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Knewton Challenge: Win a Knerd Shirt and GMAT Club Tests! [#permalink]
20 Sep 2011, 06:01
Hi all - great response on the last Knewton challenge! This time, we'll be choosing a winner at random. Your answers must be clear and well-developed in order to be included in the pool. The competition closes on Friday, so look out for a post announcing the winner. In the meantime, try your hand at this tricky data sufficiency problem:
If \(4^w = n\), what is the units digit of n?
1. w is an even integer 2. w > 0
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. (B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. (C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. (D) EACH statement ALONE is sufficient. (E) Statements (1) and (2) TOGETHER are NOT sufficient.
Looking forward to your answers!
Edit: by bb: The winner will be determined based on a random drawing. The replies will be hidden until the winner is announced. The winner will get access to the GMAT Club Tests for 1 year ($250 value)
Last edited by bb on 23 Sep 2011, 12:04, edited 3 times in total.
Re: Knewton Challenge: Win a Knerd Shirt and GMAT Club Tests! [#permalink]
20 Sep 2011, 11:01
The answer is (C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
1) w is an even integer. If we pick w to be negative then the units digit is 0 and if w is positive the answer is 6 so NOT sufficient 2) w > 0. If we pick w to be 1 we get unit as 4, w = 2 we get unit as 6. so it is NOT sufficient
If we combine them to have w is even and positive then the only possible answer is 6. so both options combined are SUFFICIENT. therefore the answer is The answer is (C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
Re: Knewton Challenge: Win a Knerd Shirt and GMAT Club Tests! [#permalink]
20 Sep 2011, 11:23
My answer would be C.
As per statement 1: w is even number. So it can be 0, 2, 4,.... For all even number except 0, unit digit of n is 6. For 0 unit digit for n is 1. so not sufficient.
As per statement 2: w is any number greater than 0. so unit digit of n changes with different value of w. for example: when w is 1 then unit digit of n is 4, when w is 2 then unit digit of n is 6, when w is 3 then unit digit of n is 4, so not sufficient.
When statement 1 and 2 are evaluated together then w can be 2,4,6,8.... For all these values of w, unit digit of n is 6.
Re: Knewton Challenge: Win a Knerd Shirt and GMAT Club Tests! [#permalink]
20 Sep 2011, 12:09
Ans according to me is C
Soln: 4^w=n
Let us simplify this statement a bit- plugging different values for w:
4^w=Units Digit(w is a positive integer) 4^1=4 4^2=6 4^3=4 4^4=6 4^5=4 4^6=6 Clearly, units digit is 4 if w is odd and units digit is 6 if w is even
Now consider 4^w=Units digit(w is a negative integer) 4^-1=0.25,units digit=0 4^-2=0.0625,units digit=0 4^-3=0.015625,units digit=0 Clearly here units digit is 0
Statement I= w is even integer:Insufficient since we dont know if w is positive or negative Statement II=w>0:Insufficient since we don't know if w is odd or even
I+II=W is a positive ,even integer Hence Units digit is 6 Therefore C
Re: Knewton Challenge: Win a Knerd Shirt and GMAT Club Tests! [#permalink]
20 Sep 2011, 21:37
ANSWER: C
4^w=n..... if no other data is given, we can determine the fact that the units digit is 4 or 6 **. OR if w=0, n=1
A negative 'w' will give fractional values of n ; whose 'unit digit' doesnt make any sense. So w has to be greater than/equal to 0. Statement 2 clears the air by stating w>0. So w is not 0. Hence n not equal to 1. n can still end in 4 0r 6. So we strike out option B whenever w is odd unit digit is 4...and whenever w is even unit digit is 6 At this point we can say that solution is 6 and strike out options A and D We needed both satements to arrive at this answer, E can be striked out Hence C is correct.
**(4^1=4,4^2=16......whenever the digit 6 is multiplied by 4...the result will have a number ending in 4(4x6=24).....which in turn when multiplie by 4 again ends in 6(4x4=16)...and so on....)=>
Re: Knewton Challenge: Win a Knerd Shirt and GMAT Club Tests! [#permalink]
21 Sep 2011, 04:05
4^w = n, what is the units digit of n?
Considering (i) When w is even integer, it can be ...,-4,-2,0,2,4,...... When w=0, 4^0 = 1, units digit is 1 When w=2, 4^2 = 16, units digit is 6 When w=-2, 4^-2 = 1/16=0.0625, units digit is 0.
We do not get a unique solution. Hence (i) is insufficient
Considering (ii) When w > 0, it can be 1,2,3,...... Also, we are not sure if it is an integer. When w=1, 4^1 = 4, units digit is 4 When w=2, 4^2 = 16, units digit is 6 When w=3, 4^3 = 64, units digit is 4.
We do not get a unique solution. Hence (ii) is insufficient
Considering (i) and (ii), we get w > 0 and w is an even integer. So, it can be 2,4,6,8,..... When w=2, 4^2 = 16, units digit is 6 When w=4, 4^4 = 256, units digit is 6
And the units digit will always be 6, as we multiply all the results by 4^2= 16 and any two numbers, whose units digit is 6, are multiplied will give a number with 6 as its unit digit.
We do get a unique solution. Hence answer is C.
--------------- ISH
Last edited by justharish on 22 Sep 2011, 07:27, edited 1 time in total.
Re: Knewton Challenge: Win a Knerd Shirt and GMAT Club Tests! [#permalink]
21 Sep 2011, 08:17
Correct Answer is Option C.
Option A says that w is an even integer. so w can be 0 also. If w is 0, the unit's digit will be 1. If w is a positive integer greater than 0, then unit's digit will be 6 and if w is a negative integer, then it will be a fraction. So option A alone is not sufficient.
Option B says that w>0, but doesn't specify whether it is an integer or not. So for different values of w, we will get different unit's digits.
Combining both A and B, w is an even integer greater than 0. so unit's place will always be 6.
Re: Knewton Challenge: Win a Knerd Shirt and GMAT Club Tests! [#permalink]
22 Sep 2011, 12:50
\(4^1=4\) \(4^2=16\) \(4^3=64\) \(4^4=256\)
Conclusion 1 \(4^e\)=units digit of 6 where e is even Conclusion 2 \(4^o=\)units digit of 4 where o is odd
1) is insufficient since n can have 2 values "6" or "0" because 0 is an even integer and \(4^0=1\) 2) is insufficient because "n" can have 2 values "6" or "4"
Together they are sufficient as it will result in only 1 value i.e. "6" Ans= C
Re: Knewton Challenge: Win a Knerd Shirt and GMAT Club Tests! [#permalink]
22 Sep 2011, 13:20
Before we even look at the statements, the sentence doesn't tell whether it is a positive exponent or if it is an integer, so we have to keep that in mind. Now, seeing statement 1, tells us that it is an integer indeed, but we still know nothing about its sign. If w < 0 then the units digit is 0. If w > 0 then the units digit is 6. Hence NOT SUFFICIENT. We keep answers BCE.
Statement 2 tells us that w is greater than zero, but it fails to address whether it is an integer or not. Therefore NOT SUFFICIENT. We keep answers CE.
Using both statements together we know exactly what we need to know. The coefficient w is greater than zero AND is an integer. We could conclude that the units digit is 6 (we don't have to, though).
The answer is C.
gmatclubot
Re: Knewton Challenge: Win a Knerd Shirt and GMAT Club Tests!
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22 Sep 2011, 13:20
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