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.

About The Author:

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.

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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.

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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! 🙂

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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

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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?

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Do you get terrified by Number Series Questions?

‘Number Series’ or ‘number sequence’ is an integral part of Quantitative Aptitude Section in various competitive examinations like IBPS Bank exams, SSC, RAIL and so on. I’ve come across many students who get petrified by looking at questions on number series. Lack of confidence subdues their logical faculty to come up with the missing number or next number in a series of numbers that are seemingly complicated. They are unable to decipher the predefined rule behind the sequence of numbers given. Once the rule is determined it’s simple to find out the next term(s) or missing term(s) in a series. It becomes simpler to solve number series questions following a step by step approach.

Different Type of Number Series

A series can be created in numerous ways. An understanding of these various ways can help us in recognizing the pattern followed in the number series. So here we go with some standard series types-

Arithmetic Series – Difference between successive terms is fixed. Subsequent terms are obtained by either adding or subtracting a fixed number. For example,

2, 5, 8, 11, 14, 17,…..                                Common Difference = 3

32, 25, 18, 11, 4,…….      Common Difference = -7

Geometric Series – Each term of the series is obtained by multiplying (or dividing) the previous number by a fixed number. Hence the ratio between any 2 consecutive terms is same. For example,

3, 6, 12, 24, 48, 96…….       Common Ratio = 2

2048, 512, 128, 32…….       Common Ratio = 1/4

Arithmetico-Geometric Series –  Each term is first added (or subtracted) by a fixed number and then multiplied (or divided) by another number to obtain the subsequent term. For example,

4, 18, 60, 186….. => 4, (4+2)x3, (18+2)x3, (60+2)x3

Geometrico-Arithmetic Series – Each term is first multiplied (or divided) by a fixed number and then added (or subtracted) by another number to obtain the subsequent term.For example,

3, 10, 24, 52…… => 3, (3×2)+4, (10×2)+4, (24×2)+4,…..

Series of Squares, Cubes, etc. – Each term is square or cube or a higher power of the previous term. For example,

3, 9, 81, 6561….                                         Each term is obtained by squaring the previous number

2, 8, 512, ………                                        Each term is obtained by cubing the previous number

Some non-standard ways in which series can be created –

Series with subsequent Differences being in Arithmetic Progression (AP)

3, 7, 13, 21, 31, 43……         The differences in subsequent terms are 4, 6, 8, 10, 12…. which are in AP

Series with Differences in Differences being in AP

336, 210, 120, 60, 24, 6, 0,….         The difference being 126, 90, 60, 36, 18, 6

The differences between differences being 36, 30, 24, 18, 12,…. and so on, which are in AP

Inter-Mingled Series – In this case any two of the above series are mixed in one. For example,

1, 3, 5, 1, 9, -3, 13, -11, 17,….

Odd terms (1, 5, 9, 13, 17,….) of the series are in AP, whereas even terms (3, 1, -3, -11,…) are in geometrico-arithmetic series in which subsequent terms are obtained by multiplying the previous term by 2 and then subtracting 5.

This list is by no means exhaustive. There can be infinite ways to make a number series. It’s not possible to think or write about them here.

I got inspired to write this article after reading Series Chapter of the awesome book named Magical Book on Quicker Maths by M.Tyra. It can be a boon for any competitive exam aspirant. We’ll talk more about series in future. You can post any question related to number series as a comment below.

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Learn to test whether a given number is prime number or not

By looking at a number do you wonder whether it’s a prime number or not?
Is it always complicated for you to figure out the answer?

If the answer to above questions is yes, go ahead and learn this method of figuring out if a number is prime or not.

To test whether any number is a prime number or not, take an integer larger than the approximate square root of that number.

To quickly find the square root of any number, you can look at finding square root without calculator and Heron’s Method of finding roots.

Let say the square root of the said number is ‘x’. Test the divisibility of the given number by every prime number less than ‘x’. If it is not divisible by any of them then it is prime number; otherwise it is a composite number (other than prime).

Example 1: Is 349 a prime number?
The square root of 349 is approximately 19. The prime numbers less than 19 are 2, 3, 5, 7, 11, 13, 17.
Clearly, 349 is not divisible by any of them. Therefore, 349 is a prime number.

Example 2: Is 881 a prime number?
The square root of 881 is approximately 30. The prime numbers less than 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
881 is not divisible by any of the above numbers. Therefore, 881 is a prime number.

If you know other ways of finding if a number is prime number or not, share it with all by posting a comment below.

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Do You Know the Sum of All Positive Integers till Infinity, be Prepared for a Shock After Knowing the Answer

This is one of the most amusing mathematics trick pulled off by someone. If you understand basics of numbers, your jaw will drop at the output of this mathematics formulation. So here we go –

What do you think is the sum of all the integers up till infinity?

1+2+3+4+5……. so on to infinity=?

Anyone who know the meaning of infinity would quite safely say the answer is infinity, but practical implications and string theory in physics tells us otherwise.

The answer is  -1/12

If you think that’s impossible, check out the detailed explanation in this video:

This proof seem counter intuitive or to some it might even sound ridiculous. However, let me clarify this equation is completely accurate and not some hoax. It is used in theoretical physics.

Do you know of more such counter intuitive Mathematics Formulations?

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Shortcut to Find the Fourth Power of any Two Digit Number

In this article we will explore a shortcut to find the fourth power of any 2 digit number. The approach will be similar to the shortcut to find the cube of any 2 digit number. I strongly suggest that you should check that first. 

Generic Form of 4th Power of 2 Digit Numbers

The generic form of fourth power of any two digit number can be algebraically expressed as:

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

We will split the above result in 2 lines. We start with 4th power of 1st digit and then keep multiplying with ratio of 2nd digit: 1st digit = b/a. The remaining will go in the next line. This can be written as –

= a4 +  a3b  +  a2b2   +  ab3   + b4   

             3a3b + 5a2b2 + 3ab3

__________________________________  Adding the above 2 lines, we get original result.

= a4 + 4a3b + 6a2b2 + 4ab3 + b4

Example of Shortcut to Find the 4th power of 2 Digit Numbers

To try this shortcut, let us consider a simple example i.e. 124

Here the first digit is 1 and its fourth power is also 1. The ratio of 2nd to the 1st digit is 2 (Ratio = b/a)

The remaining 4 terms can be obtained by multiplying each of the previous terms as shown :

124 =  1      2       4      8     16

6      20    24

_____________________
=  1    8      24     32    16

=  1    8      24     32    16

=  1    8      24     33      6

=  1    8      27      3      6

=  2    0       7      3      6

= 20736

Let us try another example. We will find out the 4th power of 91.

914 = 6561            729             81            9           1

2187           405            27

____________________________________

6561        2916             486         36           1

Keeping single digit in each step and carrying the remaining digits and adding to number on the left, we get the answer as 68574961

Tip to use the shortcut: If we memorize the fourth power of all the single digits (0-9), it will come very handy for using this trick of calculating fourth power of any two digit numbers. Find below the table of fourth power of first ten natural numbers –

Number

Fourth Power

1

1

2

16

3

81

4

256

5

625

6

1296

7

2401

8

4096

9

6561

10

10000

Do you think this shortcut can be applied for calculating fourth power of a number?

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Cyclicity

In CAT, MAT and other Competitive examinations like Bank PO, etc. you get questions where you need to find the last digit of numbers raised to large powers. It’s almost impossible to calculate the values of such numbers manually and  hence to find digit at their unit’s place.  Such problems can be solved using the concept commonly known as Cyclicity of Numbers. Here in this post I am explaining in details the concept of cyclicity and how it should be used for solving such problems.

Finding Last Digit of Any Number Raised to Any Power Read More

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