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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.