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M18-33

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Math Expert

V

Joined: 02 Sep 2009

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M18-33 [#permalink]

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16 Sep 2014, 00:04

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6

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E

Difficulty:

55% (hard)

Question Stats:

55% (00:56) correct

45% (00:58) wrong

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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) \(MN \gt 0\)

Spoiler: OA

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Joined: 02 Sep 2009

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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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Joined: 15 Apr 2015

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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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Joined: 31 Mar 2016

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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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V

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Posts: 51261

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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B

Joined: 15 Apr 2018

Posts: 15

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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B

Joined: 19 Feb 2018

Posts: 24

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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