Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: If n = (33)^43 [#permalink]
03 Oct 2012, 23:33

8

This post received KUDOS

1

This post was BOOKMARKED

Archit143 wrote:

If n = (33)^43 + (43)^33 what is the units digit of n?

0 2 4 6 8

pls suggest approach for such probs

In such type of exponents problem, first find the cyclicity of unit digit of 3. which is (3,9,7,1)

Now convert exponent value in to multiple of 4. here 43=4(10)+3 so take 3^3 as unit value of first term which is 27 ===> unit digit is 7 (3,9,7,1) Now 33=4(8)+1,so take 3^1 as unit value of second term which is 3^1====> unit digit is 3 (3,9,7,1)

Now unit digit of n = unit digit of first term + unit digit of 2nd term = 7+3=10 = unit digit is 0

Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
04 Oct 2012, 03:27

1

This post received KUDOS

Expert's post

Archit143 wrote:

If n = (33)^43 + (43)^33 what is the units digit of n?

A. 0 B. 2 C. 4 D. 6 E. 8

Topic moved to PS forum.

Check Number Theory chapter of our Math Book to know how to deal with such problems: math-number-theory-88376.html (LAST DIGIT OF A POWER part) _________________

Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
12 Oct 2013, 09:11

1

This post received KUDOS

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: If n = (33)^43 [#permalink]
09 Mar 2014, 10:43

bhavinshah5685 wrote:

Archit143 wrote:

If n = (33)^43 + (43)^33 what is the units digit of n?

0 2 4 6 8

pls suggest approach for such probs

In such type of exponents problem, first find the cyclicity of unit digit of 3. which is (3,9,7,1)

Now convert exponent value in to multiple of 4. here 43=4(10)+3 so take 3^3 as unit value of first term which is 27 ===> unit digit is 7 (3,9,7,1) Now 33=4(8)+1,so take 3^1 as unit value of second term which is 3^1====> unit digit is 3 (3,9,7,1)

Now unit digit of n = unit digit of first term + unit digit of 2nd term = 7+3=10 = unit digit is 0

I can't understand why it has to be a multiple of 4. I mean, 43 can also be expressed as (2)(21)+1 but this would be wrong. Generally in such exercises it has to be a multiple of 4, or I don't get something?

Re: If n = (33)^43 [#permalink]
09 Mar 2014, 11:56

3

This post received KUDOS

Expert's post

3

This post was BOOKMARKED

Chrysopigi89 wrote:

bhavinshah5685 wrote:

Archit143 wrote:

If n = (33)^43 + (43)^33 what is the units digit of n?

0 2 4 6 8

pls suggest approach for such probs

In such type of exponents problem, first find the cyclicity of unit digit of 3. which is (3,9,7,1)

Now convert exponent value in to multiple of 4. here 43=4(10)+3 so take 3^3 as unit value of first term which is 27 ===> unit digit is 7 (3,9,7,1) Now 33=4(8)+1,so take 3^1 as unit value of second term which is 3^1====> unit digit is 3 (3,9,7,1)

Now unit digit of n = unit digit of first term + unit digit of 2nd term = 7+3=10 = unit digit is 0

I can't understand why it has to be a multiple of 4. I mean, 43 can also be expressed as (2)(21)+1 but this would be wrong. Generally in such exercises it has to be a multiple of 4, or I don't get something?

If n = (33)^43 + (43)^33 what is the units digit of n?

A. 0 B. 2 C. 4 D. 6 E. 8

First of all, the units digit of (33)^43 is the same as that of 3^43 and the units digit of (43)^33 is the same as that of 3^33. So, we need to find the units digit of 3^43 + 3^33.

Next, the units digit of 3 in positive integer power repeats in blocks of four {3, 9, 7, 1}: 3^1=3 (the units digit is 3) 3^2=9 (the units digit is 9) 3^3=27 (the units digit is 7) 3^4=81 (the units digit is 1) 3^5=243 (the units digit is 3 again!) ...

Thus: The units digit of 3^43 is the same as the units digit of 3^3, so 7 (43 divided by the cyclicity of 4 gives the remainder of 3). The units digit of 3^33 is the same as the units digit of 3^1, so 3 (33 divided by the cyclicity of 4 gives the remainder of 1).

Therefore the units digit of (33)^43 + (43)^33 is 7 + 3 = 0.

Re: If n = (33)^43 [#permalink]
09 Mar 2014, 12:04

4

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

Bunuel wrote:

If n = (33)^43 + (43)^33 what is the units digit of n?

A. 0 B. 2 C. 4 D. 6 E. 8

First of all, the units digit of (33)^43 is the same as that of 3^43 and the units digit of (43)^33 is the same as that of 3^33. So, we need to find the units digit of 3^43 + 3^33.

Next, the units digit of 3 in positive integer power repeats in blocks of four {3, 9, 7, 1}: 3^1=3 (the units digit is 3) 3^2=9 (the units digit is 9) 3^3=27 (the units digit is 7) 3^4=81 (the units digit is 1) 3^5=243 (the units digit is 3 again!) ...

Thus: The units digit of 3^43 is the same as the units digit of 3^3, so 7 (43 divided by the cyclicity of 4 gives the remainder of 3). The units digit of 3^33 is the same as the units digit of 3^1, so 3 (33 divided by the cyclicity of 4 gives the remainder of 1).

Therefore the units digit of (33)^43 + (43)^33 is 7 + 3 = 0.

Answer: A.

For more on this check theory and problems listed below:

Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
09 Sep 2014, 06:13

Expert's post

honchos wrote:

Bunuel,

I found mistake in the solution given by you-

The Cycle here is

9, 7, 1, 3 Not 3, 9, 7 , 1 (as you stated)

Multiply 33 x 33 Once, what do we get 1089, Unit digit is 9. So the cycle should start at 9

By that logic we should add

1 + 9 = 10 Answer is still A.

But lets tweak it-

(33)^43 + (43)^34

Bunuel you will do this with your chosen cycle- 7+9 = 16, Your answer will be 6

With my chosen Cycle- 1+7 = 8.

That could be a trap if they play on the starting point of the cycle.

Please re-read the solution and follow the links given above.

You should start with power of 1: 3^1=3 (the units digit is 3) 3^2=9 (the units digit is 9) 3^3=27 (the units digit is 7) 3^4=81 (the units digit is 1) 3^5=243 (the units digit is 3 again!) ... _________________

If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
05 Oct 2014, 04:40

1

This post received KUDOS

2

This post was BOOKMARKED

Based on the Number Theory > Last digit of a power (in the GMATclub forum, I cannot link to the post)

Step 1: Find the last digit of 33^43

find the cyclicity of 33 33^1= _3 33^2= _9 33^3= _7 33^4= _1 33^5= _3 -> 33 has the cyclicity of 4

Divide the power by the cyclicity: 43/4 -> remainder is 3 (refer to the 3rd position in the cyclicity) -> the last digit of 33^43 is the same as that of 33^3 -> 7 is the last digit

Step 2: Similarly, find the last digit of 43^33 43 also has the cyclicity of 4 (ends with 3) Divide 33 by 4 -> remainder is 1 -> refer to the 1st position in the cyclicity -> last digit is 3

If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
06 Feb 2015, 16:38

I am completely lost on this question.

My (incorrect) approach was to drop the ten units digit of each number and of each exponent, so that I ended up with 3^3 + 3^3 - therefore the answer would be 7+7 = 14 or answer choice C. I'm at a complete loss as to why this is incorrect and am having trouble understanding the explanations above.

Is anyone able to kindly provide an explanation for the theory behind this, in layman's terms for this poor layman?

Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
09 Feb 2015, 00:59

Expert's post

Anonamy wrote:

I am completely lost on this question.

My (incorrect) approach was to drop the ten units digit of each number and of each exponent, so that I ended up with 3^3 + 3^3 - therefore the answer would be 7+7 = 14 or answer choice C. I'm at a complete loss as to why this is incorrect and am having trouble understanding the explanations above.

Is anyone able to kindly provide an explanation for the theory behind this, in layman's terms for this poor layman?

Thanks

Please read the solutions above and follow the links provided. _________________

Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
10 May 2015, 00:50

Bunuel wrote:

Thus: The units digit of 3^43 is the same as the units digit of 3^3, so 7 (43 divided by the cyclicity of 4 gives the remainder of 3). The units digit of 3^33 is the same as the units digit of 3^1, so 3 (33 divided by the cyclicity of 4 gives the remainder of 1).

Therefore the units digit of (33)^43 + (43)^33 is 7 + 3 = 0.

Answer: A.

I do not understand this part with remainder and cyclicity, why do i need to divide 43 by 4 to get a remainder of one which is then the units digit??

Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
11 May 2015, 02:03

1

This post received KUDOS

Expert's post

noTh1ng wrote:

Bunuel wrote:

Thus: The units digit of 3^43 is the same as the units digit of 3^3, so 7 (43 divided by the cyclicity of 4 gives the remainder of 3). The units digit of 3^33 is the same as the units digit of 3^1, so 3 (33 divided by the cyclicity of 4 gives the remainder of 1).

Therefore the units digit of (33)^43 + (43)^33 is 7 + 3 = 0.

Answer: A.

I do not understand this part with remainder and cyclicity, why do i need to divide 43 by 4 to get a remainder of one which is then the units digit??

You take the remainder obtained by dividing the power by cyclicity. For example, 3^43 --> 43 (power) divided by the cyclicity of 4 gives the remainder of 3 --> the units digit of 3^43 is the same as the units digit of 3^3.

On September 6, 2015, I started my MBA journey at London Business School. I took some pictures on my way from the airport to school, and uploaded them on...

When I was growing up, I read a story about a piccolo player. A master orchestra conductor came to town and he decided to practice with the largest orchestra...

I’ll start off with a quote from another blog post I’ve written : “not all great communicators are great leaders, but all great leaders are great communicators.” Being...