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

Question

A. 0
B. 1
C. 2
D. 3
E. 4

Senthil1981 · Accepted Answer

Answer is B : 1

We can either use the Mod formula or since the question can be simplified, let's do that.

\(3 ^{24}\) can be simplified to \(9^{12}\) and then to \(81 ^ {6}\). So if the last digit is 1 , then remainder of division by 5 will always be 1.

(Share a Kudos, if you like this explanation :-D

)

)

ScottTargetTestPrep · Answer

When solving this problem, we should recall the rule that we can determine the remainder when a number is divided by 5 by simply dividing the units digit of that number by 5. Thus, to determine the remainder when 3^24 is divided by 5, we need to first calculate the units digit of 3^24.

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 units 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 units 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.

Answer: B

Mbawarrior01 · Answer

We can also use the following method :

Since the divisor is 5, the number that ends in 0 or 5 will be fully divisible. You can deduce the remainder utilising the units digit of the number.

Now, 3^24

Using the concept of cyclicity => 3^24 will have "1" in units digit.

Thus, 1/5 => Remainder is 1.

This concept may help when the power is a bigger number.

Thanks

adiagr · Answer

3^{24}  can be written as  (3^4)^6

(81)^6

This no. will have last digit as 1. On dividing by 5, remainder will always be 1. (241/ 5, remainder will be 1; 81/5: remainder will be 1)

B is the answer.

OptimusPrepJanielle · Answer

Remainder (3^24/5) = Remainder (81^6/5) = 1
Correct Option: B

Another way:
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

Ndkms · Answer

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

A. 0
B. 1
C. 2
D. 3
E. 4

Whenever you see ridiculous power there two approaches 1) Simplification or 2) Find the pattern. For this question we are falling on the second category so:

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81 ... by now you should be quite suspicious that the patter will repeat within a reasonable number of calc. steps
3^5 = 24 3  ... indeed last digit is 3

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

SO YOU KNOW that the cycle repeats every 4 steps. The number 4 is a factor of 24 hence 24/ 4 = 6 cycles  . Now why are we doing all this? Take a step back and think

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 THAT ( 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 therefore B

In my solution I show too much detail but all this has to be very intuitive. Excellent question as it combines many topics.

Sumitj2016 · Answer

this is a cyclic problem: 3^0 / 5. rem = 1 3^1 / 5. rem = 3 3^2 / 5. rem = 4 3^3/5 . rem = 2 3^4 / 5. rem = 1 cycle will repeat at 4 => 3^24 = (3^4)^6 rem = 1

Ndkms · Answer

Scot nice solution but I fully disagree with your wording I have in bold. There are so many rules in GMAT that you have to remember so is beneficial to eliminate as many as you can.

Indeed the rule you mentioned is true and well spotted although test-takers they need to be able to derive it on the fly...

The robust "on the fly" approach (GMAT is not for mathCrunchers)

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81 
3^5 = 243
---------------------
3^6 = 729
3^7 = 2187 etc etc

Point 1) The question asks 3^24 and with confidence we can say that 3^24 it will be a number that will have more than two digits - common sense

Point 2) Lets rearrange the numbers little bit

3^1 = 3
3^2 = 9
3^3 = 27 =  20 + 7 
3^4 = 81 =  80 + 1  
3^5 = 243 =  240 + 3 
---------------------
3^6 = 729 =  720  + 9
3^7 = 2187 =  2180  + 7 etc etc

So all results can be written as mul(5) + Integer_x. Thus 3^24 / 5 = mul(5)/5 + integer_x/5 = integer_y + integer_x/5

As you can see integer_y is just an integer value because mul(5) / 5 it will give a nice round value

Integer_x is the number 1 because of the pattern and hence integer_x / 5 = Remainder / 5

yezz · Answer

another way is to rewrite

3^24 = (5-2)^24 ... this is a polynomial function and all terms but the last will be divisible by 5 ( e.g : (5-3)^2 = (25-30+9)) the last term is (2^8)^3 = (256)^3 i.e. ending in 6 thus when divided by 5 will yield a remainder of 1

B

Abhishek009 · Answer

\frac{3^4}{5} = \frac{81}{5} = Remainder \ 1

3^{24} = 3^{4*6}

Thus, remainder will be 1

Hence, answer will be (B) 1

Ndkms · Answer

Can you please elaborate what someone has to pickup with the statement above?

Fair enough 3^4 = 81 but then you basically saying that 3^24 = 3^(4*6) = (81) ^ 6

and then what?

ScottTargetTestPrep · Answer

Hi Ndkms -

Thanks for the feedback. Looking at our two solutions, we actually followed a similar path: find the pattern of units digits of 3, and then divide the units digit of 3^24 by 5 to get the answer. 
 
Regarding "eliminating rules" on the GMAT, I have to disagree with your stance. The GMAT is an exam that forces students to answer questions in a limited amount of time. Thus, ANY rule/method that can be easily memorized that will allow students to efficiently and effectively attack a question, without having to create a scenario for its solution, is a valuable rule/method because it will save them valuable time on the GMAT. Does that mean that students should go crazy and memorize every rule/method ever written--no, of course not.

Of course the GMAT is not a number-crunching test; I don’t think you’ll find many people who will disagree with you on that point. The test is clearly a critical-thinking test. However, when you examine the research of those on the forefront of studying rational thought and decision-making, people such as Daniel Kahneman and Keith Stanovich, you see that without rules/methods to follow, critical thinking becomes difficult. Just as we need software on our computers to do spreadsheet analysis, humans require “mindware” in their brains to reason logically. Mindware is simply a term that describes all of the content, facts, and rules that one needs to logically reason through a problem. In general, the more pertinent mindware a person has regarding a certain problem, the better positioned that person will be, all else equal, to begin thinking logically about that problem.

Furthermore, the rule/method that I referenced in my solution simply goes one step further from the rule of units digit patterns, which is a commonly known GMAT rule. Thus, I don't think it would take too much effort to memorize and effectively use it.

But, the beauty of math, and of the GMAT, is that there are many ways to solve most problems. And, one beauty of this forum is that we can all collaborate and help each other, respectfully. So, thank you for the input.

Happy Studying.

Scott

sobby · Answer

One way to do this is

when a number is divided by 5) the pattern 3^n follows (unit digit pattern):3,9,1,3,9,1.... So at 3^24 , we will get 1 as unit digit ..so dividing by 5 ..the remainder will be 1.

daviddaviddavid · Answer

hey I've got a questions =)

could we just assume that we´re dealing with 3^23; and then 23/4(cyclicality) = remainder of 3 therefore 3^3 = 2 7  but actually its 3^23 so the next number in our cycle 3^4 = 8 1  ??

is that a valid approach ??

hope its clear

jetmat · Answer

Hello,

This is not obvious for me.
It works because the factor here is five: 11/5=>1; 21/5=>1; etc. It wouldn't work for the factor 4, for which the remainder should be 1, not in its power cyclicity.

Am I wrong? Please elaborate.
Thx

Prasannathawait · Answer

3^ 21 will come with units digit 1 as cyclicity of 3 is 4, and 4th power term has 1 in the units digit.

1 is the right answer

MHIKER · Answer

Bunuel  I need little help

Cyclicity of 2, 3, 7, 8 is 4, so will not I divide 24 by 4? If I do that the remainder is "0".

So, 3^0=1

\frac{1}{5}= remainder \ is 1

The answer is B.

Is that process correct?

100mitra · Answer

(a^n) when divided by d, 
will always give remainders which will have a pattern and 
will move in cycles of r such that r is less than or equal to d
- Integer ending with  0 ,  1 ,  5  or  6 , in the integer power k>0, has the same last digit as the base.
- Integers ending with  2 ,  3 ,  7  and  8  have a cyclicity of  4 .
3^24 divided by 5
3^1 =  3  (1/5 = Reminder 1)
3^2 = 9
3^3 = 7
3^4 =  1  (81/5 = Reminder 1)
3^5 = 3
Similarly - 3^24 divided by 5 = Reminder 1 =  Option B

dkorir · Answer

Honestly, I have no idea what you guys are getting 1. can you share more information on this?

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

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

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