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t and w are distinct integers between 105 and 100, not inclusive. Whic 20 Jul 2017, 21:52
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34% (02:54) correct 66% (02:59) wrong based on 366 sessions

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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 \(n = (t^w - w^t)^w?\)

A. 0
B. 1
C. 2
D. 3
E. 4
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A
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t and w are distinct integers between 105 and 100, not inclusive. Whic 25 Jul 2017, 13:14
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Bunuel
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 \(n = (t^w - w^t)^w?\)

A. 0
B. 1
C. 2
D. 3
E. 4
Show more


Ans (A)
Interesting problem.
From Qn we can deduce the following values for t & w:
t w
103 101 ==> (103^101 - 101^103)^101 ----- case(1)
104 102 ==> (104^102 - 102^104)^102 ----- case(2)
104 101 ==> (104^101 - 101^104)^101 ----- case(3)

Now, before proceeding further, we must know two important number theory rules:
Rule#1 : Cyclicity for units digits of all natural numbers is 4 i.e. after every four occurrences, units digit must repeat.
Based on this rule, above cases will give the following results:
Case(1) : 3-1 = 2 (units digit)
Case(2) : 6-6 = 0 (units digit)
Case(3) : 4-1 = 3 (units digit)

Rule#2: For scenario like a^b - b^a, it will give negative result as we increase the value of a and b. Lets check this in a pattern using numbers:
2^1 - 1^2 = 2-1 = 1
3^2 - 2^3 = 9-8 = 1 -------For numbers after this, we will get negative values.
4^3 - 3^4 = 64-81 = -17
7^5 - 5^7 = 16807-78125 = Big negative no
.
.
Now, going back to our cases, for case(1) & Case(3) :
(103^101 - 101^103)^101 = A negative no odd no of times will yield a negative no
(104^101 - 101^104)^101 = A negative no odd no of times will yield a negative no

But for Case(2)
(104^102 - 102^104)^102 = A negative no even no of times will yield a positive no [ The question statement clear mentioned the positive value of n]

Thanks!
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t and w are distinct integers between 105 and 100, not inclusive. Whic 21 Jul 2017, 01:02
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Bunuel - is there any mistake in the question/answers? i tried to solve it but something doesn't work for me. Here is my analysis.

As t and w are between 100 to 105 not inclusive, and as t>w+1, the options for the values of t and w are limited (three options): (1) t=103 and w=101; (2) t=104 and w=101; (3) t=104 and w=102. Now we can check what will be the units digit in each of those cases. In the first scenario we have (103^101 - 101^103)^101. The units digit of 103^101 is 3 and the units digit of 101^103 is 1, so the units digit of the expression in brackets is 2 (3-1). While we raise the number in brackets in a power of 101 we get a units digit of 2. In the second case we have (104^101 - 101^104)^101. The units digit of 104^1 will be 4 and the units digit of 101^104 will be 1, so the units digit of the expression in brackets will be equal to 3 (4-1). When we raise the number in a power of 101, the units digit suppose to be 3. In the last case the units digit of 104^102 is 6 and the units digit of 102^104 is 6, so the units digit of the expression (104^102-102^104) will be zero. We'll get the same units digit after raising the number in power of 102.

So it seems to me that there is more than one correct answer. But maybe i got wrong in my solution :/
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t and w are distinct integers between 105 and 100, not inclusive. Whic 25 Jul 2017, 11:59
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I also thought of approximation
taking t =104 and w = 102
So I plug in: (104^102 - 102^104)^102 => going by the last digit rule => (4^2 - 2^4)^102 = zero
However using 103 and 101, answer is 2 in units digit
:(
Will wait for the detailed OA
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t and w are distinct integers between 105 and 100, not inclusive. Whic 28 Jul 2017, 12:07
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praveen8047
Bunuel
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 \(n = (t^w - w^t)^w?\)

A. 0
B. 1
C. 2
D. 3
E. 4


Ans (A)
Interesting problem.
From Qn we can deduce the following values for t & w:
t w
103 101 ==> (103^101 - 101^103)^101 ----- case(1)
104 102 ==> (104^102 - 102^104)^102 ----- case(2)
104 101 ==> (104^101 - 101^104)^101 ----- case(3)

Now, before proceeding further, we must know two important number theory rules:
Rule#1 : Cyclicity for units digits of all natural numbers is 4 i.e. after every four occurrences, units digit must repeat.
Based on this rule, above cases will give the following results:
Case(1) : 3-1 = 2 (units digit)
Case(2) : 6-6 = 0 (units digit)
Case(3) : 4-1 = 3 (units digit)

Rule#2: For scenario like a^b - b^a, it will give negative result as we increase the value of a and b. Lets check this in a pattern using numbers:
2^1 - 1^2 = 2-1 = 1
3^2 - 2^3 = 9-8 = 1 -------For numbers after this, we will get negative values.
4^3 - 3^4 = 64-81 = -17
7^5 - 5^7 = 16807-78125 = Big negative no
.
.
Now, going back to our cases, for case(1) & Case(3) :
(103^101 - 101^103)^101 = A negative no odd no of times will yield a negative no
(104^101 - 101^104)^101 = A negative no odd no of times will yield a negative no

But for Case(2)
(104^102 - 102^104)^102 = A negative no even no of times will yield a positive no [ The question statement clear mentioned the positive value of n]

Thanks!
Show more

Hi Praveen,

Quick question, In the cases 1 and 3, how can we be sure that the units digit it 2 and 3 respectively when as stated by Rule #2 the first number will be smaller than the second?

Thanks!
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t and w are distinct integers between 105 and 100, not inclusive. Whic 28 Jul 2017, 12:21
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Bunuel
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 \(n = (t^w - w^t)^w?\)

A. 0
B. 1
C. 2
D. 3
E. 4
Show more

Good question to check concepts for cyclicity. This question is a "could be" and not "must be" and thus is tells us that picking 1 set of values for t and w should do the trick. DO note that t>w+1 such that 100< t,w<105.

The only possible sets for (t,w) are (104,102), (103,101) and (104,101). I will take (104,102) as I like to play around with even numbers and because I know 2 and 4 have 4 and 2 as cyclicity.

Taking that into account, n = \((104^{102} - 102^{104})^{102}\)

Treating them separately, units digit of 104^102 will be the same as that for 4^2 = 16 = 6 as the unit's digit.

Similarly, Units digit for 102^104 will be the same as that for 2^4 (as 104 is divisible by 4, cyclicity of 2). 2^4 = 16 and as such the unit's digit will be =6.

Now realize that as both the unit's digits (for 104^102 and 102^104)=6, the difference (t^w-w^t) = 0 and 0 ^ (any power) = 0.

A is thus the correct answer.

Hope this helps.
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t and w are distinct integers between 105 and 100, not inclusive. Whic 28 Jul 2017, 23:02
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praveen8047
Bunuel
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 \(n = (t^w - w^t)^w?\)

A. 0
B. 1
C. 2
D. 3
E. 4


Ans (A)
Interesting problem.
From Qn we can deduce the following values for t & w:
t w
103 101 ==> (103^101 - 101^103)^101 ----- case(1)
104 102 ==> (104^102 - 102^104)^102 ----- case(2)
104 101 ==> (104^101 - 101^104)^101 ----- case(3)

Now, before proceeding further, we must know two important number theory rules:
Rule#1 : Cyclicity for units digits of all natural numbers is 4 i.e. after every four occurrences, units digit must repeat.
Based on this rule, above cases will give the following results:
Case(1) : 3-1 = 2 (units digit)
Case(2) : 6-6 = 0 (units digit)
Case(3) : 4-1 = 3 (units digit)

Rule#2: For scenario like a^b - b^a, it will give negative result as we increase the value of a and b. Lets check this in a pattern using numbers:
2^1 - 1^2 = 2-1 = 1
3^2 - 2^3 = 9-8 = 1 -------For numbers after this, we will get negative values.
4^3 - 3^4 = 64-81 = -17
7^5 - 5^7 = 16807-78125 = Big negative no
.
.
Now, going back to our cases, for case(1) & Case(3) :
(103^101 - 101^103)^101 = A negative no odd no of times will yield a negative no
(104^101 - 101^104)^101 = A negative no odd no of times will yield a negative no

But for Case(2)
(104^102 - 102^104)^102 = A negative no even no of times will yield a positive no [ The question statement clear mentioned the positive value of n]

Thanks!

Hi Praveen,

Quick question, In the cases 1 and 3, how can we be sure that the units digit it 2 and 3 respectively when as stated by Rule #2 the first number will be smaller than the second?

Thanks!
Show more

Hi DKingdom,
Cyclicity for numbers is calculated as described below, but before that we should know that the cyclicity of all natural numbers is 4. That means after every 4th occurrence, pattern will repeat.
For ex: 3^1 = 3 (Units Digit=3)
3^2 = 9 (U D=9)
3^3 = 27 (U D=7)
3^4 = 81 (U D=1)
3^5 = 243 (U D=3)
3^6 = 729 (U D=9)...... So on.

Here the pattern is 3,9,7,1,3,9,...... Hence we can see that Units Digit repeat after every 4th occurrence. This is true for all the natural numbers
Now, to calculate 3^101 = 3^(100+1) = 3^(4*25 + 1) = 3^1 = 1. Here 101 is broken into 100 + 1 as 100 is closest multiple of 4.

Hence, 103^101 = 3^101 = 3(Units Digit)
101^103 = 1^103 = 1(Units Digit)
Hence, 103^101 - 101^103 = 3-1 =2. Similarly, we can get the vales for other cases also.

Using above method, we can calculate the Units Digit for all the cases. But the question is asking for positive result.
Now, we can use the Rule#2, to discard negative numbers.

I hope, this was helpful.

Thanks!
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t and w are distinct integers between 105 and 100, not inclusive. Whic 19 Oct 2017, 19:52
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Engr2012
Bunuel
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 \(n = (t^w - w^t)^w?\)

A. 0
B. 1
C. 2
D. 3
E. 4

Good question to check concepts for cyclicity. This question is a "could be" and not "must be" and thus is tells us that picking 1 set of values for t and w should do the trick. DO note that t>w+1 such that 100< t,w<105.

The only possible sets for (t,w) are (104,102), (103,101) and (104,101). I will take (104,102) as I like to play around with even numbers and because I know 2 and 4 have 4 and 2 as cyclicity.

Taking that into account, n = \((104^{102} - 102^{104})^{102}\)

Treating them separately, units digit of 104^102 will be the same as that for 4^2 = 16 = 6 as the unit's digit.

Similarly, Units digit for 102^104 will be the same as that for 2^4 (as 104 is divisible by 4, cyclicity of 2). 2^4 = 16 and as such the unit's digit will be =6.

Now realize that as both the unit's digits (for 104^102 and 102^104)=6, the difference (t^w-w^t) = 0 and 0 ^ (any power) = 0.

A is thus the correct answer.

Hope this helps.
Show more

Excellent explanation!!
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t and w are distinct integers between 105 and 100, not inclusive. Whic 05 Dec 2017, 14:39
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Was revisiting this question - I am still now sure why the 101,103 pair is being rejected.
Cyclicity wise, 104,102 makes sense, but a different and positive answer is given by [103^101 - 101^103] ^101

Would be glad to know any explanation on this
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t and w are distinct integers between 105 and 100, not inclusive. Whic 29 Aug 2022, 05:33
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I think it is a poor quality question.

Consider t = 103 and w= 101.
Is the unit digit of the equation not 2?

What am I missing here?

Posted from my mobile device
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t and w are distinct integers between 105 and 100, not inclusive. Whic 31 Aug 2022, 02:40
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poorvi125
I think it is a poor quality question.

Consider t = 103 and w= 101.
Is the unit digit of the equation not 2?

What am I missing here?

Posted from my mobile device
Show more

Same with me, anyone can explain why 2 is wrong?
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t and w are distinct integers between 105 and 100, not inclusive. Whic 31 Aug 2022, 03:22
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poorvi125
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 \(n = (t^w - w^t)^w?\)

A. 0
B. 1
C. 2
D. 3
E. 4

I think it is a poor quality question.

Consider t = 103 and w= 101.
Is the unit digit of the equation not 2?

What am I missing here?

Posted from my mobile device

Same with me, anyone can explain why 2 is wrong?
Show more


    If t = 103, and w = 101, then the units digit of \((103^{101} - 101^{103})^{101}\) is 2;

    If t = 104, and w = 101, then the units digit of \((104^{101} - 101^{104})^{101}\) is 3;

    If t = 104, and w = 102, then the units digit of \((104^{102} - 102^{104})^{102}\) is 0.

The thing is, we need such t and w that \(n = (t^w - w^t)^w\) is POSITIVE!!! Check the stem:

    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 \(n = (t^w - w^t)^w?\)

Here, both \((103^{101} - 101^{103})^{101}\) (t = 103, and w = 101) and \((104^{101} - 101^{104})^{101}\) (t = 104, and w = 101) are NEGATIVE, so we are left with only t = 104, and w = 102, which gives the units digit of 0.

Hope it's clear.
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t and w are distinct integers between 105 and 100, not inclusive. Whic 06 Apr 2025, 02:36
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