There is a sequence, 4(10^n), 4(10^(n-1)), ..., 4(10^(n-m)), for posit

The average (arithmetic mean) of the sequence’s terms to be an integer , we should have the number of terms to be either 2 or 4k where k > 1 or 10, 20 ....

So that we can we have average to be an integer.

Statement 1
m < 6

we don't have the values of n


If m = 5 and n = 12 , then the average of the terms will be non integer

If m = 3 and n = 12 , then then average of the terms will be integer

This statement is not sufficient.


Statement 2

n = 12

we don't have the values of m

If m = 5 and n = 12 , then the average of the terms will be non integer

If m = 3 and n = 12 , then then average of the terms will be integer

This statement is not sufficient.


Combining statement 1 and 2

If m = 5 and n = 12 , then the average of the terms will be non integer

If m = 3 and n = 12 , then then average of the terms will be integer



Those statement are not sufficient.


Answer is E

Given: sequence, 4(10^n), 4(10^(n-1)), ..., 4(10^(n-m)), for positive integers n, m (n>m).

(1) m < 6
and n>m
Eg (1) Let n=6 and m=1

Average = 4 [(10^6)+4(10^(5)] /6
= 4*10^(5) [10+1] /6
= 22*10^(5)/3------- Not an Integer

Eg (2) Let n=6 and m=2

Average = 4 [(10^6)+4(10^(5)+4(10^(4)] /6
= 4*10^(4) [100+10+1] /6
= 4*10^(4) (111)/3*2
= 74*10^(4)-------- Integer

Hence, (1) is insufficient

(2) n = 12
if m= 1
Average = 4 [(10^12)+4(10^(11)/12
= 4*10^(11) [10+1]/(3*4)
= 11*10^(11)/3----- Not an integer
If m=2 then
=4*10^(10) [100+10+1]/(3*4)
=37*10^(10) ----- Integer

(1)+(2) As given in the above example is not sufficient to prove if Average is an integer or not. Hence,E

option E.

number of terms in the given sequence is m+1.
Thus mean = 4[ 10^n + 10(^n-1) + 10(^n-2) + ... 10^(n-m)]/ m+1.

Now we need to find if this mean is an integer or not. For this we need the values of m and n.
Statement 1: m<6. Now as per this m can take multiple values from 1 to 5. But since value of n is not provided this is insufficient.
Statement 2 : n=12. Now since value of m not provided => insufficient.

Combining statement 1 and 2 : m<6 and n=12. still this is not sufficient. Lets say m=5 and n=12 then mean will not be an integer. and for values m=4 and n=12, mean will be an integer.

Thus both together are not sufficient. => option E.

Answer is E.

4(10^n), 4(10^(n-1)), ..., 4(10^(n-m)) ;

sum of these numbers can be written as

4*10^n + 4*(10^n)/10 + 4*(10^m)/10^2+... Simplifying.

4*10^n(1+1/10+1/10^2+... 1/10^m)...

4*10^n ((10^m+10^m-1+10^m-2+...+1)/10^m))...

Average of the above =

(4*10^n ((10^m+10^m-1+10^m-2+...+1)/10^m)))/m .. simplifyig,

A*(4*10^n)/(m*10^m) where A = (10^m+10^m-1+10^m-2+...+1). Here basically will be either 1 or 11 or 111 or 1111, etc., depending on m.

or A*(4*10^(n-m))/m

1) m<6; try multiple combinations m=5, n=6 -> integer' m=4, n=5 -> integer, m=3, n=4 (not integer as denominator is 3 and value of A is 1111 which is not divisible by 3). NOT SUFFICIENT.

2) n=12, try n=12, m=11 -> mostly not integer as denominator is 11, try n=12, m=10 -> integer, try again m=3 and n=12, -> not integer as denominator is again 3 and A is 1111, hence NOT SUFFICIENT.

Combine 1 & 2.

Try n=12, m=5 -> Integer but we know n=12 and m=3 is not integer. Hence NOT SUFFICIENT.

C
4(10^n)+4(10^n-1)+.....+4(10^n-m) has m+1 elements
Sum = 4(10^n(1-(1/10)^m+1))/(9/10)= 4(10^n+1(1-(1/10)^m+1))/9 = 4(10^some positive integral power *(1111...m+1 1's))
Mean = sum/m+1
I is insufficient as we do not have any information about n. 5 > n < ~infinity
II is insufficient as if we take m=6,n=12 we see that 7 1's is not divisible by 7 but for m=3,n=12 the sum is divisible by the number of items i.e 4.

I and II are sufficient
the pairs are(as m is a positive integer less than 6):
n = 12 m = 5
n = 12 m = 4
n = 12 m = 3
n = 12 m = 2
n = 12 m = 1
For all the above scenarios the sum is divisible by the number of elements
For m =5,no of elements=6,no if 1's = 6(6 1's are divisible by 3 and we have a 2)
For m =2,no of elements=3,,no if 1's = 3(3 1's are divisible by 3)

There is a sequence, 4(10^n), 4(10^(n-1)), ..., 4(10^(n-m)), for positive integers n, m (n>m). Is the average (arithmetic mean) of the sequence’s terms an integer?
Rewriting the sequence as
4 * 10^(n-m), 4 * 10^(n-m+1)....... 4 * 10^(n-1), 4 * 10^n

(1) m < 6
=> m = 1, 2, 3, 4 or 5
The sequence becomes (for m=5)
4*10^(n-5), 4*10^(n-4), 4*10^(n-3), 4*10^(n-2), 4*10^(n-1), 4*10^n

4*10^n { 1+.1+.01+.001+.0001+.00001} = 4*10^n (1.11111)
Since n>m => the value of the decimals(m) will be multiplied and converted into an integer and the resulting will be of the form 4*10^(n-m) * 111111

And, for all the values from m=1 to 5, ie; 6 terms, its average will be an integer sufficient


(2) n = 12
=> m = 1 to 11
The Sequence is (for m=1)
4*10^12, 4*10^11....\(4*10^2, 4*10^1\)

The average taken of this sequence for value of m=1 to 11 will also be integer

Eg:
let m=5; n-5=7
=> \(4*10^7 [ 1+10+100+1000+10000+100000] => 4*10^7 * 111111\)
and 7 divides 111111

Hence n=12 is sufficient

Since I and II separately are sufficient, Answer is D

4(10^n), 4(10^(n-1)), ..., 4(10^(n-m)) , n>m
(1) m < 6
(2) n = 12

This is an interesting sequence such that whatever may be the value of n (such that n>m given in equation) and for m<6, it is always divisible by its arithmetic mean. For instance, we can take any value of n , say n =7 and m<6 from first choice, say m=5

that gives 4444440/6 = which is integer.
Whatever may be the value of n, and m<6 it will always be divisble by its arithmetic mean.
But this fails when m =7. Lets take n=12 and m =7 which is 4444444000000/7= not an integer.

Therefore n can be any value as long as m< 6. Therefore m must be known to answer this question for sure.

Hence Choice A is sufficient, Choice B is not sufficient as we dont know the value of m. It can be 7 or anything.

Hence answer choice A.

Given: Sequence is 4*10^n [1/(10)^0 + 1/(10)^1 + . . . 1/(10)^m ] where n > m and both are +ve integers.

(1) m < 6 -> m = 1,2,...5

if m = 5, then sum of the sequence is:

= 4*10^n [1/(10)^0 + 1/(10)^1 + 1/(10)^2 +1/(10)^3 + 1/(10)^4 + 1/(10)^5 ] = 4*(10^n/10^5)*(111111)

number of terms will be m+1 i.e. 6 because sequence started from 0. As n > m so lets say n is 6

Average = [4*(10^6/10^5)*(111111)]/6 = 4*10*111111/(3*2) = integer (111111 is divisible by 3)

we notice that average in any case when m = 1,2,3,4,5,6 will be an integer because numerator will be divisible by any of the numbers such as 1,2,3,4,5,6.

Sufficient.

(2) n = 12
hence m = 1,2,3.....11. from statement-1 we know for m=1,2...5 average is integer.

if m = 6, then sum of the sequence is:
= 4*10^12 [1/(10)^0 + 1/(10)^1 + 1/(10)^2 +1/(10)^3 + 1/(10)^4 + 1/(10)^5 + 1/(10)^6] = 4*(10^12/10^6)*(1111111)

number of terms will be m+1 i.e. 7 because sequence started from 0.

Average = [4*(10^12/10^5)*(1111111)]/6 = 4*10^7*1111111/7 = NOT an INTEGER (1111111 is NOT divisible by 7)

we see, in few cases Average is an integer and in others it isn't. Answer is not definite.

In-Sufficient

Option A is the correct answer.

MATH REVOLUTION OFFICIAL SOLUTION:

Since we have 2 variables (x, y) in the ori_condi, we also need 2 equations to match the number of variables and the number of equations. Since we need both 1) and 2), the correct answer is likely C. Using both con 1) & 2), the average=[4(10^(n-m))+…+4(10^n)]/[n-(n-m)+1]=4(10^(n-m))(4^(m+1)-1)/3(m+1). If we substitute, we get 4(1012-m)[4m+1-1]/3(m+1).

If, M=1, 4(10^(12-1))(4^(1+1)-1)/3(1+1)=2(10^11)5=integer and this is a “yes”.
If, M=2, 4(10^(12-2))(4^(2+1)-1)/3(2+1)=4(10^10)7=integer and this is a “yes”.
If, M=3, 4(10^(12-3))(4^(3+1)-1)/3(3+1)=(10^9)85=integer and this is a “yes”.
If, M=4, 4(10^(12-4))(4^(4+1)-1)/3(4+1)=4(2^8*5^7)341=integer and this is a “yes”.
If, M=5, 4(10^(12-5))(4^(5+1)-1)/3(5+1)=2(10^7)455=integer and this is a “yes” and is sufficient. Therefore the answer is C. However, since this is an “integer” question, which is one of the key questions, we should apply Common Mistake Type 4(A).

In case of con 1),
If M=1, 4(10^(n-1))(4^(1+1)-1)/3(1+1)=2(10^(n-1))5=integer and this is a “yes”.
If M=2, 4(10^(n-2))(4^(2+1)-1)/3(2+1)=4(10^n-2))7=integer and this is a “yes”.
If M=3, 4(10^(n-3))(4^(3+1)-1)/3(3+1)=(10^(n-3))85=integer and this is a “yes”.
If M=4, 4(10^(n-4))(4^(4+1)-1)/3(4+1)=4(2^(n-4)*5^(n-5))341=integer and this is a “yes”.
If M=5, 4(10^(n-5))(4^(5+1)-1)/3(5+1)=2(10^(n-5))455=integer and this is a “yes” and is sufficient.

In case of con 2),
If M=1, 4(10^(12-1))(4^(1+1)-1)/3(1+1)=2(10^11)5=integer and this is a yes.
If M=6, 4(10^(12-6))(4^(6+1)-1)/3(6+1)≠integer and this is a “no” and not sufficient.

Therefore, the correct answer is A.

If both C and A are correct answers, then A is the final correct answer. This type of question appears for a perfect 51.

NOTE: Also, solving this type of question usually takes over 5 minutes during the actual exam. However, if you understand the relationship between Variable Approach Method and Common Mistake Types, you will be able to solve this type of question in just about 2 minutes.

I think problem and answer statements are little off.

Problem gives a series as 4*10^n,4*10^(n-1) and so on...
But answer is solved for series 10*4^n,10*4^(n-1) and so on...
Can you please check if my understanding is correct?

If series is 4*10^n,4*10^(n-1) and so on... then answer will be E
Sum is 4*10^(n-m) * (10^m-1)/9
and average will be 4*10^(n-m) * (10^m-1)/((m+1)*9) which is an integer for m = 1 but not for m=2
but if it 10*4^n,10*4^n-1 and so on... then answer may be B, I have not solved it for this

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