Number System Byju-1

7/22/2019 Number System Byju-1

http://slidepdf.com/reader/full/number-system-byju-1 1/6

BYJU's Classes -

Byju's classes

Call: +91.9880031619

Home

About UsCAT

UPSCEngg/Med

CRTOnline Course

Tablets

RegisterLogin

Quantative

Aptitude DIand LR VerbalAbility

Demo

Tutorials - Arithmetic

Algebra Arithmetic Geometry

Pages: 1 2 3 4 5

1) POWER CYCLE

Main Application to find the last digit of a power expression

The last digit of a number of the form ab falls in a particular sequence or orderdepending on the unit digit of the number (a) and the power the number is

raised to (b). The power cycle of a number thus depends on its unit digit.

Consider the power cycle of 2

21=2, 2

2=4, 2

3=8, 2

4=16, 2

5=32, 2

6=64, 2

7=128, 2

8=256

As it can be observed, the unit digit gets repeated after every 4th

power of 2. Hence, we can say that 2 has a power cycle of 2,4,8,6 with

frequency 4.

This means that, a number of the form

24k+1

will have the last digit as 2

24k+2


24k+3


24k+4

will have the last digit as 6 (where k=0, 1, 2, 3…)

This is applicable not only for 2, but for all numbers ending in 2. ( eg 1232

,3452123

)

Therefore to find the last digit of a number raised to any power, we just need to know the power cycle of digits from 0 to 9, which is given below

Unit digit Power cycle Frequency

0 0 1

1 1 1

U's Classes -Arithmetic » BYJU's Classes - http://byjusclasses.com/a

24-09-20



2 2,4,8,6 4

3 3,9,7,1 4

4 4,6 2

5 5 1

6 6 1

7 7,9,3,1 4

8 8,4,2,6 4

9 9,1 2

APPLICATIONS OF POWER CYCLE

I. DIRECT QUESTIONS TO FIND THE UNIT DIGIT OF A POWER EXPRESSION OF THE FORM ab

Eg 1) Find the last digit of 455

55 can be written as 4k+3

In the cycle of 4, in the third position we get 4 itself. Last digit of the expression is 4 itself

Eg 2) Find the last digit of 12345734

Since 7 is the units digit write 34=4k+2

In the cycle of 7, in the 2nd

position = 9. last digit of the expression=9

II. REMAINDER QUESTIONS WHERE THE DIVISIBILITY OF THE DIVISOR DEPENDS ONLY ON ITS UNIT DIGIT (eg: when the divisorsare 2,5,10)

Eg 3) Find the remainder when 375

is divided by 5.

1) Express the power in the form, 4k+x where x=1, 2, 3, 4. In this case 75 = 4k+3.

2) Take the power cycle of the base (3) which is 3,9,7,1. Since the form is 4k+3, take the third digit in the cycle, which is 7.

Any number divided by 5, the remainder will be that of the unit digit divided by 5. Hence the remainder is 2.

Eg 4) Find the remainder when 4466

is divided by 10?

1) Express the power in the form, 4k+x where x=1, 2, 3, 4. In this case 66 = 4k+2.

2) Take the power cycle of the base (4) which is 4,6,4,6. Since the form is 4k+2, take the second digit in the cycle, which is 6.

Any number divided by 10, the remainder will be that of the unit digit divided by 10. Hence the remainder is 6.

Sometimes, you may get a question in the term of variables, where you need to substitute values to get the answer in the fastest way possible.

For example,

Eg 5) Find the unit digit of 73^4n

Put n=1, the problem reduces to 73^4

, which is 781.

Since 81=4k+1, take the first digit in the power cycle of 7, which is 7.

What is the logic behind this? Since “n” is a variable, whichever value of “n” you substitute; you have to get the same answer.

We substitute n=1 for ease of calculation. You can try substituting n=2 as well, you will get the answer as 7.

III. POWER CYCLE QUESTIONS OCCURRING IN GROUPS

In some questions, given a sequence of numbers; there will be a pattern in the unit digit which will be followed with a particular frequency. Basedon this frequency, we can form groups of the sequence of numbers. In those cases, we need to find out the frequency and we can easily

evaluate the unit digit of the entire group based on one single group.

Let us understand this concept with the help of a typical example

Eg 6) The sum of the third powers of the first 100 natural numbers will have a unit’s digit of

a) 3 b) 6 c) 9 d) 0

The question asks us to find 13+2

3+3

3+…….100

3


24-09-20



If you notice, the unit digit of 13+2

3+..10

3is the same as the unit digit of 11

3+12

3….+20

3and so on till 91

3+92

3+….100

3

Hence it is is sufficient to find the unit digit of the first group and then multiply by 10 to get the unit digit of 13+2

3+3

3+…….100

3

Let us find out the unit digit of each element in the first group. Using the concept of power cycle, it can be found as

Element Unit digit

13

1

23

8

33

7

43

4

53

5

63

6

73

3

832

93

9

103

0

? ..5

Therefore units digit of 13….100

3= ..5*10= 0

Shortcut:- since there are 10 groups involved, the unit digit of the first group has to be multiplied by 10 to get the answer. Thereis absolutely no need to even find the unit digit of the first group as any number multiplied by 10 gives 0 as the unit digit!!

IV. TO FIND THE RIGHT MOST NON ZERO INTEGER

Eg 7) Find the rightmost non-zero integer of the expression 1430343

+1470367

?

a) 3 b) 9 c) 7 d) 1

It can be easily observed that 1470367

has more number of zeroes as compared to 1430343

. hence, the rightmost non-zero

integer will depend on the unit digit of only 1430343

, which in turn will depend on the unit digit of 3343

343 can written as 4k+3

In the power cycle of 3 (3,9,7,1), the third digit is 7. Hence, the last non-zero integer in the expression is 7.

Useful technique to find the last 2 digits of any expression of the form ab

Depending on the last digit of the number in question, we can find the last two digits of that number.

We can classify the technique to be applied into

A) Odd Numbers

B) Even Numbers

A) Odd Numbers

TYPE METHOD EXAMPLES

1) Numbers ending in 1 The last digit is always 1. The 2nd

last digit = product

of tens digit of base * unit digit of the power.

2167

=__41 (2 * 7=__4)

9) 4187

=__ 81


24-09-20



For example, In 2167

; 2 is the tens digit of base and 7 is the unit digit of power

( 4 *7= __8)

10) 1261167

=__21

( 6 * 7= __2)

11) 31124

=__ 21

( 3 * 4=__2)

2) Number ending with 5 Tens digit of base power Last digit Second last digit

Odd odd 5 7

Odd Even 5 2

Even Odd 5 2

even Even 5 2

12) 155534

= __ 25

3) Numbers ending in 3, 7,

9

Change the power so that the base ends with 1 and

then use the same technique as for those numbers

ending with 1. eg) 34, 7

4&9

2all will end in1.

image004

B) For even Number

It is based on the cycle of 210

i.e. 210

raised to even power ends with 76 and

210

raised to odd power ends with 24.

This can be used for all even numbers

4) For even numbers (2,4,6,8) Use the pattern of the number 1024

=210

i.e.*210

raised to even powerends with 76 and

* 210raised to odd power ends with 24.

image004

Note:-

76 x 04=04

76 x 08=08

76 x 16 =16

76 x 32= 32

76 x 64= 64

76 x 28= 28

76 x 56 = 56……

ILLUSTRATIONS

Eg 14) Find the last two digits of 8234

8234

= 23(234)

= 2702

(210

)70

*22

= 76 x 04 = __04

Eg 15) What are the last two digits of 47523

?

a) 25 b) 50 c) 75 d) None of these


24-09-20



Ans – (c )

If the digit in the tens place is odd and the power is odd, then any number ending with 5, always have its last two digits as 75.

Eg 16) Find the ten’s digit of 74288

(a) 3 (b) 6 (c) 7 (d) 8

Ans. (c)

74288

= (37×2)288

37288

x2288

(374)

72x (2)

288

(372x37

2)

–5672x(__56)

(69 x 69)72

x(__56)

(__61)72

x(__56)

(__21)x(__56) = __76. Answer is 7

Shortcut of last two digit of squares:- in the above question to get the last two digits of 372

as 69;we do not need to calculate

manually. There is a pattern which will be very useful here

The last two digits of x2, (50-x)

2, (50+x)

2, (100-x)

2will always be the same. For example 12

2, 38

2, 62

2, 88

2, 112

2…. will all be

the same. That is, the last two digits of these numbers will always be 44

Also, last two digits of 112=39

2=61

2=89

2=111

2=139

2=161

2=189

2and so on

Eg 17) Find the remainder when (148)1084

is divided by 100.

a) 06 b) 66 c) 56 d) 76

When the question is about finding the remainder when you divide a number by 100, what we need to f ind out is the last twodigits

(148)1084

= (37)1084

x (4)1084

= (374) 271 x (2) 2168

=(372

x 372)

271x …56

(The last two digits of 372

will be the same as the last two digits of 132=69)

=(..69 x ..69)271

x …56

=(61)271

x …56 = …21 x …56= …76

Pages: 1 2 3 4 5

Home

About Us

CATUPSC

Engg/Med

CRT

Online CourseTablets

©ByjusClasses.com

General

Home

About us

CareerContact Us


24-09-20



Courses

Engg/Med Entrance

Offerings Through

Regular Class

VSAT

Tablet

Centres

Karnataka

Andhra PradeshKerala

Maharashtra

GujaratTamil Nadu

NCR

West BengalPunjab/Haryana

Bihar/Jharkhand

Online Test

GRE

test email

demotabs

Like

follow


Number System Byju-1

Documents

Transcript of Number System Byju-1