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M18-33

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New post 16 Sep 2014, 01:04
1
6
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

55% (00:58) correct 45% (01:03) wrong based on 268 sessions

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Math Expert
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New post 16 Sep 2014, 01:04
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New post 27 Aug 2015, 10:10
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I think this the explanation isn't clear enough, please elaborate. does non negative nature of m and n not disclude one from applying zero as values to m and n?
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New post 27 Aug 2015, 12:18
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New post 25 Jul 2016, 06:52
I think this is a high-quality question and I agree with explanation.
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New post 17 Jul 2017, 11:49
Suppose i substitute values :
1. M+N>0 say M = 0, N=2 (0+2>0)
5^0+6^2=1+36=37
So Statement 1 should be sufficient right?
Please correct me if my approach is wrong
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New post 17 Jul 2017, 11:56
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New post 21 Jun 2018, 07:41
Bunuel wrote:
If \(M\) and \(N\) are non-negative integers, what is the units digit of \(5^M + 6^N\) ?


(1) \(M + N \gt 2\)

(2) \(MN \gt 0\)


It is perhaps more difficult than 600-700 tag.

M or N can be 0 or any +ve integer.

if any if them is zero -> sum could be anything so can be the unit digit.
if none of them is zero -> unit digit is certainly 1.(5+6=11).

hence
statement 1 is of no use here.
statement 2 eliminates the first conditions.-> now unit digit can only be 1.

hence B
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New post 14 Aug 2018, 18:51
Quote:
If M and N are non-negative integers

Tricky! Non-negative integer includes zero and other +ve integers.
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Re: M18-33   [#permalink] 14 Aug 2018, 18:51
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