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# What is the units digit of the above expression?

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Intern
Joined: 18 Nov 2012
Posts: 4

Kudos [?]: 5 [0], given: 3

Location: Greece
Concentration: Finance, Accounting
GMAT 1: 650 Q45 V34
What is the units digit of the above expression? [#permalink]

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18 Nov 2012, 03:02
(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, 03:31, edited 1 time in total.
Renamed the topic and edited the question.

Kudos [?]: 5 [0], given: 3

VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1375

Kudos [?]: 1721 [1], given: 62

Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75
Re: What is the units digit of the above expression? [#permalink]

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18 Nov 2012, 06:42
1
KUDOS
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.
_________________

Kudos [?]: 1721 [1], given: 62

Re: What is the units digit of the above expression?   [#permalink] 18 Nov 2012, 06:42
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# What is the units digit of the above expression?

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Moderator: Bunuel

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