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# p/s digits

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Manager
Joined: 04 Feb 2011
Posts: 62
Location: US
Followers: 2

Kudos [?]: 130 [0], given: 42

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07 Mar 2011, 09:29
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60% (01:26) correct 40% (00:34) wrong based on 7 sessions

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What is the units digit of (13^4)(17^2)(29^3)
(A) 9
(B) 7
(C) 5
(D) 3
(E) 1
Manager
Joined: 14 Feb 2011
Posts: 194
Followers: 4

Kudos [?]: 135 [1] , given: 3

Re: p/s digits [#permalink]

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07 Mar 2011, 09:38
1
KUDOS
Lolaergasheva wrote:
What is the units digit of (13^4)(17^2)(29^3)
(A) 9
(B) 7
(C) 5
(D) 3
(E) 1

Unit's digit would be same as product of unit's digit of 13^4, 17^2 and 29^3

For 13^4 , it is 3^4 which has unit digit 1, similarly for 17^2, it is 9 and for 29^3 it is 9, so we have 1*9*9, so units digit is 1. Answer E
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Joined: 20 Dec 2010
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Kudos [?]: 1822 [1] , given: 376

Re: p/s digits [#permalink]

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07 Mar 2011, 09:39
1
KUDOS

last-digit-of-a-power-70624.html#p520621

We just see the unit's digit;
13^4 = 3^4 = 1
17^2 = 7^2 = 7*7 = 49 = Last digit 9
29^3 = 9^3 = 9 (last digit of all powers of 9 are either 9 or 1. 9 for odd power and 1 for even. Here power is 3; so 9)

Now multiply;
1*9*9 = 9^2 = 1;

Last digit is 1.

Ans: "E"
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Director
Joined: 01 Feb 2011
Posts: 755
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Kudos [?]: 124 [0], given: 42

Re: p/s digits [#permalink]

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07 Mar 2011, 20:01
13^4 * 17^2 * 29^3

we all have to find is the unit digit of a product, then we can just concentrate on the unit digit

unit digit of 13^4 * 17^2 * 29^3

= unit digit of 1 * 9 *9 = 1

SVP
Joined: 16 Nov 2010
Posts: 1666
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 34

Kudos [?]: 533 [0], given: 36

Re: p/s digits [#permalink]

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07 Mar 2011, 20:30
Basically, we have to find the last digit of :

1 * 9 * 9

which is 1, so answer is E.
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Re: p/s digits   [#permalink] 07 Mar 2011, 20:30
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# p/s digits

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