VeritasPrepKarishma wrote:
idreesma wrote:
i did it the following way (just making sure that its not by luck that i got the right answer)
43^86 /5 ==> since we are looking for a remainder that is 40+3 ... 3^86 [3 repeats in the following manner
3^1=3, 3^2=9, 3^3=7, 3^4=1 [only unit digits]
86/4 ==> Remainder is 2 .. which means that unit digit is going to be 9 (9/5 ==> gives a remainder of 4 as E)
Your logic worked because we are discussing divisibility by 5 here. The last digit decides the remainder when a number is divided by 5.
Remainder when ****7 is divided by 5 will always be 2. Remainder when *****4 is divided by 5 will always be 4. This is so because every number that ends in 0 or 5 is divisible by 5 and only numbers ending in 0 or 5 are divisible by 5. Last digit works
only for 2 and 5.
If you consider divisibility by say 3 or 7 etc last digit logic doesn't work so
be careful.
Dear Karishma,
Please could you just quickly look through on this:
"Second, you have to know that when it comes to these kinds of questions, the only digit that matters is the
UNITS DIGIT of the number."
So is it wrong OR is it true only for 2 and 5?
Can we use cyclicity approach finding the remainder
ONLY when dividing by 2, 5, or 10?
Does the last digit work
ONLY for 2 and 5?
If it's so, I prefer to use ONE general method in order not to mess everything up and to avoid mistakes, what do you think about it, am I right?
[Since 40 is completely divisible by 5, you only have to think about 3^86;
3^86=9^43=(10−1)^43AND as in your post "A Tricky Question on Negative Remainders"
Question: What is the remainder when 3^(7^11) is divided by 5? (here, 3 is raised to the power (7^11))
7^(11) = (8 – 1)^(11)]
- Could I also ask you what the remainder is if for example 3^77/5?
I tried to solve it this way: 3^1*3^76 / 5 = 3*(3^2)^38 / 5 = 3*9^38 / 5 = 3*(10-1)^38 / 5 --> -1^38 = +1*3 = 3 – REMAINDER.
Is my answer correct?
- Is it possible to solve this question by defining the last digit:
As we know 3 has a cyclicity of 4 values in succession: 3 9 7 1 3 9 7 1 3 etc
By applying this method we need 77/4 = 1 – remainder, 1 more than 4, which is 3-->13-(
5*2)=3 – ANSWER
Thank you!
Wishing u all the best