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# a, b, c, d and e are five consecutive positive integers. What is the

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Manager
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a, b, c, d and e are five consecutive positive integers. What is the  [#permalink]

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Updated on: 09 Apr 2018, 00:16
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65% (hard)

Question Stats:

62% (01:38) correct 38% (02:01) wrong based on 64 sessions

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a, b, c, d and e are five consecutive positive integers. What is the unit's digit of $$a^b*c^d * e$$?

(1) c = 5
(2) a + b is six less than e + d.

source- time4education

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Originally posted by Nixondutta on 08 Apr 2018, 22:39.
Last edited by Bunuel on 09 Apr 2018, 00:16, edited 1 time in total.
Renamed the topic and edited the question.
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Re: a, b, c, d and e are five consecutive positive integers. What is the  [#permalink]

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08 Apr 2018, 23:06
Nixondutta wrote:
a, b, c, d and e are five consecutive positive integers. What is the unit's digit of $$a^b$$ * $$c^d$$ * e?
source- time4education

(1) c = 5
(2) a + b is six less than e + d.

From1:

if c=5 then sequence is 3,4,5,6,7

we can find the unit digit so sufficient.

From2:

seq1 can be 3,4,5,6,7
seq2 can be 4,5,6,7,8

so not sufficient.

Option A
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Re: a, b, c, d and e are five consecutive positive integers. What is the  [#permalink]

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26 Jul 2018, 01:16
Nixondutta wrote:
a, b, c, d and e are five consecutive positive integers. What is the unit's digit of $$a^b*c^d * e$$?

(1) c = 5
(2) a + b is six less than e + d.

source- time4education

Wait. If we are just told that a,b,c,d and e and consecutive positive integers, but no further info., why should I make the assumption that the letters are in increasing order - a<b<c<d<e - the order can be otherwise. Meaning that c could be the greatest of consecutive integers, the least and in the middle, who knows. Thus our answer A will not be sufficient. Explain someone pls, and correct me if I'm somewhere wrong.
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a, b, c, d and e are five consecutive positive integers. What is the  [#permalink]

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26 Jul 2018, 03:31
1
hovhannesmkrtchyan wrote:
Nixondutta wrote:
a, b, c, d and e are five consecutive positive integers. What is the unit's digit of $$a^b*c^d * e$$?

(1) c = 5
(2) a + b is six less than e + d.

source- time4education

Wait. If we are just told that a,b,c,d and e and consecutive positive integers, but no further info., why should I make the assumption that the letters are in increasing order - a<b<c<d<e - the order can be otherwise. Meaning that c could be the greatest of consecutive integers, the least and in the middle, who knows. Thus our answer A will not be sufficient. Explain someone pls, and correct me if I'm somewhere wrong.

Hi hovhannesmkrtchyan,
I am in your club with the same reasoning. Order is immaterial to us. Okay.

Let's analyze.

Statement-1:- c=5

We have 3 cases(say).
1. c is the minimum value in the set ,set={c=5,a=6,d=7,b=8,e=9}. So unit digit of $$a^b*c^d * e$$=$$6^8*5^7*9=6*5*9$$=270 (unit digit $$6^8$$:-6, unit digit $$5^7$$:-5)
2. c is the median(middle value) in the set, set={a=3,e=4,c=5,d=6,b=7} So unit digit of $$a^b*c^d * e$$=$$3^7*5^6*4=7*5*4$$=140 (unit digit $$3^7$$:-7, unit digit $$5^6$$:-5)
3. c is the highest value in the set, set={b=1,a=2,d=3,e=4,c=5} So unit digit of $$a^b*c^d * e$$=$$2^1*5^3*4=2*5*4$$=40 (unit digit $$2^1$$:-2, unit digit $$5^3$$:-5)
So, in any case unit digit of the product is 0.
So, st1 is sufficient.

P.S:-In each case, we have unit digit 5 multiplied by an even digit, which yields unit digit zero.

Hope it helps.
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Regards,

PKN

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a, b, c, d and e are five consecutive positive integers. What is the   [#permalink] 26 Jul 2018, 03:31
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