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To solve this kind of problems, it saves time to remember the following regarding the units digit:

.............................^1....^2.....^3.....^4.....^5......^6 0 has a cycle of 1.......0......0........0 1 has a cycle of 1.......1......1........1 2 has a cycle of 4.......2......4......8......16.......32.....64 3 has a cycle of 4.......3......9......27.....81......243....729 4 has a cycle of 2.......4......16......64....256 5 has a cycle of 1.......5......25......125 6 has a cycle of 1.......6......36......216 7 has a cycle of 4.......7......49......341...2401...16807..117649 8 has a cycle of 4.......8......64......512...4096..32768..262144 9 has a cycle of 2.......9......81......729....6561

This pattern of 1-1-4-4-2 goes on and on into infinity: 10 has a cycle of 1 11 has a cycle of 1 12 has a cycle of 4 13 has a cycle of 4 14 has a cycle of 2
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I didnt know you had to divide by 4 and to pick the number of the remainder..

How about the case for 3^12 where has zero remainder...?

When remainder is zero you should take the base in the power of cyclisity. Last digit of 3^12 = last digit of 3^4, as the cyclisity of 3 if 4 and 12 divided by 4 leaves remainder zero.

For more see the Number Theory chapter in Math Book.
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the ans will be 3. for this type of question which has small no as 2, 3 ,4. you can calculate it by multiplying only the unit digit if the nos are countable on fingers, else you should use the above method explained by Gmat Aspirants.
_________________

kudos me if you like my post.

Attitude determine everything. all the best and God bless you.

the ans will be 3. for this type of question which has small no as 2, 3 ,4. you can calculate it by multiplying only the unit digit if the nos are countable on fingers, else you should use the above method explained by Gmat Aspirants.

OA for this question is D (7).

What is the last digit of \(3^{3^3}\)? (A) 1 (B) 3 (C) 6 (D) 7 (E) 9

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: \(a^m^n=a^{(m^n)}\) and not \((a^m)^n\), which on the other hand equals to \(a^{mn}\).

So: \((a^m)^n=a^{mn}\);

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\).

According to above:

\(3^{3^3}=3^{(3^3)}=3^{27}\)

Cyclicity of 3 in positive integer power is four (the last digit of 3 in positive integer power repeats in the following patter {3-9-7-1}-{3-9-7-1}-...) --> the units digit of \(3^{27}\) is the same as for 3^3 (27=4*6+3) --> 7.

the ans will be 3. for this type of question which has small no as 2, 3 ,4. you can calculate it by multiplying only the unit digit if the nos are countable on fingers, else you should use the above method explained by Gmat Aspirants.

OA for this question is D (7).

What is the last digit of \(3^{3^3}\)? (A) 1 (B) 3 (C) 6 (D) 7 (E) 9

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: \(a^m^n=a^{(m^n)}\) and not \((a^m)^n\), which on the other hand equals to \(a^{mn}\).

So: \((a^m)^n=a^{mn}\);

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\).

According to above:

\(3^{3^3}=3^{(3^3)}=3^{27}\)

Cyclicity of 3 in positive integer power is four (the last digit of 3 in positive integer power repeats in the following patter {3-9-7-1}-{3-9-7-1}-...) --> the units digit of \(3^{27}\) is the same as for 3^3 (27=4*6+3) --> 7.

Answer: D (7).

ohh ! i got that question wrong. the ans will be 7. above explanation is appreciable.
_________________

kudos me if you like my post.

Attitude determine everything. all the best and God bless you.

I didnt know you had to divide by 4 and to pick the number of the remainder..

How about the case for 3^12 where has zero remainder...?

When remainder is zero you should take the base in the power of cyclisity. Last digit of 3^12 = last digit of 3^4, as the cyclisity of 3 if 4 and 12 divided by 4 leaves remainder zero.

For more see the Number Theory chapter in Math Book.

Good point. I did everything like you guys except I chose the wrong starting point. I divided 27 by 4 and got 6 remainder 3 but thought 3^1 was the point of reference and chose 1 as the answer and not 7.

Thanks for clearing that up Bunuel!
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I'm trying to not just answer the problem but to explain how I came up with my answer. If I am incorrect or you have a better method please PM me your thoughts. Thanks!

3^3^3 = 3^27 Now we need to find a pattern for last digits of different powers of 3 --> 1 2 3 4 3 9 7 1

after every 4 different unit digits the pattern repeates itself. Hence for 4*6 = 24th power the units digit will be 1 27 - 4 = 3 the third unit's digit in the pattern above is 7

i broke it down to 3^27 => then 27 is divided by 3, 9 times. 3^3 = 27 then 7^9. break it down to pairs 7*7 = 49. there will be four 9's and a remaining 7. 1*1*7 = 7
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I had not find a topic for this question, so I have created this one. The thing is - 3^3^2 = 3^27 in the official answer. Is not it - 3^3^2 = 3^9 ? Thank you for the attention.

Attachments

d63d1fe678552ac0338e3bdaa5102421.png [ 261.88 KiB | Viewed 4951 times ]

I had not find a topic for this question, so I have created this one. The thing is - 3^3^2 = 3^27 in the official answer. Is not it - 3^3^2 = 3^9 ? Thank you for the attention.

Merging topics.

It's \(3^{3^3}=3^{(3^3)}=3^{27}\), not \(3^{3^2}\).