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What is the units digit of the above expression?

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Joined: 18 Nov 2012
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Kudos [?]: 5 [0], given: 3

What is the units digit of the above expression? [#permalink]

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18 Nov 2012, 04:02
(2^18)*(3^17)*(7^14)*(5^20)

What is the units digit of the above expression?

A. 0
B. 2
C. 5
D. 7
E. 9

Last edited by Bunuel on 18 Nov 2012, 04:31, edited 1 time in total.
Renamed the topic and edited the question.
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Kudos [?]: 1098 [1] , given: 62

Re: What is the units digit of the above expression? [#permalink]

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18 Nov 2012, 07:42
1
KUDOS
Expert's post
stef91 wrote:
(2^18)*(3^17)*(7^14)*(5^20)

What is the units digit of the above expression?

A. 0
B. 2
C. 5
D. 7
E. 9

cyclicity of 2 is 4
cyclicity of 3 is 4
cyclicity of 7 is 4
cyclicity of 5 is 1.
So divide the respective powers by the integer's own cyclicity.
So the above expression can be written as:
$$2^2 * 3^3 * 7^2 * 5$$
It becomes $$4*27*49*5$$
Since 4*5=20,
so any number, if multiplied by 0 will yield 0 as the last digit.

Hence A.
_________________
Re: What is the units digit of the above expression?   [#permalink] 18 Nov 2012, 07:42
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What is the units digit of the above expression?

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