ziyuen wrote:
For positive integers n and m \(n>m\), is the average (arithmetic mean) of \(4*10^n\), \(4*10^{n-1}\), ......, and \(4*10^{n-m}\) an integer?
1) \(m<6\)
2) \(n=10\)
Hi
Few important points to solve this Q..
1) n>m, so lowest term \(4*10^{n-m}\) 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
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