Re: What is the remainder when
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05 Feb 2020, 09:12
One more way to solve
calculate the remainder for each 3^1,3^2,3^3,3^4,3^5 when divided by 5
For 3^1 remainder = 3
For 3^2 remainder = 4
For 3^3 remainder = 2
For 3^4 remainder = 1
For 3^5 remainder = 3
we can see that the series of the remainder is repeating after 3^4
so, we can conclude that the remainders of the upcoming numbers will be in this same order i.e 3,4,2,1
3^5 will have a remainder = 3
3^6 will have a remainder of 4
3^7 will have a remainder of 2
...and so on
thus we can conclude that after every 4 places we will get 1 as the remainder
After this, we select a number less than 35(cause we are asked for the value of 3^35) which is divisible by 4(cause we know that for every 4 places we will get 1 as a remainder)
such number is 32
from the above conclusion, we can say that the remainder of 3^32 will be 1
Now, we continue our series of the remainder(3,4,2,1) for 3 more places(cause we need to find the value of 35 and 35 is after 3 places from 32)so we get 1....3,4,2
thus we can conclude that the remainder of 3^35 will be 2
I know this is a lengthy description but once you understand the concept it will be damn faster!