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:

If \(\Large{@}\large{Y}\Large{@} = \frac{Y^Y}{2Y^2}\) , what is the last digit of \(\Large{@}\large{(}\Large{@}\large{4}\Large{@}\large{)}\Large{@}\) ?

OA: Based on the formula, the value of $4$ is which is or .

However, in the numerator will be a large number with the last digit being 4 - you can get it just multiplying and getting 4 as a last digit for every returns 4 as a last digit. So, 4 divided by 2 is 2.

The correct answer is D.

End of OA. -----------------------------------------------------------------------------------------------------------------

My problem is that when the OA says " returns 4 as a last digit. So, 4 divided by 2 is 2" it assumes that the last digit of a quotient is the last digit of the division between the last digits of the numerator and the denominator which is not always true... in this example it worked since 64/2=32 and 4/2=2... but what about 14/2?? The result is 7, not 2....

OA: Based on the formula, the value of $4$ is which is or .

However, in the numerator will be a large number with the last digit being 4 - you can get it just multiplying and getting 4 as a last digit for every returns 4 as a last digit. So, 4 divided by 2 is 2.

The correct answer is D.

End of OA. -----------------------------------------------------------------------------------------------------------------

My problem is that when the OA says " returns 4 as a last digit. So, 4 divided by 2 is 2" it assumes that the last digit of a quotient is the last digit of the division between the last digits of the numerator and the denominator which is not always true... in this example it worked since 64/2=32 and 4/2=2... but what about 14/2?? The result is 7, not 2....

I don't think that's relevant. In this question, you are specifically asked for the last digit of $($4$)$. _________________

If \(\Large{@}\large{Y}\Large{@} = \frac{Y^Y}{2Y^2}\) , what is the last digit of \(\Large{@}\large{(}\Large{@}\large{4}\Large{@}\large{)}\Large{@}\) ?

OA: Based on the formula, the value of $4$ is which is or .

However, in the numerator will be a large number with the last digit being 4 - you can get it just multiplying and getting 4 as a last digit for every returns 4 as a last digit. So, 4 divided by 2 is 2.

The correct answer is D.

End of OA.

Hello BB,

I got the same answer but with a longer method.

can you please explain "the numerator will be a large number with the last digit being 4 - you can get it just multiplying and getting 4 as a last digit for every"

Thanks in advance. _________________

_________________ If you like my post, consider giving me a kudos. THANKS!

My method, i think is rather lengthy: (4^4)/2(4)^2 = 16/2 = 8 (8^8)/(2(8^2) = 8^6/2 = 2^17 2^17 = (2^10)(2^7) = 1024 x 128 last digit -from: 4x8 = 32 that is 2. OA = D _________________

KUDOS me if you feel my contribution has helped you.

If \(\Large{@}\large{Y}\Large{@} = \frac{Y^Y}{2Y^2}\) , what is the last digit of \(\Large{@}\large{(}\Large{@}\large{4}\Large{@}\large{)}\Large{@}\) ?

OA: Based on the formula, the value of $4$ is which is or .

However, in the numerator will be a large number with the last digit being 4 - you can get it just multiplying and getting 4 as a last digit for every returns 4 as a last digit. So, 4 divided by 2 is 2.

The correct answer is D.

End of OA. -----------------------------------------------------------------------------------------------------------------

My problem is that when the OA says " returns 4 as a last digit. So, 4 divided by 2 is 2" it assumes that the last digit of a quotient is the last digit of the division between the last digits of the numerator and the denominator which is not always true... in this example it worked since 64/2=32 and 4/2=2... but what about 14/2?? The result is 7, not 2....

Consider this way:- (8^k)/2 never has a factor i.e. odd except for 1. So in the given expression, 1, and 7 are not possible. Now remain - 2, 4, 6 and 8. (8^k) has never 2 in unit digit when k is even so 6 is also not possible. Similarly 8 is also not possible when k is not equal to 2^2n where n is an integer. After a but further simplification, 2 remains between 2 and 4.

I have a question what is that all @ signs? I do not get questions? What is being asked?

Don't be more worried with '@'. It could be any other symbol also, in place of @. All we need to be focusing is, what to be substituted in place of 'Y' in the given equation. Given a equation for @Y@ = \(Y^Y/2Y^2\), they asking for @(@4@)@. So, first you calculate @4@, and second, calculate the same for the result of @4@.

1. @4@ = 8 2. @8@ = some number ending with '2'. We dont need to spend too much time in calculating the whole #, as question is looking only for the last digit.

I think this is a good approach. It is worth to remember the clyclicity of some numbers. For instance, numbers 2, 3 and 7 has the same cyclicity: 4. That is, the units digit of the powers of 2, 3 and 7 are repeated after every 4 powers. Remembering this, you could save a couple of seconds during the test. _________________

If \(#x= \frac{x^x}{2x^2}-2\), what is the units digit of \(#(#4)\) ?

A. 1 B. 3 C. 4 D. 6 E. 8

First of all \(\frac{x^x}{2x^2}-2= \frac{x^{x-2}}{2}-2\), so \(#4=\frac{4^{4-2}}{2}-2=6\);

Next, \(#6=\frac{6^{6-2}}{2}-2=\frac{6^{4}}{2}-2=\frac{6*6^{3}}{2}-2=3*6^3-2\). Now, the units digit of 6^3 is 6, thus the units digit of 3*6^3 is 8 (3*6=18), so the units digit of \(3*6^3-2\) is 8-2=6.