Question

The unit digit of the number \[{{\left( 27 \right)}^{50}}+{{\left( 18 \right)}^{50}}\] is:
(a) 1
(b) 3
(c) 5
(d) 7

Answer 1

Hint: To solve the given question, we will determine the pattern in which the unit digits of \[{{\left( 27 \right)}^{n}}\] and \[{{\left( 18 \right)}^{n}}\] respectively will vary when we will put different values of n. Once we have determined the variation of unit digit with respect to n, we will put n = 50 and determine the unit digit of \[{{27}^{50}}\] and \[{{18}^{50}}\] and then we will add their last digits to get the answer.

Complete step by step solution:
To start with, we will first find out the initial powers of 7 as 27 ends with 7, and the unit digit of \[{{\left( 27 \right)}^{n}}\] will depend on \[{{\left( 7 \right)}^{n}}.\] Thus, now we will write the initial powers of 7 to get the pattern.
When n = 1, we have \[{{\left( 7 \right)}^{1}}=7.\]
When n = 2, we have \[{{\left( 7 \right)}^{2}}=49.\]
When n = 3, we have \[{{\left( 7 \right)}^{3}}=343.\]
When n = 4, we have \[{{\left( 7 \right)}^{4}}=2401.\]
When n = 5, we have \[{{\left( 7 \right)}^{5}}=16807.\]
When n = 6, we have \[{{\left( 7 \right)}^{6}}=117649.\]
Here, we can see that the unit digit of the power of 7 starts repeating at n = 5. Thus, the unit digits will be the same when n = 1, 5, 9….. and it will be equal to 7. When n = 2, 6, 10…. the unit digit will be equal to 9. When n = 3, 7, 11, 15…. The unit digit will be equal to 3. When n = 4, 8, 12, ….. the unit digit will be equal to 1. Now, we have to find the unit digit of \[{{\left( 27 \right)}^{50}}.\] For this, we will find the unit digit of \[{{\left( 7 \right)}^{50}}.\] Thus, the value of n = 50.
Now, 50 is of the form (4r + 2), and from the above discussion, the terms n = 2, 6, 10 are in the form (4r + 2). So, the unit digit will be 9. Thus, the unit digit of \[{{\left( 27 \right)}^{50}}=9.\]
Now, we will do the same thing for \[{{\left( 18 \right)}^{50}}.\] As 18 ends with 8 we will find the initial powers of 8.
When n = 1, we have \[{{\left( 8 \right)}^{1}}=8.\]
When n = 2, we have \[{{\left( 8 \right)}^{2}}=64.\]
When n = 3, we have \[{{\left( 8 \right)}^{3}}=512.\]
When n = 4, we have \[{{\left( 8 \right)}^{4}}=4096.\]
When n = 5, we have \[{{\left( 8 \right)}^{5}}=32768.\]
When n = 6, we have \[{{\left( 8 \right)}^{6}}=262144.\]
Here, we can see that when n is of the form (4r + 1), the unit digit will be 8.
When n is of the form (4r + 2), the unit digit will be 4.
When n is of the form (4r + 3), the unit digit will be 2.
When n is of the form (4r + 4), the unit digit will be 6.
Now, \[{{\left( 8 \right)}^{50}}\] will have the unit digit 4 because 50 is of the form (4r + 2). Thus, the unit digit of \[{{\left( 18 \right)}^{50}}=4.\]
Thus, the sum of the unit digits of \[{{\left( 27 \right)}^{50}}+{{\left( 18 \right)}^{50}}=9+4=13.\]
Now, 13 has the unit digit as 3. So, \[{{\left( 27 \right)}^{50}}+{{\left( 18 \right)}^{50}}\] has unit digit 3.
Therefore, option (b) is the right answer.

Note: The answer would have been the same even if we had \[{{\left( 77 \right)}^{50}}+{{\left( 28 \right)}^{50}}.\] Thus it is because the unit digit of these numbers is the same. As long as the unit digits are the same, we will get the same answer. Thus, we can say that, \[{{\left( 10a+7 \right)}^{50}}+{{\left( 10b+8 \right)}^{50}}\] will have the unit digit as 13 (a and b are whole numbers).