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

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

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