In the CAT quantitative aptitude section, questions based on the power cycle concept or cyclicity of numbers are common. Every year few questions are based on this topic and thus, it is crucial to be thorough with this topic. These questions are easily solvable and can improve the overall CAT scores. Please refer BYJU’S CAT College Predictor tool.
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”. Thus, the power cycle of a number depends on its’ unit digit.
Consider the power cycle of 2:
23=8 27 =128
24=16 28 =256
It can be observed that 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
Note:- You don’t need to remember the frequencies, as in every case, the frequency of 4 is valid.
There are various applications based questions asked in the CAT exam on this topic. Visit applications of power cycle to know various types of questions with solved examples.
One of the application questions is given below.
Find the remainder when 375 is divided by 5.
Step 1: Express the power in the form, 4k+x where x=1, 2, 3, 4. In this case 75 = 4k+3.
Step 2: Take the power cycle of 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.
Note: 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.