 64 = 100 + 160 + 64= 324
19^2 = 118 81 = 100 + 180 + 81= 361
20^2 = 111^2 = 1 20 100 = 100 + 200 + 100 = 400
21^2 = 111^2 = 1 22 121 = 100 + 220 +121 = 441
22^2 = 112^2 = 1 24 144 = 100 + 240 +240 = 484
23^2 = 113^2 = 1 26 169 = 100 + 260+169 = 529
24^2 = 114^2 = 1 28 196 = 100 + 280+196 =576
25^2 = 115^2 = 1 30 225 = 100 +300 +225 =625
26^2 = 116^2 = 132 256 = 100 + 320 +256 =676
27^2 = 117^2 = 134 289 = 100 + 340 + 289 =729
28^2 = 118^2 =136 324 = 100 + 360 + 324 = 784
29^2 = 119^2 = 138 361 = 100 + 380 + 361 = 841
30^2 = 120^2 = 1 40 400 = 100 + 400 +400 =900
31^2 = 121^2= 121^2 = 1 42(111^2) =1 42 (1 22 121) =100 + 420 + 441 =961
41^2 = 131^2 = 1 62 (1 21^2) = 1 62 (1 42) (1 11^2) = 1 62 (1 42) (1 22 121) = 1 62 (961) = 100 + 620 + 961 = 1681

This is time saving and the best part is we do not have to mug up additional formulae or large algorithms of processes.

Piyush Goel, born on 10th February, 1967. He has a Diploma in Mechanical Engineering and a Diploma in Vastu Shastra. Always wanting to do something new, Mr. Goel has written 15 Spiritual and World Famous Books with his own hands in Mirror Image in different ways. He is now known as “Mirror Image Man”. He also has World’s First Hand-written Needle Book “Madhushala” to his credit.

Edited by Preeti Patawari

How I chanced upon Factorial Function

Today we will see how Piyush Goel(the contributor of this article) discovered something while playing with numbers. Here is the story contributed by him:

One day while sitting idle, and having nothing important to do I was just scribbling on paper. I was writing random numbers and thinking how I could link them. To my amazement I discovered this:

I wrote down 0, 1, 2, 3, 4, 5.

Next to each number I wrote their respective squares, viz. 0,1, 4, 9, 16, 25.

Then start subtracting each successive square from the next bigger square.

It looks something like this: (1-0), (4-1), (9-4), (16-9) and (25-16).

What is the result?

I got 1, 3, 5, 7, 9.

Now again subtract each successive number from the next one in the order.

We are performing this: (3-1), (5-3), (7-5) and (9-7).

Surprisingly, I got (2, 2, 2, 2). A 2 in each case. I was amazed and decided to take a step further. This time with cubes.

So here go my numbers 0, 1, 2, 3 ,4, 5, 6.

And their respective cubes: 0, 1, 8, 27, 64, 125, 216.

Time to perform first round of successive subtraction, i.e. (1-0), (8-1), (27-8), (64-27), (125-64) and (216-125). The result: 1, 7, 19, 37, 61, 91.

Second round of successive subtraction as performed earlier: (7-1), (19-7), (37-19), (61-37) and (91-61).

And this is what I got: 6, 12, 18, 24, 30.

Successive subtraction performed the third time: (12-6), (18-12), (24-18) and (30-24).

And this time I got the result (6, 6, 6, 6).

You see, squaring means ‘to the power 2’ and I got the result 2 after 2 successive subtractions, which is the value of 2!

Cubing means ‘to the power 3’ and after performing 3 successive subtraction I got 6 in each case which is nothing but the value of 3!

I similarly performed with 4th and 5th powers of the first few whole numbers. I was surprised to find the final value of 24 and 120 which are of course the values of 4! and 5!.

It was all right up to 5, but I wanted to go a notch higher and check for 6 just for my personal satisfaction. While doing for 6th power the calculations were massive and numbers so huge that it took me a couple of hours to do my calculations. Just to be sure I also did it for numbers upto 15.

Put 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 and 15 and their respective 6th powers:

0,1,64,729, 4096, 15625, 46356, 11649, 262144, 531441, 1000000, 1771561, 2985984,4826809, 7829536 and 11390625.

I kept on doing successive subtractions till to my contentment I finally got 720 for all which is the value of 6!

With little efforts, we can work out algebraic relationship for the above but I’ll leave it for some other time. Do share your comments.

Piyush Goel, born on 10th February, 1967. He has a Diploma in Mechanical Engineering and a Diploma in Vastu Shastra. Always wanting to do something new, Mr. Goel has written 15 Spiritual and World Famous Books with his own hands in Mirror Image in different ways. He is now known as “Mirror Image Man”. He also has World’s First Hand-written Needle Book “Madhushala” to his credit.

How to find the largest power of a number that divides a factorial number?

Factorial

Factorial of a number or n! is the product of n consecutive natural numbers starting from 1 to n. Hence, 4! is nothing but 1x2x3x4 = 24

Common Type of Questions on Factorials

Two types of questions are very common on factorial.

One, finding out the number of zeros at the end of a factorial expansion. You can find the detailed explanation on how to deal with such questions here.

Two, finding out the largest power of any number that divides any factorial number.
Today, I am going to explain how to approach questions like

Find the largest power of 5 contained in 124!
Find the highest power of 7 that can exactly divide 777!

Let us get started.

Let’s say we have to find out the largest power of 5 contained in 25!
1!, 2!, 3! and 4! are not divisible by 5 because 5 is not a factor in these factorial numbers.

Now if I consider any factorial number greater than 4!, every number consists of 5 or the higher powers of 5.
5!= 1x2x3x4x5
6!= 1x2x3x4x5x6
… …. … …. …
10!= 1x2x3x4x5x6x7x8x9x10
…. …. …. …. …. ….
25!= 1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17x18x19x20x21x22x23x24x25

Multiples of 5 in the expansion of 25! are 5, 10, 15, 20 and 25.

Basically, 25 divided by 5 equals 5. So there are 5 multiples of 5 from 1 to 25.

But, 25=5X5. This means that 25 is giving me an extra 5 which I need to account for. With this logic every multiple of 25 will give me an extra 5.

25 divided by 25 equals 1. So there is 1 multiple of 25 from 1 to 25.

Therefore highest power of 5s in 25! = 25/5+25/25= 5+1 = 6

The same reasoning extends to larger numbers. Suppose we had to find the highest power of 5 in 1000!

There are 1000/5=200 multiples of 5 between 1 and 1000.

The next power of 5 is 25 and there are 1000/25=40 multiples of 25 between 1 and 1000.

The next power of 5 is 53=125 and each such multiple of 125 gives me still one more extra 5 as compared to other multiples of 25. There are 1000 /125 = 8 such multiples of 125 between 1 and 1000.

The next power of 5 is 54=625 and there is only one multiple of 625 between 1 and 1000, 625 itself.
Hence the total number of 5s in 1000! = 200 + 40 + 8 + 1 = 249.

You keep on dividing with higher powers of the given number till you get 1 as the whole number part. And of course, while adding up we consider only the whole number part and ignore the fractional part.

If you observe, this is nothing but the greatest integer function.
We have to find out the highest power of k that can exactly divide n!.

We divide n by k, n by k2, n by k3.. and so on till we get n by kx equals to 1 or 1 and some decimal part ;(where [P] means the greatest integer less than or equal to P) and then add up.

Highest power of prime number k in n! =[n/k]+[n/k2]+[n/k3]+…….+[n/kx]; where [P] means the greatest integer less than or equal to P.

Example: Find the highest power of 2 in 50!
Solution: Highest power of 2 in 50! = [50/2]+[50/4]+[50/8]+[50/16]+[50/32]= 25+12+6+3+1= 47

Example: Find the highest power of 30 in 40!
Solution: Express 30 as product of its prime factors. 30=2x3x5. So to make a 30 you need each of 2, 3 and 5. Now in 40! there will be more 2s compared to 3s and more 3s compared to 5s. Hence to extract each 30 out of 40!, we should be bothered only about number of 5s in 40!.

Memory trick: Find number of the largest prime factor of 30 i.e. 5 in 40!.

Highest power of 5 in 40! = [40/5]+[40/25] = 8+1=9.

What if I am asked to find the highest power of numbers like 17, 19 or say 23? Then I will have to deal with numbers like 173,194, 235, etc. Finding the higher powers of such divisors and the subsequent division can be a real pain. So is there an alternate method? Luckily, yes.

Short-cut to find the highest power of a number in a factorial

By now it is clear that we consider only the integral part of the quotient while performing the series of divisions. So we can do this calculation in an easier method. We just divide the number and divide the succeeding quotients by the same number till we get 0. Finally add up all the quotients.

In this method we get rid of finding higher powers of the divisor.
So if we have to find the highest power of 5 in 1000!

200+40+8+1= 249. Isn’t this just awesome. I no longer have to worry about the values of higher powers of 5. Also division by 5 is relatively easier than division by 25, 125 or 625.

Share your feedback and let me know if you liked this.

How many zeros are there at the end of 100 factorial?

How many zeros at the end of 100 factorial? How many trailing zeros in 1000 factorial? Do you get baffled by questions like these?

How useful is calculator in solving this problem?

Sadly, calculator is not of much help because the answer will be in scientific notation. I will get the first few digits but what I am really bothered about in this case is the end of the multiplied expansion which the calculator will not be able to display.

This trailing zeros in a factorial is fairly simple to handle once you are equipped with the right way to handle it and then you will be able to handle 3468796! with equal ease as 10!.

There are many interesting derivation from factorials.

How to quickly find number of zeros at the end of factorial of any number or say ‘n!’?

Let us get a few facts straight.

First, any number will have a zero at the end if that number is a multiple of 10; or in other words the number has 10 as a factor. So basically I have to find out how many 10s are there in the expansion of n!

Since 5X2=10, so I will have to consider all the products of 5 and 2 because each such product will give me a 10.

Obviously there are way more multiples of 2 than multiples of 5 in any given factorial expansion. This is because every 5th number is a multiple of 5 whereas every 2nd number is a multiple of 2.

Therefore, to find the number of 10s in the expansion of n!, what I am really bothered about is how many times 5 is a factor in all the numbers between 1 and n.

If the above logic is clear, we can go through an example for further clarity.

Figuring out the number of 5s in a factorial

Find the number of trailing zeros in 17!

17! = 17x16x15x14x13x12x11x10x9x8x7x6x5x4x3x2x1

Basically I have to take all the numbers with 5 as a factor which when clubbed with a multiple of 2 will yield a 10. multiples of 5 are 15, 10 and 5.

Hence, 17! has three trailing zeros.

Let us now look at a few more examples.

How many zeros at the end of 100 factorial?

Multiples of 5 in the expansion of 100! are 5, 10, 15, 20,….., 100.

You can count it this way and waste a lot of time or consider this.

100 divided by 5 equals 20. So there are 20 multiples of 5 from 1 to 100.

But wait, 25=5X5. This means that 25 is giving me an extra 5 which I need to account for. With this logic every multiple of 25 will give me an extra 5.

100 divided by 25 equals 4. So there are 4 multiples of 25 from 1 to 100.

Therefore number of trailing zeros in 100! = 100/5+100/25= 20+4 = 24

The same reasoning extends to larger numbers.

How many Zero at the end of 1000 factorial?

There are 1000/5=200 multiples of 5 between 1 and 1000.

The next power of 5 is 25 and there are 1000/25=40 multiples of 25 between 1 and 1000.

The next power of 5 is 53=125 and each such multiple of 125 gives me still one more extra 5 as compared to other multiples of 25. There are 1000/125 = 8 such multiples of 125 between 1 and 1000.

The next power of 5 is 54=625 and there is only one multiple of 625 between 1 and 1000, 625 itself.

Hence the total number of 5s in the expansion of 1000! = 200 + 40 + 8 + 1 = 249 and thus 249 trailing zeros in 1000!

How to find zeros in any factorial?

Take the number.

Divide by 5; if you get a decimal consider only the whole number part.

Divide by 52=25; if you get a decimal consider only the whole number part.

Divide by 53=125; if you get a decimal consider only the whole number part.

Divide by 54=625; if you get a decimal consider only the whole number part.

Continue this division with higher powers of 5 until your division results in a number less than one that is only a fraction with no whole number part.

Add up all the whole numbers you got in the series of divisions you performed. This is the number of trailing zeros. Since you’ve learned the method of finding trailing zeros for any factorial, it will be easy for you to find the largest power of a number that divides a factorial number

One last example before I wind up.

Find the number of trailing zeros in 3426!

While performing the divisions, I am only considering the whole number part and conveniently ignoring the decimal or fraction part.

51: 3426/5= 685

52: 3426/25= 137

53: 3426/125= 27

54: 3426/625= 5

55: 3426/3125= 1

So the total number of trailing zeros is 685+137+27+5+1 = 855.

By the way you can get the same result by dividing the number repeatedly by 5 till you get 1 as quotient and then adding up all the results. This saves a lot of time as you don’t have to deal with division by higher powers of 5.

3426 ÷ 5 = 685

685 ÷ 5 = 137

137 ÷ 5 = 27

27÷ 5 = 5

5 ÷ 5 = 1

685 + 137 + 27 + 5 + 1 = 855 which is the number of trailing zeros in 3426!

Hope this article was of help to you. See you with many more such ways to tackle seemingly difficult problems. You can post your queries in the comment box below or post it on Quickermaths.com/questions. Till then, happy solving! 🙂

How to Convert from Other Number Bases to Decimal System?

Questions on conversion of numbers in some base to some other base is very common in competitive examination. Here in this post I present a simple technique to help you do such conversions.

First, let us understand what do we mean by number bases or systems. In our decimal number system, the rightmost position represents the “ones” column, the next position represents the “tens” column, the next position represents “hundreds”, etc. Therefore, the number 123 represents 1 hundred and 2 tens and 3 ones, whereas the number 321 represents 3 hundreds and 2 tens and 1 one.

The values of each position correspond to powers of the base of the number system. So for our decimal number system, which uses base 10, the place values correspond to powers of 10:
… 1000    100       10           1

… 103        102      101         100 Read More

Solving Number Series Questions Can Be So Simple…Learn How?

In the last article, I had discussed various types of series. Today I will discuss how to approach a series problem once you have identified what kind it is. Let me be honest. There is no algorithm for solving number series questions appearing in quantitative section of any competitive exam because each question is different and there are innumerable ways in which a series can be generated. Cracking a series question involves a lot of practice and the intuition that comes with practice.

Series Questions – Samples

• 2, 4, 8, 16, 32, 64
• 729, 512, 343, 216, 125, 64
• 1, 1, 2, 4, 6, 10, 16, 26, 42, 68… (Fibonacci Series)
• 4, 9, 6, 18, 9, 27, 13, 36

I will today broadly discuss steps how to build this intuition so that you can solve series questions easily and quickly.

Step 1: Screening

First check the series by giving a cursory look at it. Many a times a careful first look may be enough to tell the next term. Have a look at these examples:

1. 3, -9, 27, -81, 243, ?
2. 1, 3, 7, 13, 21, 31, ?
3. 1, 3, 6, 10, 15, 21, ?
4. 2, 4, 8, 16, 32, 64, ?

Solution:

1. Each successive term is multiplied by -3. Next term will be 243 X -3= -729.
2. The series is +2, +4, +6, +8, +10, +12. So next term will be 31 + 12 = 43.
3. The series is +2, +3, +4, +5,+6, +7. Next term will be 21 + 7 = 28.
4. Each term is multiplied by 2. Next term will be 64 X2 = 128.

Step 2: check the pattern: increasing/ decreasing/ alternating

In case you fail to decipher the rule of the series by just preliminary series, try to understand the trend of the series. Look at the pattern. Is it increasing or decreasing? Is it following an alternating pattern?

1. 1, 4, 9, 16, 25, 36, 49.
2. 4, -8, 16, -32, 64, -128.
3.  729, 512, 343, 216, 125, 64.
4. 5, 10, 13, 26, 29, 58.

Clearly, i and iv are following an increasing pattern. ii is following an alternating pattern while iii is a decreasing pattern series .

Step 3: if the series is increasing or decreasing, find the rate of increase or decrease

Start with the first term and move onto the next. Gauge whether the series proceeds arithmetically or geometrically or alternately. You have to feel whether this rise or fall is slow or fast. In an arithmetic progression, the increase or decrease of terms is by virtue of addition or subtraction. So the rise or fall will appear slow. In contrast, in a geometric progression the increase or decrease of terms is by virtue of multiplication or division. So naturally the rise or fall will be very fast. If the geometric progression involves squaring or cubing, this rise or fall will be even sharper. I hope you understand how you need to feel the rate of increase or decrease.

In the above examples now it is clear that 1, 4, 9, 16, 25, 36, 49,… is an arithmetic type of series. The trend being +3, +5, +7, +9 and so on. Clearly, 729, 512, 343, 216, 125, 64,… is a geometric type of series involving cubes of 9, 8, 7, 6, 5, 4 and so on.

Now look at 1, 5, 14, 30, 55, 91. It can easily be figured out that the rise is rather sharp. So it is a geometric progression. On trial you will see that just successive multiplication is not involved here. Checking for addition of squared numbers or cubed numbers we see that this is actually 12, 12+22, 12+22+32 and so on.

Basically, a geometric increase or decrease can take place in two ways –

• Multiplication or division by terms
• Addition or subtraction of squared or cubed terms

Obvious question is how does one differentiate between the two?

Look at the trend of the increase. If the increase is because of addition of squared or cubed numbers, the increase will be very sharp initially but not very sharp in the later terms say 5th or 6th term onwards.

Watch the earlier example:  1, 5, 14, 30, 55, 91.

1, 1X5=5, 5X2.8=14, 14X2.1428=30, 30X1.8333=55, 55X1.6545= 91.

Notice how the multiplication factor is gradually decreasing from 5 in the initial step to 1.6545 in the last step. Hence, where the rise is very sharp initially but gradually slows down, it should strike in your mind that you have to look for a pattern involving addition of squared or cubed numbers. You needn’t do the above calculations of multiplication factor, just try to build the intuition of understanding the rate of increase or decrease.

Consider 4, 5, 12, 39, 121, 610. This series also rises very sharply. Hence this must be a geometric progression. However the rate of increase does not slow down in later terms. In fact it picks up as the series progresses. Therefore we can conclude that the series must be of the first kind i.e. formed by multiplication. A little more exercise will tell us that the series is : x1+1, x2+2, x3+3, x4+4, x5+5 and so on.

Now let me show you an example of alternating increase. Two possibilities will exist here:

• Two different series may be intermixed
• Two different kind of operations may be being performed on successive terms

Consider the series 1, 4, 5, 9, 14, 20, 30, 43. This series increases gradually but the increase is rather haphazard. Actually it is a mix of two series!

1, 5, 14, 30, 55, 91 which is  a series: 1, 1+22, 1+22+32, 1+22+32+42 ; and 4, 9, 20, 43, 90 which is a series: x2+1, x2+2, x2+3, x2+4.

Again, look at 3, 13, 18, 76, 81, 409, 414. Here two different operations are being performed alternately: the first operation is that of multiplication by 3, 4 and 5 successively and adding a constant number 4 and the second operation is adding a constant number 5. Hence the series is x3+4, +5, x4+4, +5, x5+4, +5.

Step 4: Check if the series is neither increasing nor decreasing but alternating

For an alternating series also you should check whether it is a mix of two series or two different operations are being performed alternately.

For example, 4, 9, 6, 18, 9, 27, 13, 36. It is a mix of two series 4, 6, 9, 13 and 9, 18, 27, 36.

In 200, 600, 1200, 1600, 3200, 3600, 7200 two operations are going on; one addition of 400 and second multiplication by 2.

With practice these steps can help you build the intuition required to solve problems related to series. Did you ever face any difficulty in solving number series questions?