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Tutorials - Arithmetic
Algebra Arithmetic GeometryPages: 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 order depending 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, 22=4, 23=8, 24=16, 25=32, 26=64, 27=128, 28=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 will have the last digit as 4 24k+3 will have the last digit as 8
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
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 divisors are 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. Based on 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
If you notice, the unit digit of 13+23+..103 is the same as the unit digit of 113+123….+203 and so on till 913+923+….1003 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+23+33+…….1003 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 83 2 93 9 103 0 ? ..5
Therefore units digit of 13….1003= ..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. There is 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
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, 74 &92 all 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 210raised 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 power ends 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
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 37288x2288 (374) 72 x (2) 288 (372x372) –5672 x(__56) (69 x 69)72x(__56) (__61)72x(__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)2 will always be the same. For example 122, 382, 622, 882, 1122…. will all be the same. That is, the last two digits of these numbers will always be 44
Also, last two digits of 112=392=612=892 =1112=1392=1612=1892 and 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 find out is the last two digits
(148)1084= (37) 1084 x (4) 1084 = (374) 271 x (2) 2168
=(372 x 372) 271 x …56
(The last two digits of 372 will be the same as the last two digits of 132=69) =(..69 x ..69) 271x …56 =(61) 271 x …56 = …21 x …56= …76 Pages: 1 2 3 4 5 Home About Us CAT UPSC Engg/Med CRT Online Course Tablets ©ByjusClasses.com
