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# M09-32

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Math Expert
Joined: 02 Sep 2009
Posts: 50619

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15 Sep 2014, 23:40
2
1
00:00

Difficulty:

45% (medium)

Question Stats:

50% (00:22) correct 50% (00:22) wrong based on 122 sessions

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What is the last digit of $$3^{3^3}$$?

A. 1
B. 3
C. 6
D. 7
E. 9

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 50619

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15 Sep 2014, 23:40
1
1
Official Solution:

What is the last digit of $$3^{3^3}$$?

A. 1
B. 3
C. 6
D. 7
E. 9

THEORY

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 $$(a^m)^n$$.

BACK TO THE QUESTION

According to the above: $$3^{3^3}=3^{(3^3)}=3^{27}$$

Next, the units digit of 3 in positive integer power repeats in pattern of 4: {3, 9, 7, 1}. So, the units digit of $$3^{27}$$ (27=4*6+3) is the same as the units digit of $$3^3$$, which is 7.

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Intern
Joined: 05 Nov 2012
Posts: 46

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22 Feb 2016, 00:29
Bunuel wrote:
Official Solution:

What is the last digit of $$3^{3^3}$$?

A. 1
B. 3
C. 6
D. 7
E. 9

THEORY

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 $$(a^m)^n$$.

BACK TO THE QUESTION

According to the above: $$3^{3^3}=3^{(3^3)}=3^{27}$$

Next, the units digit of 3 in positive integer power repeats in pattern of 4: {3, 9, 7, 1}. So, the units digit of $$3^{27}$$ (27=4*6+3) is the same as the units digit of $$3^3$$, which is 7.

Hi Bunuel,

I wanted to know can't we proceed with this question in the following way: (3^3)^3 = (27)^3 = units digit = 3. Please explain where I went wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 50619

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22 Feb 2016, 00:37
1
nishatfarhat87 wrote:
Bunuel wrote:
Official Solution:

What is the last digit of $$3^{3^3}$$?

A. 1
B. 3
C. 6
D. 7
E. 9

THEORY

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 $$(a^m)^n$$.

BACK TO THE QUESTION

According to the above: $$3^{3^3}=3^{(3^3)}=3^{27}$$

Next, the units digit of 3 in positive integer power repeats in pattern of 4: {3, 9, 7, 1}. So, the units digit of $$3^{27}$$ (27=4*6+3) is the same as the units digit of $$3^3$$, which is 7.

Hi Bunuel,

I wanted to know can't we proceed with this question in the following way: (3^3)^3 = (27)^3 = units digit = 3. Please explain where I went wrong.

_________________
Intern
Joined: 04 Mar 2015
Posts: 6

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09 Apr 2016, 23:17
(27=4*6+3) so you add 4 + 3 = 7 (Please more details)
Math Expert
Joined: 02 Sep 2009
Posts: 50619

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10 Apr 2016, 03:26
Manager
Joined: 15 Feb 2018
Posts: 80

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06 Sep 2018, 00:16
I think this is a high-quality question and I agree with explanation. You have it as a Fractions/Ratios/Decimals question when it should be Exponents and Roots?
Re M09-32 &nbs [#permalink] 06 Sep 2018, 00:16
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# M09-32

Moderators: chetan2u, Bunuel

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