Question: What is the remainder when 3^24 is divided by 5?

1. 0
2. 1
3. 2
4. 3
5. 4

Correct Answer: B

Solution and Explanation:
Approach Solution 1:

Let’s start by evaluating the pattern of the units digits of 3^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 3. When writing out the pattern, notice that we are ONLY concerned with the units digit of 3 raised to each power.

3^1 = 3
3^2 = 9
3^3 = 7
3^4 = 1
3^5 = 3

The pattern of the unit digit of powers of 3 repeats every 4 exponents. The pattern is 3–9–7–1. In this pattern, all positive exponents that are multiples of 4 will produce a 1 as their units digit. Thus:
3^24 has a unit digit of 1.
Finally, since the remainder is 1 when 1 is divided by 5, the remainder is 1 when 3^24 is divided by 5.

Approach Solution 2:

Remainder (3^24/5) = Remainder (81^6/5) = 1

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243

Hence the cyclicity of 3 = 4
Therefore 3^24 will have the last digit = 1
Remainder when divided by 5 = 1

Approach Solution 3:

Whenever there is ridiculous power there are two approaches 1) Simplification or 2) Find the pattern. For this question the second category falls so:

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81 ...
3^5 = 243 ... indeed last digit is 3

Hence the cycle of the last digit is 3 , 9 , 27, 81 | 3 , 9 ....

The number 4 is a factor of 24 hence 24/ 4 = 6 cycles.
3^24 = GazilionXyZ i.e. it cannot be calculated under GMAT condition unless you are a prodigy or a numberCruncher. BUT by trying to find the pattern you revealed another clue which is:

3^4 = 81 = 80 +1 : 80 is a mul(10) + 1 Therefore 3 raised to any power mul(4) effectively is a mul(10) + 1
For the sake of verification lets do 3^8 in order to prove it. So 3^8 = 6561 = 6560 + 1 Indeed 6560 = mul(10) and therefore 6561 = mul(10) + 1
This means ( 3^24 ) / 5 = ( mul(10) + 1 ) / 5 = INT + 1/5

Basic GMAT theory suggests that Dividend / Divisor = Quotient + Remainder/Divisor and by doing the mapping

Remainder = 1

Approach Solution 4:

The problem statement asks to find out the remainder when 3^24 is divided by 5.

We can solve this question by using the method of cyclicity.
As given in the question, the divisor is 5 hence the number which ends with 0 or 5 will be fully divisible.
The remainder can be taken out by using the unit digit of the number.

According to the cyclicity concept, 3^24 will have 1 in the unit’s digit.
Thus, the remainder will be 1 as 1/5 = 1
Hence, when 3^24 is divided by 5 the remainder will also be 1.

Approach Solution 5:
The problem statement asks to find out the remainder when 3^24 is divided by 5.

This is a sort of cyclic problem.

When 3^0 is divided by 5, the remainder will be = 1
When 3^1 is divided by 5, the remainder will be = 3
When 3^2 is divided by 5, the remainder will be = 4
When 3^3 is divided by 5, the remainder will be = 2
When 3^4 is divided by 5, the remainder will be = 1
….and so on.
Hence, when 3^24 is divided by 5, which implies (3^4) ^6, the remainder will be 1.

“What is the remainder when 3^24 is divided by 5?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Official Guide". To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. GMAT Quant practice papers improve the mathematical knowledge of the candidates as it represents multiple sorts of quantitative problems.

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