Author 
Message 
TAGS:

Hide Tags

Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1348
Location: Malaysia

For positive integers n and m (n>m), is the average (arithmetic mean)
[#permalink]
Show Tags
25 Mar 2017, 15:41
Question Stats:
16% (02:29) correct 84% (02:28) wrong based on 220 sessions
HideShow timer Statistics
For positive integers n and m \(n>m\), is the average (arithmetic mean) of \(4*10^n\), \(4*10^{n1}\), ......, and \(4*10^{nm}\) an integer? 1) \(m<6\) 2) \(n=10\)
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
"Be challenged at EVERY MOMENT."“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”"Each stage of the journey is crucial to attaining new heights of knowledge."Rules for posting in verbal forum  Please DO NOT post short answer in your post! Advanced Search : https://gmatclub.com/forum/advancedsearch/




Math Expert
Joined: 02 Aug 2009
Posts: 8023

Re: For positive integers n and m (n>m), is the average (arithmetic mean)
[#permalink]
Show Tags
25 Mar 2017, 22:48
ziyuen wrote: For positive integers n and m \(n>m\), is the average (arithmetic mean) of \(4*10^n\), \(4*10^{n1}\), ......, and \(4*10^{nm}\) an integer?
1) \(m<6\) 2) \(n=10\) Hi Few important points to solve this Q.. 1) n>m, so lowest term \(4*10^{nm}\) will be ATLEAST 4*10 or 40 2) Therefore each term will be multiple of 40 or 2,4,5,8,10,20,40 3) Average of the terms is " INTEGER or Not" depends on number of elements. 4) number of elements depends on 'm' and is m+1 5) since each term multiple of 40, so if m is multiple of the factors of 40, answer is YES, otherwise it has to be checkedLet's see the statements: 1) m<6 Therefore m can be 1 to 5 So number of terms can be 1 to 5+1.. We already know if number of elements are factors of 40, and is YES.. So 1,2,4,5 will have answer as YES.. Let's check for 3 and 6.. Each number in series consists of digits 4 and 0.... 40,400,4000 etc So if there are 3 or 6 numbers, the SUM of digits of total of these numbers will be 3*4 or 6*4, which will be divisible by 3, thus we will have an integer here too.. Example: 40+400+4000=4440.. sum of digits is 12, so number is div by 3 Or 400000+40000+4000=444000, again SUM of digits is 12 6 numbers will have total div by 3 and the numbers being EVEN, total will be div by 6, thus an integer again Ans is YES always Sufficient 2) n=10.. We are concerned about m only Insufficient A
_________________




Manager
Joined: 22 Mar 2014
Posts: 99
Location: United States
Concentration: Finance, Operations
GPA: 3.91
WE: Information Technology (Computer Software)

Re: For positive integers n and m (n>m), is the average (arithmetic mean)
[#permalink]
Show Tags
26 Mar 2017, 13:23
Hi Chetan,
I did not understand your explanation.



Intern
Joined: 10 Jul 2017
Posts: 30

For positive integers n and m (n>m), is the average (arithmetic mean)
[#permalink]
Show Tags
19 Aug 2018, 19:16
Ques: Arithmetic mean an integer? we can simply take common from the expression 4∗10^n , 4∗10^n−1, ......, and 4∗10^n−m => 4*10^n (1+1/10 +1/10^2+......+1/10^m)(1)
so here we can see that the value of 'n' doesn't matter...it can be 40,400,40000..... The only information you need is the value of 'm' , hence A sufficient !
Further: you can take the LCM of equation 1 4*10^nm (10^m+10^m1.....+1) => 4*10^nm (11111.....m times)



Senior Manager
Joined: 15 Jan 2017
Posts: 340

Re: For positive integers n and m (n>m), is the average (arithmetic mean)
[#permalink]
Show Tags
12 Sep 2018, 01:06
How did we assume m to be between 1 to 5? if M< 6 then can't m be negative, like say m = 6?



Manager
Joined: 21 Jun 2017
Posts: 234
Concentration: Finance, Economics
WE: Corporate Finance (Commercial Banking)

Re: For positive integers n and m (n>m), is the average (arithmetic mean)
[#permalink]
Show Tags
08 Nov 2018, 07:32
Madhavi1990 wrote: How did we assume m to be between 1 to 5? if M< 6 then can't m be negative, like say m = 6? Stem mentions n and m are positive integers
_________________
Even if it takes me 30 attempts, I am determined enough to score 740+ in my 31st attempt. This is it, this is what I have been waiting for, now is the time to get up and fight, for my life is 100% my responsibility.
Dil ye Ziddi hai !!!
GMAT 1  620 .... Disappointed for 6 months. Im back Im back. Bhai dera tera COMEBACK !!!



SVP
Joined: 03 Jun 2019
Posts: 1755
Location: India

For positive integers n and m (n>m), is the average (arithmetic mean)
[#permalink]
Show Tags
25 Aug 2019, 09:54
hazelnut wrote: For positive integers n and m \(n>m\), is the average (arithmetic mean) of \(4*10^n\), \(4*10^{n1}\), ......, and \(4*10^{nm}\) an integer?
1) \(m<6\) 2) \(n=10\) Asked: For positive integers n and m \(n>m\), is the average (arithmetic mean) of \(4*10^n\), \(4*10^{n1}\), ......, and \(4*10^{nm}\) an integer? Number of terms = n  (nm) +1 = m+1 the average (arithmetic mean) of \(4*10^n\), \(4*10^{n1}\), ......, and \(4*10^{nm}\) = \(\frac{4*10^n + 4*10^{n1} + ......+4*10^{nm}}{(m+1)}\) 1) \(m<6\) Let us take m=0 \(4*10^n/1 = 4*10^n\) Integer Let us take m=1 \(\frac{4*10^n + 4*10^{n1}}{2} = \frac{4*10^{n1} ( 1+10)}{2} =\frac{4*10^{n1} *11}{2}\) Integer Let us take m=2 \(\frac{4*10^n + 4*10^{n1} + 4*10^{n2}}{3} = \frac{4*10^{n2} ( 1+10+100)}{3} =\frac{4*10^{n2} *111}{3}\) Integer Let us take m=3 \(\frac{4*10^n + 4*10^{n1} + 4*10^{n2} + 4*10^{n3}}{4} = \frac{4*10^{n3} ( 1+10+100+1000)}{4} = \frac{4*10^{n3} *1111}{4}\) Integer Let us take m=4 \(\frac{4*10^n + 4*10^{n1} + 4*10^{n2} + 4*10^{n3} + 4*10^{n4}}{5} = \frac{4*10^{n4} ( 1+10+100+1000+10000)}{5} = \frac{4*10^{n4} *11111}{5}\) Integer Let us take m=5 \(\frac{4*10^n + 4*10^{n1} + 4*10^{n2} + 4*10^{n3} + 4*10^{n4} + 4*10^{n5}}{6} = \frac{4*10^{n5} ( 1+10+100+1000+10000 +10^5)}{6} =\frac{4*10^{n5} *111111}{6}\) Integer SUFFICIENT 2) \(n=10\) Since if we take n=10 and m=6 The expression will not be divisible by 7 NOT SUFFICIENT IMO A
_________________
"Success is not final; failure is not fatal: It is the courage to continue that counts." Please provide kudos if you like my post. Kudos encourage active discussions. My GMAT Resources:  Efficient LearningAll you need to know about GMAT quantTele: +911140396815 Mobile : +919910661622 Email : kinshook.chaturvedi@gmail.com



Intern
Joined: 14 Jun 2015
Posts: 15

Re: For positive integers n and m (n>m), is the average (arithmetic mean)
[#permalink]
Show Tags
28 Aug 2019, 20:16
hazelnut wrote: For positive integers n and m \(n>m\), is the average (arithmetic mean) of \(4*10^n\), \(4*10^{n1}\), ......, and \(4*10^{nm}\) an integer?
1) \(m<6\) 2) \(n=10\) Hi..... Not convinced with the OA..... Let us consider n=10, m=3, which gives Avg as an Integer. However, if we consider n=10, m=5, its Avg will not be an integer. I had opted for Optn E.... Please clarify. Bunuel chetan2u request your help




Re: For positive integers n and m (n>m), is the average (arithmetic mean)
[#permalink]
28 Aug 2019, 20:16






