M07-20

Question

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

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

Bunuel · Accepted Answer

Official Solution:

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

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

Firstly, we can simplify the given expression as follows:  \frac{x^x}{2x^2}-2 = \frac{x^{x-2}}{2}-2 .
 
Next, we can evaluate  f(4)  using this expression:  f(4)=\frac{4^{4-2}}{2}-2=6 .
 
To find  f(f(4)) , we substitute  f(4)  into the expression for  f(x)  and simplify: 
 
 f(f(4)) = f(6) = \frac{6^{6-2}}{2}-2=\frac{6^{4}}{2}-2=\frac{6*6^{3}}{2}-2=3*6^3-2 .
 
Since the units digit of  6^3  is 6, we know that the units digit of  3*6^3  is 8. Therefore, the units digit of  3*6^3-2  is  8-2=6 .

Answer: D

mvictor · Answer

4^4   /   2*4^2 - 2 
 2^8/2^5 -2 
 2^3 = 8 -2 = 6 
 6^6/2*6^2 
 2^6*3^6 / 2^3 * 3^2 -2 
 2^3*3^4 -2 
 2^3 = 8 
 3^4 = 81 
81*8 = 648 -2 = 646  so units digit 6.

devbond · Answer

We find out that @4 = 6

now while solving @6 : (6^4)/2 - 2

6^1= units digit 6
6^2 = units digit 6
..
..
6^4= units digit 6

Now, when units digit 6 is divided by 2 we get units digit 3.

Units digit 3 -2=1

What did i do wrong?

Bunuel · Answer

6/2 = 3, so yes the units digit is 3 but 6^2/2 = 18, the units digit is 8.

SUNILAA · Answer

.

how u got this:
  \frac{x^x}{2x^2}-2 = \frac{x^{x-2}}{2}-2

Bunuel · Answer

\frac{a^y}{a^y}=a^{x-y} , so  \frac{x^x}{x^2}=x^{x-2}

Sumit1703@4 · Answer

I wrote 6^4/2 -2 as (6^4 - 4) /2

Hence, unit digit of 6^4 is 6 : 6-4 =2 and 2/2=1 . What did I do wrong?

Posted from my mobile device

Bunuel · Answer

The point is that if the units digits of a number is 2, the units digit of that number divided by 2 won't always be 1, it could also be 6. For example, 22/2 = 11 but 12/2 = 6.

HWPO · Answer

Bunuel , could I solve it this way:

x^x * 2x^-2 -2?

Because   2x^2  means that the number was  2x^-2  before.

** I am trying to use the shortcuts to make 2 to the negative 2 power. I am sorry for how it looks like

Bunuel · Answer

Notice that only x is squared in the denominator not (2x). So,  \frac{x^x}{2x^2}-2= \frac{x^x*x^{-2}}{2}-2

gh2021 · Answer

This sum confused the hell out of me, would have been better if it actually said function f(x) rather than @. For the life of me, I couldn't figure out that it was a function (I thought it was a 2 digit number).
We actually have to find units digit of f(f(4))

So, we have to find f(4) = 6 first.

And then get to the units digit of f(6) = Units digit of 6^4/2 - 2 is 6.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Go650GMAT · Answer

I think this is a  high-quality  question.

sayan640 · Answer

KarishmaB   MartyMurray  , Is there any specific concept that we should keep in mind so that we can avoid mistakes in these kind of sums where in dividing by 2 while determining the unit's digit is a big no no ?

MartyMurray · Answer

You can just keep in mind that, in many cases, half of a multi-digit number is not a number produced by dividing each digit of the number by 2.

For instance, it's true that half of 248 is 124.

However, half of 18 does not end in 4; it's 9. Half of 36 does not end in 3; it's 18.

In general, if the tens digits of a number is odd, the units digit of half the number won't be half the units digit of the number.

renzosm93 · Answer

I was trying to do: (4^4)/(2*4^4) = 1/2. Why can't we just divide 4^4/4^4? = 1?

Bunuel · Answer

f(x) = \frac{x^x}{2x^2}-2 
 
So:
 
 f(4) = \frac{4^4}{2*4^2}-2=

= \frac{4^2}{2}-2=

= 8-2=

=6 
 
Then we should find the value of f(6):
 
 f(6) = \frac{6^6}{2*6^2}-2=

= \frac{6^4}{2}-2=

= 648-2=

=646 
 
Thus, the inits digit is 6.

bumpbot · Answer

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