Find the remainder when 3^98 is divided by 5.?

Question

Find the remainder when 3^98 is divided by 5.

Answer 1

I use a shorthand Multiple_of_5 for any multiple of 5. So for example Multiple_of_5 + Multiple_of_5 = Multiple_of_5 (when you add two multiples of 5, you get a multiple of 5)


81 = 5 * 16 + 1
81^2 = (5 * 16 + 1 ) * (5 * 16 + 1) = Multiple_of_5 + 1
81^3 = (5 * 16 + 1 ) * (Multiple_of_5 + 1) = Multiple_of_5 + 1

So is easy to see that in general 81^N, where N is an integer, is Multiple_of_5 + 1.
(If you wanted a formal proof you would use induction)

3^98 = 3^96 * 3^2
= (3^4)^24 * 9
= 81^24 * 9
= (multiple_of_5 + 1) * 9
= multiple_of_5 + 9
= multiple_of_5 + 5 + 4
= multiple_of_5 + 4

So the answer is 4

After submitting the above I see that there are a bunch of other answers that all get 4. They are all correct but not really proofs; just because you see a pattern looking at the first 8 or so powers does not show that it will repeat for ever. I think with what I have above you have ironclad proof

Answer 2

Ok, let's look at a pattern here. The remainder when N is divided by M is called N mod M. So you are asking, what is 3^98 mod 5. Two things we know -- it must be a whole number between 0 and 4, and it cannot be 0 because 3^98 has no 5's as factors. So let's look at some lower powers of 3 mod 5 and see if there is a pattern.
3^0 mod 5 = 1
3^1 mod 5 = 3
3^2 mod 5 = 4
3^3 mod 5 = 2
3^4 mod 5 = 1
3^5 mod 5 = 3
3^6 mod 5 = 4
3^7 mod 5 = 2
3^8 mod 5 = 1
So you see there is a cyclic pattern here, 3,4,2,1. You can define this pattern in terms of mod 4. Extening it to 98, you get 3*98 mod 5 = 4.
Interesting question here.

Answer 3

The answer is 4. You can tell because the last number of each power of three progresses as follows: 3, 9, 7, 1, 3, 9, 7, 1,...

So, you can divide 98 by four to find out where in the cycle you lie. You get something with a remainder of 2.... and that plants you on 9. So, your remainder is 4.

I should have gone into math.

Mike

Answer 4

The answer is 4 becuase the remainder of 3^1 is 3, 3^2 is 4, 3^3 is 2, 3^4 is 1, 3^5 is 3, 3^6 is 4, and the pattern continues after every 4 exponents. Therefore, 98-96, which is the highest multiple of 4 that is lower than it, is two, and the remainder of 3^2 divided by 5 is 4

Answer 5

2.
-->3^98 is that same as starting with 1 and multiplying it by 3 98 times. Start with 1*3:
3 3/5 = 0 rem 3
3*3 = 9 9/5 = 1 rem 4
3*3*3 = 27 27/5 = 5 rem 2
3*3*3*3 = 81 81/5 = 16 rem 1
3*3*3*3*3 = 243 243/5 = 48 rem 3
3*3*3*3*3*3 = 729 729/5 = 145 rem 4

the remainder repeat every 4th time (3,4,2,1),
98/4 = 24 remainder 2
So, the remainder repeat as 3,4,2,1 24 times before 2 on the 98th time you multiply by 3.

Answer 6

You divide 398 by using 5 like regularly taking place so 5 doesnt go into 3, so divide 39 by using 5. so your first variety is 7, so there's a 4 left. next deliver down the 8, so it really is 40 8 divided by using 5 so your 2d variety is 9. yet there remains a three left over, it is your the relax

Answer 7

3^98=(3^4)^24 * 9 = 81^24 * 9

The last digit for all powers of 81 is allways 1.
(It is obvious but it can be prooved)

We have to proove that for every n there exists an
integer k so that 81^n=10*k+1

a) For n=1 81^1=81 --> 8*10+1 (ok!)
b) Accept for n=r it is TRUE so 81^r=10m+1
c) For n=r+1 we have
81^(r+1)=81*81^r=81*(10m+1)=(80+1)*(10m+1)
=800m+80+10m+1 = 10*(81m+8) + 1 = 10*S+1


If we mulptiply this number by 9 we get a number with last
digit the number 9 (you can proove it as above!!) so the remainder is 4.

It is not mathematical corect to see a sequence of the remainder
and be sure that it continues infinetely without prooving it!
(Of course in this case the sequence does continue infinitely)

Answer 8

3^0 = 1, 1mod5 = 1
3^1 = 3, 3mod5 = 3
3^2 = 9, 9mod5 = 4
3^3 = 27, 27mod5 = 2
3^4 = 81, 81mod5 = 1
3^5 = 243,343mod5 = 3
so there is repetition every 4th power.
98mod4 = 2, so
3^98mod5 = 3^2mod5 = 4