Applications Of Power Cycle

There are various applications of the power cycle in the CAT quantitative aptitude questions. Some of the important applications and related questions are given here. These examples can help the candidates to get well acquainted with the power cycle and its applications in different questions.

 

1. Power Expression of the Form ab

In this, direct questions are asked to find the unit digit of a power expression of the form ab

Example 1:

Find the last digit of 455.

Solution:

55 can be written as 4k+3. In the cycle of 4, in the third position, we get 4 itself. So, the last digit of the expression is 4 itself.

Example 2:

Find the last digit of 12345734

Solution:

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.

2. Remainder Questions 

In these questions, the remainder based questions are asked where the divisibility of the divisor depends only on its unit digit (like when the divisors are 2,5,10).

Example 3:

Find the remainder when 375 is divided by 5.

Solution:

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.

 

Example 4:

Find the remainder when 4466 is divided by 10.

Solution:

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 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. An example is given below.

Example 5:

Find the unit digit of 734n, where n is a positive integer.

Solution:

Put n=1, the problem reduces to 734, 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.

 

3. 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 by 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.

Example 6:

The sum of the third powers of the first 100 natural numbers will have a units digit of?

a) 3

b) 6

c) 9

d) 0

Solution:

The question asks us to find 13+23+33+…….1003 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 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’s find out the unit digit of each element in the first group. Using the concept of power cycle, it can be found as

Applications Of Power Cycle

Therefore units digit of 13-1003= ..5×10= ..0

Logical 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.

 

4. To Find The Right Most Non-Zero Integer

Example 6:

Find the rightmost non-zero integer of the expression 1430343+1470367?

a) 3

b) 9

c) 7

d) 1

Solution:

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 be 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.

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