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# What is the units digit of 2222^(333)*3333^(222) ?

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Intern
Joined: 06 Sep 2011
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What is the units digit of 2222^(333)*3333^(222) ?  [#permalink]

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28 Jan 2012, 11:55
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72% (01:16) correct 28% (01:29) wrong based on 741 sessions

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What is the units digit of $$2222^{333}*3333^{222}$$ ?

A. 0
B. 2
C. 4
D. 6
E. 8
Math Expert
Joined: 02 Sep 2009
Posts: 49858
What is the units digit of 2222^(333)*3333^(222) ?  [#permalink]

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28 Jan 2012, 12:09
What is the units digit of $$2222^{333}*3333^{222}$$ ?

A. 0
B. 2
C. 4
D. 6
E. 8

The units digit of $$2222^{333}$$ is the same as that of $$2^{333}$$;
The units digit of $$3333^{222}$$ is the same as that of $$3^{222}$$;
Hence, the units digit of $$2222^{333}*333^{222}$$ is the same as that of $$2^{333}*3^{222}$$;

Now, the units digits of both 2 and 3 in positive integer power repeat in patterns of 4. For 2 it's {2, 4, 8, 6} and for 3 it's {3, 9, 7, 1}.

The units digit of $$2^{333}$$ will be the same as that of $$2^1$$, so 2 (as 333 divided by cyclicity of 4 yields remainder of 1, which means that the units digit is first # from pattern);
The units digit of $$3^{222}$$ will be the same as that of $$3^2$$, so 9 (as 222 divided by cyclicity of 4 yields remainder of 2, which means that the units digit is second # from pattern);

Finally, 2*9=18 --> the units digit is 8.

For more on this check Number Theory chapter of Math Book: http://gmatclub.com/forum/math-number-theory-88376.html

Hope it helps.
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Re: What is the units digit of 2222^(333)*3333^(222) ?  [#permalink]

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28 Jan 2012, 12:18
I used the method described by Bunuel.
but not to be repetitive, I will post another solution -

(2222^333)*(3333^222)=2222^111*(2222^222*3333^222)

here please pay attention to the fact that the unit digit of multiplication of 2222 and 3333 is 6 (2222^222*3333^222). since 6 powered in any number more than 0 results in 6 as a units digit, as a result we have-
6*2222^111

2 has a cycle of 4 . 111=27*4+3 . 2^3=8
6*8=48
so the units digit is 8, and the answer is E
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Re: What is the units digit of 2222^(333)*3333^(222) ?  [#permalink]

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09 Mar 2014, 13:13
1
For more on this kind of questions check Units digits, exponents, remainders problems collection.
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Re: What is the units digit of 2222^(333)*3333^(222) ?  [#permalink]

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26 Feb 2015, 21:22
Use cyclicity rule here ,
2^1=2,
2^2=4,
2^3=8
2^4=6
2^5=2
We can see here unit digit repeated after 4 powers so cyclicity of 2 is 4 . devide power by 4 and check above .
Ans E
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Re: What is the units digit of 2222^(333)*3333^(222) ?  [#permalink]

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26 Feb 2015, 21:43
Hi All,

Each of the other explanations to this question has properly explained that you need to break down the calculation into pieces and figure out the repeating "pattern" of the units digits.

Here's another way to organize the information.

We're given [(2222)^333][(3333)^222]

We can 'combine' some of the pieces and rewrite this product as....
([(2222)(3333)]^222) [(2222)^111]

(2222)(3333) = a big number that ends in a 6

Taking a number that ends in a 6 and raising it to a power creates a nice pattern:
6^1 = 6
6^2 = 36
6^3 = 216
Etc.
Thus, we know that ([(2222)(3333)]^222) will be a gigantic number that ends in a 6.

2^111 requires us to figure out the "cycle" of the units digit...

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16

2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256

So, every 4 "powers", the pattern of the units digits repeats (2, 4, 8, 6.....2, 4, 8, 6....).

111 = 27 sets of 4 with a remainder of 3....

This means that 2^111 = a big number that ends in an 8

So we have to multiply a big number that ends in a 6 and a big number that ends in an 8.

(6)(8) = 48, so the final product will be a gigantic number that ends in an 8.

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Re: What is the units digit of 2222^(333)*3333^(222) ?  [#permalink]

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22 May 2018, 02:57
belagerfeld wrote:
What is the units digit of $$2222^{333}*3333^{222}$$ ?

A. 0
B. 2
C. 4
D. 6
E. 8

This type of sums can also be solved using Fermet's Theorem approach.

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Re: What is the units digit of 2222^(333)*3333^(222) ? &nbs [#permalink] 22 May 2018, 02:57
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