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 500,000 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:

218. If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10? (1) x = 25. (2) y = 1.

If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10?

Last digit of \(3^{positive \ integer}\) repeats in blocks of 4: {3, 9, 7, 1} - {3, 9, 7, 1} - ... So cyclicity of the last digit of 3 in power is 4. Now, \(3^{4+4x}=3^{4(1+x)}\) will have the same last digit as \(3^4\) which is 1 (remainder upon division the power 4+4x by cyclicity 4 is 0, which means that \(3^{(4+4x)}\) will have the same last digit as 3^4).

Now \(3^{(4 + 4x)}+9^y\) will be divisible by 10 if \(y=odd\), in this case the last digit of \(9^{odd}\) will be 9 (9 has a cyclicity of 2: {9, 1} - {9, 1} - ...) so the last digit of \(3^{(4 + 4x)}+9^y\) will be 1+9=0.

(1) x = 25. Not sufficient. (2) y = 1 --> \(y=odd\) -> remainder=0. Sufficient.

remainder upon division the power 4+4x by cyclicity 4 is 0, which means that 3^{(4+4x)} will have the same last digit as 3^4)

Regards, Subhash

Theory:

First of all note that the last digit of xyz^(positive integer) is the same as that of z^(positive integer);

• Integers ending with 0, 1, 5 or 6, in the positive integer power, have the same last digit as the base (cyclicity of 1): xyz0^(positive integer) ends with 0; xyz1^(positive integer) ends with 1; xyz5^(positive integer) ends with 5; xyz6^(positive integer) ends with 6;

• Integers ending with 2, 3, 7 and 8 have a cyclicity of 4; For example last digit of \(xyz3^{positive \ integer}\) repeats in blocks of 4: {3, 9, 7, 1} - {3, 9, 7, 1} - ... So cyclicity of the last digit of 3 in power is 4.

• Integers ending with 4 and 9 have a cyclicity of 2.

Now, last digit of \(xyz3^{positive \ integer}\) repeats in blocks of 4: {3, 9, 7, 1} - {3, 9, 7, 1} - ... means that: 3^1, 3^5, 3^9, 3^13, ..., 3^(4x+1), all will have the same last digit as 3^1 so 1; 3^2, 3^6, 3^10, 3^14, ..., 3^(4x+2), all will have the same last digit as 3^2 so 9; 3^3, 3^7, 3^11, 3^15, ..., 3^(4x+3), all will have the same last digit as 3^3 so 7; 3^4, 3^8, 3^12, 3^16, ..., 3^(4x), all will have the same last digit as 3^4 so 1;

So to get the last digit of 3^x, (where x is a positive integer) you should divide x by 4 (cylcility) and look at the remainder: If remainder is 1 then the last digit will be the same as for 3^1 (so the first digit from the pattern {3, 9, 7, 1}); If remainder is 2 then the last digit will be the same as for 3^2 (so the second digit from the pattern {3, 9, 7, 1}); If remainder is 3 then the last digit will be the same as for 3^3 (so the third digit from the pattern {3, 9, 7, 1}); If remainder is 0 then the last digit will be the same as for 3^4 (so the fourth digit from the pattern {3, 9, 7, 1});

Next, as 4+4x (the power of 3^(4+4x)) is clearly divisible by 4 (remainder 0) then the last digit of 3^(4+4x) is the same as the last digit of 3^4 so 1.

You can apply this to integers ending with other digits as well (with necessary modification of pattern).

Re: If x and y are positive integers, what is the remainder when [#permalink]

Show Tags

04 Feb 2014, 06:49

banksy wrote:

If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10?

(1) x = 25. (2) y = 1.

Its a value question and hence we need a definate value.

3^(4+4x) will be 3^8, 3^12, 3^16....when x = 1, 2, 3..... So we can safely conclude that 3^(4 + 4x) will always have remainder of 1(based on cyclicity) . So value of X does not matter.

Hence 1 is not sufficient - or does not help us to respond uniquely.

Second statement tells us that Y=1, which is exactly what need to reply. We dont have to go into the details of calculations to find out the remainder. Answer would be B
_________________

“Confidence comes not from always being right but from not fearing to be wrong.”

Re: If x and y are positive integers, what is the remainder when [#permalink]

Show Tags

20 Oct 2014, 02:31

Bunuel wrote:

banksy wrote:

218. If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10? (1) x = 25. (2) y = 1.

If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10?

Last digit of \(3^{positive \ integer}\) repeats in blocks of 4: {3, 9, 7, 1} - {3, 9, 7, 1} - ... So cyclicity of the last digit of 3 in power is 4. Now, \(3^{4+4x}=3^{4(1+x)}\) will have the same last digit as \(3^4\) which is 1 (remainder upon division the power 4+4x by cyclicity 4 is 0, which means that \(3^{(4+4x)}\) will have the same last digit as 3^4).

Now \(3^{(4 + 4x)}+9^y\) will be divisible by 10 if \(y=odd\), in this case the last digit of \(9^{odd}\) will be 9 (9 has a cyclicity of 2: {9, 1} - {9, 1} - ...) so the last digit of \(3^{(4 + 4x)}+9^y\) will be 1+9=0.

(1) x = 25. Not sufficient. (2) y = 1 --> \(y=odd\) -> remainder=0. Sufficient.

I didn't get the second point, y=odd. If we are provided the value of, irrespective of the fact that whether the value is even or odd, we can calculate the remainder, since we know the cyclicity.

Please correct me if I am wrong.
_________________

218. If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10? (1) x = 25. (2) y = 1.

If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10?

Last digit of \(3^{positive \ integer}\) repeats in blocks of 4: {3, 9, 7, 1} - {3, 9, 7, 1} - ... So cyclicity of the last digit of 3 in power is 4. Now, \(3^{4+4x}=3^{4(1+x)}\) will have the same last digit as \(3^4\) which is 1 (remainder upon division the power 4+4x by cyclicity 4 is 0, which means that \(3^{(4+4x)}\) will have the same last digit as 3^4).

Now \(3^{(4 + 4x)}+9^y\) will be divisible by 10 if \(y=odd\), in this case the last digit of \(9^{odd}\) will be 9 (9 has a cyclicity of 2: {9, 1} - {9, 1} - ...) so the last digit of \(3^{(4 + 4x)}+9^y\) will be 1+9=0.

(1) x = 25. Not sufficient. (2) y = 1 --> \(y=odd\) -> remainder=0. Sufficient.

I didn't get the second point, y=odd. If we are provided the value of, irrespective of the fact that whether the value is even or odd, we can calculate the remainder, since we know the cyclicity.

Please correct me if I am wrong.

Sorry, but I don;t understand what you mean there...

Anyway, y=odd, means that the units digit of 9^y is 9. Since we established that the units digit of 3^(4 + 4x) is 1, irrespective of x, then the units digit of the sum is 1+9=0, which ensures the divisibility by 10.
_________________

The above equation implies that, the value of x is of no use since we'll get remainder 1 from that part of the equation. The second part of the equation which involves y, will be of value to us.

1 - Insufficient 2 - y = 1. The remainder will be -1 from here, so the final remainder will be \(1 - 1 = 0\)
_________________

Re: If x and y are positive integers, what is the remainder when [#permalink]

Show Tags

22 Feb 2016, 09:03

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 x and y are positive integers, what is the remainder when [#permalink]

Show Tags

19 May 2017, 21:48

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.
_________________

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...