Question

How to find unit digit of single number over multiple powers?

Answer 1

These numbers can be broadly classified into three categories for this purpose:

Digits 0, 1, 5 & 6: When we observe the behaviour of these digits, they all have the same unit's digit as the number itself when raised to any power, i.e. 0^n = 0, 1^n =1, 5^n = 5, 6^n = 6. Let's apply this concept to the following example.

Example: Find the unit digit of following numbers:

185563
Answer= 5

2716987
Answer= 1

15625369 
Answer= 6

190654789321 
Answer= 0

Digits 4 & 9: Both these numbers have a cyclicity of only two different digits as their unit's digit.

Let us take a look at how the powers of 4 operate: 41 = 4, 
42 = 16, 
43 = 64, and so on.
Hence, the power cycle of 4 contains only 2 numbers 4 & 6, which appear in case of odd and even powers respectively.

Likewise, the powers of 9 operate as follows:
91 = 9, 
92 = 81, 
93 = 729, and so on.
Hence, the power cycle of 9 also contains only 2 numbers 9 & 1, which appear in case of odd and even powers respectively.

So, broadly these can be remembered in even and odd only, i.e. 4odd = 4 and 4even = 6. Likewise, 9odd = 9 and 9even = 1.

Example: Find the unit digit of following numbers:

189562589743 
Answer = 9 (since power is odd)

279698745832 
Answer = 1(since power is even)

154258741369 
Answer = 4 (since power is odd)

19465478932 
Answer = 6 (since power is even)

Digits 2, 3, 7 & 8: These numbers have a power cycle of 4 different numbers.

21 = 2, 22 = 4, 23 = 8 & 24 = 16 and after that it starts repeating.
So, the cyclicity of 2 has 4 different numbers 2, 4, 8, 6.

31 = 3, 32 = 9, 33 = 27 & 34 = 81 and after that it starts repeating.
So, the cyclicity of 3 has 4 different numbers 3, 9, 7, 1.

7 and 8 follow similar logic.

So these four digits i.e. 2, 3, 7 and 8 have a unit digit cyclicity of four steps.