### Finding the digit in the unit’s place of a big number (with power)

This is something I found out when I was in college, we had to find the digit of a bigger number with power (eg:12345^12346) with in a minute, the usual **mod **process is slow, this method (that I’m gonna tell) can solve any bigger problem in less than 10 seconds.

## If the base number ends with 0,1,5,6

Any number with ending number 0/1/5/6 will have 0/1/5/6 in its unit digit respectively no matter how much you rise its power to

Look at these examples:

10^2=10**0**, 20^3=800**0**; 12345670^12345678= ………**0**

11^2=12**1**, 11^3=133**1**, 11^4=1464**1**;

5^2=2**5**; 5^12**5**; 5^4=62**5**

6^2=3**6**;6^3=21**6**;6^4=129**6**

this should be enough to agree

Try and test out something huge on your own

## If the base number ends with 4,9

First let’s look at number ending with 4,9 and how their powers behave

4^1=4, 4^2=16, 4^3=64, 4^4=256

9^1=9, 9^2=81, 9^3=729, 9^4=6561

from observation (of these and many more examples) we can say numbers ending with 4 will have 4 or 6 in their unit’s place, depending on its power, if a number ending with 4 is raised to to an odd degree (eg:1234^25) it will end with 4, if it is raised to an even number (eg:12354^48) it will end with 6

from the examples of numbers ending with 9 above it is also evident that nine behaves in similar fashion, a number ending with 9 raised to odd degree (eg:9^35) will have 9 in its unit place, and 1 in its unit place if raised to even degree

## If the base number ends with 2,3,7,8

2,3,7,8 behave similarly in their own way, now lets look at examples

2^1=2; 2^2=4;2^3=8, 2^4=16, 2^5=32, 2^6=64, 2^7=128, 2^8=256, 2^9=512

observe that the numbers 2,4,8,16,32,64,128,256,512,1024 have a pattern, their ending numbers are 2,4,8,6,2,4,8,6…., 2,4,8,6 i.e there is cycle for every four degree, this is an important fact to remember, any big number can be solved just by looking what is left after we divide the power by 4.

2^1201 will have 2 in unit’s place (since 1200 is divisible by 4 and 1 is the remainder)

2^1202 will have 4

2^1203 will have 8

2^1204 will have 6

3,7,8 behave similarly so I’m only gonna list those 4 numbers which repeat again and again for each number

Number ending with 2 will have pattern of 2,4,8,6 in its unit place as we go on raising the power

Number ending with 3 will have pattern of 3,9,7,1 in its unit place as we go on raising the power

Number ending with 7 will have pattern of 7,9,3,1 in its unit place as we go on raising the power

Number ending with 8 will have pattern of 8,4,2,6 in its unit place as we go on raising the power

January 25th, 2012 at 5:32 am

thanks a lot guys, you saved a lot of time for me

July 31st, 2012 at 3:37 am

Great job…..