Today, we will discuss the question we left you with last week. It involves a lot of different concepts – remainder on division by 5, cyclicity and negative remainders. Since we did not get any replies with the solution, we are assuming that it turned out to be a little hard.

It actually is a little harder than your standard GMAT questions but the point is that it can be easily solved using all concepts relevant to GMAT. Hence it certainly makes sense to understand how to solve it.

Question: What is the remainder when 3^(7^11) is divided by 5? (here, 3 is raised to the power (7^11))

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

Solution: As we said last week, this question can easily be solved using cyclicity and negative remainders. What is the remainder when a number is divided by 5? Say, what is the remainder when 2387646 is divided by 5? Are you going to do this division to find the remainder? No! Note that every number ending in 5 or 0 is divisible by 5.

2387646 = 2387645 + 1

i.e. the given number is 1 more than a multiple of 5. Obviously then, when the number is divided by 5, the remainder will be 1. Hence the last digit of a number decides what the remainder is when the number is divided by 5.

On the same lines,

What is the remainder when 36793 is divided by 5? It is 3 (since it is 3 more than 36790 – a multiple of 5).

What is the remainder when 46^8 is divided by 5? It is 1. Why? Because 46 to any power will always end with 6 so it will always be 1 more than a multiple of 5.

On the same lines, if we can find the last digit of 3^(7^11), we will be able to find the remainder when it is divided by 5.

Recall from the discussion in your books, 3 has a cyclicity of 4 i.e. the last digit of 3 to any power takes one of 4 values in succession.

3^1 = 3

3^2 = 9

3^3 = 27

3^4 = 81

3^5 = 243

3^6 = 729

and so on… The last digits of powers of 3 are 3, 9, 7, 1, 3, 9, 7, 1 … Every time the power is a multiple of 4, the last digit is 1. If it is 1 more than a multiple of 4, the last digit is 3. If it is 2 more than a multiple of 4, the last digit is 9 and if it 3 more than a multiple of 4, the last digit is 7.

What about the power here 7^(11)? Is it a multiple of 4, 1 more than a multiple of 4, 2 more than a multiple of 4 or 3 more than a multiple of 4? We need to find the remainder when 7^(11) is divided by 4 to know that.

Do you remember the binomial theorem concept we discussed many weeks back? If no, check it out here.

7^(11) = (8 – 1)^(11)

When this is divided by 4, the remainder will be the last term of this expansion which will be (-1)^11. A remainder of -1 means a positive remainder of 3 (if you are not sure why this is so, check last week’s post here). Mind you, you are not to mark the answer as (D) here and move on! The solution is not complete yet. 3 is just the remainder when 7^(11) is divided by 4.

So 7^(11) is 3 more than a multiple of 4.

Review what we just discussed above: If the power of 3 is 3 more than a multiple of 4, the last digit of 3^(power) will be 7.

So the last digit of 3^(7^11) is 7.

If the last digit of a number is 7, when it is divided by 5, the remainder will be 2. Now we got the answer!

Answer (C)

Interesting question, isn’t it?

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

Beautiful question!

Thanks for you insights.

This is how I approached the problem:

Cyclicality of 3 and its units digit give:

3, 9, 27, {…}1,{…} 3,{…} 9….

Dividing each number by 5 gives a remainder, in the following order:

2, 4, 2, 4, 2, 4…….

What is interesting is that the remainder when 3^x is divided by 5 depends on whether x is even or odd. When x is odd, the remainder is 2 and when x is even, the remainder is 4.

Thus, since x=7^11 is a odd number (an odd base to any power is odd), the remainder is 2.

Therefore, answer choice C, as required.