CYCLICITY OF NUMBERS (SHORT CUT TECHNIQUE)

For the concept of identifying the unit digit, we have to first familiarize with the concept of cyclicity. Cyclicity of any number is about the last digit and how they appear in a certain defined manner. Let’s take an example to clear this thing: The cyclicity chart of 2 is: 2¹ =2 2² =4 2³ =8 2⁴=16 2⁵=32 Have a close look at the above. You would see that as 2 is multiplied every-time with its own self, the last digit changes. On the 4^th multiplication, 2⁵ has the same unit digit as 2¹. This shows us the cyclicity of 2 is 4, that is after every fourth multiplication, the unit digit will be two. Cyclicity table: The cyclicity table for numbers is given as below: How did we figure out the above? Multiply and see for yourself. It’s good practice. Now let us use the concept of cyclicity to calculate the Unit digit of a number. What is the unit digit of the expression 4⁹⁹³? Now we have two methods to solve this but we choose the best way to solve it i.e. through cyclicity We know the cyclicity of 4 is 2 Have a look: 4¹ =4 4² =16 4³ =64 4⁴ =256 From above it is clear that the cyclicity of 4 is 2.  Now with the cyclicity number i.e. with 2 divide the given power i.e. 993 by 2 what will be the remainder the remainder will be 1 so the answer when 4 raised to the power one is 4.So the unit digit in this case is 4. For checking whether you have learned the topic, think of any number like this, calculate its unit digit and then check it with the help of a calculator. Note : If the remainder becomes zero in any case then the unit digit will be the last digit of  a^{cyclicity number}  where a is the given number and cyclicity number is shown in above figure. Lets solve another example: The digit in the unit place of the number 7²⁹⁵ X 3¹⁵⁸ is A.  7 B.  2 C.  6 D.  4 Solution The Cyclicity table for 7 is as follows: 7¹ =7 7² =49 7³ = 343 7⁴ = 2401 Let’s divide 295 by 4 and the remainder is 3. Thus, the last digit of 7²⁹⁵ is equal to the last digit of 7³ i.e. 3. The Cyclicity table for 3 is as follows: 3¹ =3 3² =9 3³ = 27 3⁴ = 81 3⁵ = 243 Let’s divide 158 by 4, the remainder is 2. Hence the last digit will be 9. Therefore, unit’s digit of (7⁹²⁵ X 3¹⁵⁸) is unit’s digit of product of digit at unit’s place of 7⁹²⁵ and 3¹⁵⁸ = 3 * 9 = 27. Hence option 1 is the answer. Try this one: What is the units digit of 13^35? A) 1 B) 3 C) 5 D) 7 E) 9 Let’s begin by looking for a pattern as we increase the exponent. 13^1 = 13 (units digit is 3) 13^2 = 169 (units digit is 9) 13^3 = 2197 (units digit is 7) Aside: As you can see, the powers increase quickly! So, it’s helpful to observe that we need only consider the units digit when evaluating large powers. For example, the units digit of 13^2 is the same as the units digit of 3^2, the units digit of 13^5 is the same as the units digit of 3^5, and so on. Continuing along, we get: 13^1 has units digit 3 13^2 has units digit 9 13^3 has units digit 7 13^4 has units digit 1 13^5 has units digit 3 13^6 has units digit 9 13^7 has units digit 7 13^8 has units digit 1 Notice that a nice pattern emerges. We get: 3-9-7-1-3-9-7-1-3-9-7-1-… As you can see, the pattern repeats itself every 4 powers. I like to say that the “cycle” equals 4 Now that we know the cycle is 4, we can make a very important observation: Whenever n is a multiple of 4, the units digit of 13^n is 1 That is, 13^4 has units digit 1 13^8 has units digit 1 13^12 has units digit 1 13^16 has units digit 1 . . . etc. At this point, we can find the units digit of 13^35 Since 32 is a multiple of 4, 13^32 must have units digit 1. From here, we’ll just continue the pattern: 13^32 has units digit 1 13^33 has units digit 3 13^34 has units digit 9 13^35 has units digit 7 The units digit of 13^35 is 7, which means D is the answer to the original question. For additional practice try these two questions: 1. Find the units digit of 57^30 2. Find the units digit of 34^33 Answers below . . . Answers to above questions: 1. 9 2. 4