t and w are distinct integers between 105 and 100, not inclusive. Which of the following could be the units digit of positive integer ‘n’, if t > w + 1, where ?

- 0
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9

As always, check back on Thursday for the explanation!

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t and w are distinct integers between 105 and 100, not inclusive. Which of the following could be the units digit of positive integer ‘n’, if t > w + 1, where ?

- 0
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9

As always, check back on Thursday for the explanation!

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Can u please explain that “n=(t^w-w^t)^w” in detail with steps if possible,its kinda eating my head that why t=103 and w=101 is not applicable.

Hi Aravind 🙂

We’re told that t and w are two integers between 100 and 105, not inclusive. That means that the possible values of t and w (before considering any other restrictions) are 101, 102, 103, and 104. We then told that t > w + 1; t is greater than (w+1). So, initially, t = 103 and w = 101 is a pair that we should test. However, it turns out that the difference (103^101 – 101^103) is

negative. For a thorough explanation of how we can determine that this difference is negative, please take a look at the explanation for this brain twister: “Big Numbers Explanation”. 🙂Once we recognize that the quantity inside the parenthesis is negative, we need to focus our attention to the exponent, w. When we have an odd exponent, as is the case when w = 101, the sign of the value raised to the exponent remains the same. In other words, a positive number raised to an odd exponent is positive and a negative number raised to an odd exponent is negative. In the case that w = 101 and t = 103, we have a negative number raised to an odd exponent, which means the final value will be negative. However, the prompt states that n is a positive number. This means that we cannot have w = 101, t = 103. Rather, we need w to be even so that n is positive in this question.

I hope this helps! 🙂

The answer is 0, as other options will produce -2 and -3.

Yes, that is correct!

The two possible outcomes are 0 and 2 (If n is an integer) . But here n is a postive integer so the answer is 0.

Yep, that’s it!

The correct answer choice is 0.

You got it!

0 is the answer;only possible values of w are 101 and 102; for w=101 t will be either 103 or 104 in both cases equation will give a negative value and the power is odd so the total value will be negative. In case of w=102 the only possible value for t is 104 even the term within bracket is negative but the power is even so the unit value of this term should only be calculated and it comes out to be 0.

Yep, that’s it!

case 1: w=102 & t=104

since w is even hence n will be positive integer

the unit digit for n would be 0

case 2,3: w=101 & t=103 or 104

both cases w is odd hence for n to be positive

(t^w – w^t ) > 0

=> (104^101) – (101^104) > 0

=> ((101 + 3)^101) – (101^104) > 0

~ ((100 + 3)^101) – (100^104) > 0

the extra 100^3 would add far more weight hence this would be false i.e. n would be negative hence the answer should be just option A (0)

Hi Sheetanshu,

I’m not sure if you answered the question correctly, and did so without looking at the hints. If so, you are the only one–out of at least a dozen people–to do so correctly. Good job 🙂

Ans A

Yes, that’s it!

The distinct integers between 105 and 100, not inclusive, are 104, 103, 102 and 101.

Now, in order to satisfy the given condition, t and w can have the following values: 104 and 102; 103 and 102; 104 and 101.

Substituting the above pairs for n value:

((104)^102 – (102)^104)^102 – unit digits -> (4^102 – 2^104)^102 => (6-6)^102 => 0

((103)^101 – (101)^103)^101 – unit digits -> (3^101 – 1^103)^101 => (3-1)^101 => 2

((104)^101 – (101)^104)^101 – unit digits -> (4^101 – 1^104)^101 => (4-1)^101 => 3

(Since unit digits for powers of 2 are 2,4,8,6 recurring and 3 are 3,9,7,1 recurring)

It seems that nobody has answered this question correctly yet. I think everyone is overlooking the phrase “positive integer n” part.

Take another shot 🙂

Ouch! the minute detail.. 😀

Well, Now I think the answer is just 0.

Since the power it (((104)^102 – (102)^104)^102) is raised to is even, it has to be positive.

Now coming to the second and third choices, the polarity of the number is based on the inner part, i.e, for ((103)^101 – (101)^103)^101, it can be written as a^100 * a. (a decides if the number is + or -)

103^101 comes out as a number of 203 digits, while 101^103 is of 207 digits. So, this is clearly gone.

Similarly, 104^101 is a 203 digit number and 101^104 is a 209 digit number. Clearly, this is also negative.

The only answer is 0. (Hope i got it right this time 🙂 )

(Phew! That was definitely “fiendishly diabolical” 😀 )

Well you got it this time around! It seems that everybody missed that positive number part. This coming week’s Brain Twister may be more diabolical yet 🙂

Well, my exam is on this monday, so I think this is my last brain twister (before exam) but I will continue solving them even after the exam. Love this blog 😀

2,3,0 for possible values of w=101 , t=103,104

w=102 , t=104

There is one little detail you are overlooking that changes the outcome :). It’s something about ‘n’ being a positive number. Hope that helps!

2,3,4

Not sure the process you arrived to get there, but those aren’t the answers. Have another go at it (hint: ‘n’ is a positive integer).

4, 0

0, 2

Both of your answers are close :). I think there may be one little thing you are missing. Hint: n has to be positive.

Good luck!

P.S. nobody has answered it correctly yet.

0,2,3

One of those is correct :). Hint: ‘n’ has to be a positive integer.

Good luck!