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# M18-33

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Math Expert
Joined: 02 Sep 2009
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M18-33  [#permalink]

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16 Sep 2014, 00:04
1
6
00:00

Difficulty:

55% (hard)

Question Stats:

55% (00:56) correct 45% (00:58) wrong based on 236 sessions

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If $$M$$ and $$N$$ are non-negative integers, what is the units digit of $$5^M + 6^N$$ ?

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

(2) $$MN \gt 0$$

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Re M18-33  [#permalink]

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16 Sep 2014, 00:04
1
Official Solution:

If $$M$$ and $$N$$ are both positive, $$5^M + 6^N$$ ends with 1 (because $$5^M$$ ends with 5 and $$6^N$$ ends with 6).

Statement (1) by itself is insufficient. S1 does not preclude $$M$$ (or $$N$$) from being 0.

Statement (2) by itself is sufficient. S2 gives that both $$M$$ and $$N$$ are positive.

Answer: B
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Re M18-33  [#permalink]

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27 Aug 2015, 09:10
1
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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M18-33  [#permalink]

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27 Aug 2015, 11:18
SoumiyaGoutham wrote:
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?

Please read carefully!

It's not given that m and n are negative, they are given to be non-negative...
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Re M18-33  [#permalink]

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25 Jul 2016, 05:52
I think this is a high-quality question and I agree with explanation.
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Joined: 03 Jun 2017
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Re: M18-33  [#permalink]

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17 Jul 2017, 10:49
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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Re: M18-33  [#permalink]

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17 Jul 2017, 10:56
namrata88 wrote:
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

But what if m = 1 and n = 2? What would be the answer in this case?
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Re: M18-33  [#permalink]

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21 Jun 2018, 06:41
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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Re: M18-33  [#permalink]

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14 Aug 2018, 17:51
Quote:
If M and N are non-negative integers

Tricky! Non-negative integer includes zero and other +ve integers.
Re: M18-33 &nbs [#permalink] 14 Aug 2018, 17:51
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# M18-33

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