We often approximate the numbers for convenience and the digits to the rightmost side are neglected the most. I wanted to see of there is any pattern in the ignored right digits. For this problem, I have considered the increasing powers of a number. For example, have a look at this sequence.

2^{1} = 2

2^{2} = 4

2^{3} = 8

2^{4} = 16

2^{5} = 32

2^{6} = 64

2^{7} = 128

Have you found any pattern ?

The last digits are repeating as 2,4,8,6,2,4,8.

What if this is done not just for 2 but for the next numbers too ?

Let us see it first before plotting.

3^{1} = 3

3^{2} = 9

3^{3} = 27

3^{4} = 81

3^{5} = 243

3^{6} = 729

3^{7} = 2187

It is more apparent now isn’t it. The sequence is 3,9,7,1,3,9,7………

This being the case, I have taken it a few more steps ahead. I have run the same sequence with last two digits, three digits, four digits and so on. After proceeding to the higher numbers, *I was taken aback by the visual patters that were emerging from the seemingly mundane and neglected string of integers.*

For plotting, what has been done is C^{n} for increasing n (which means increasing powers of a constant number)

2^{n} for last three digits

2^{n} for last four digits

I have skipped the other one digit numbers for convenience but there were similar patterns in them too. The most intriguing ones are shown below.

### last 5 digits of 11^{n}

### last 6 digits of 11^{n}

### last 5 digits of 13^{n}

### last 6 digits of 14^{n}

### last 7 digits of 14^{n}

### last 6 digits of 16^{n}

### last 5 digits of 18^{n}

### last 6 digits of 18^{n}

### last 4 digits of 19^{n}

### last 5 digits of 19^{n}

### last 4 digits of 21^{n}

#### last 5 digits of 21^{n}

This post will be updated when other similar patterns are discovered.

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Wonderful work, your patience is appreciable.

Thanks buddy.

Amazing representation of pattern. Happy New Year Friend 🙂

Thanks man. Happy New year.