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There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for posit
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20 Dec 2015, 06:37
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Math Revolution and GMAT Club Contest Starts! QUESTION #16:There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for positive integers n, m (n>m). Is the average (arithmetic mean) of the sequence’s terms an integer? (1) m < 6 (2) n = 12 Check conditions below: Math Revolution and GMAT Club ContestThe Contest Starts November 28th in Quant Forum We are happy to announce a Math Revolution and GMAT Club Contest For the following four (!) weekends we'll be publishing 4 FRESH math questions per weekend (2 on Saturday and 2 on Sunday). To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific. Then a week later, the forum moderator will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6months access to GMAT Club Tests. PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize: PS + DS course with 502 videos that is worth $299! All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time. NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.
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Re: There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for posit
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22 Dec 2015, 10:52
(1) m < 6, then m can be 1,2,3,4,5. \(mean = \frac{4.10^n + 4.10^{n1}+...+4.10^{nm}}{(m+1)}=\frac{4.10^{nm}(10^{m}+10^{m1}+...+1)}{(m+1)}=\frac{4.10^{nm}(11...1)}{(m+1)}\) (The number of 1 is m+1) The mean will be an integer if the numerator is divisible by (m+1). m = 1 > m + 1 = 2 : Obviously YES because 4 is divisible by 2 m = 2 > m + 1 = 3 : YES because 111 is divisible by 3 m = 3 > m + 1 = 4 : YES because we have 4 in the numerator. m = 4 > m + 1 = 5 : YES because n > m then the numerator is divisible by 10 and of course by 5. m = 5 > m + 1 = 6 : YES because 111111 is divisible by 3 and 4 is divisible by 2 > the numerator is divisible by 6. => (1) is sufficient to conclude that the mean is an integer. (2) n = 12 if m = 1 then the mean is an integer. if m = 6 then mean = \(\frac{4.10^5.1111111}{7}\) is not an integer => (2) is insufficient Answer A.
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Re: There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for posit
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20 Dec 2015, 07:59
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



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Re: There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for posit
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20 Dec 2015, 08:09
Given: sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), 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



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Re: There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for posit
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20 Dec 2015, 11:19
option E. number of terms in the given sequence is m+1. Thus mean = 4[ 10^n + 10(^n1) + 10(^n2) + ... 10^(nm)]/ 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.
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Re: There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for posit
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21 Dec 2015, 18:48
Answer is E. 4(10^n), 4(10^(n1)), ..., 4(10^(nm)) ; 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^m1+10^m2+...+1)/10^m))... Average of the above = (4*10^n ((10^m+10^m1+10^m2+...+1)/10^m)))/m .. simplifyig, A*(4*10^n)/(m*10^m) where A = (10^m+10^m1+10^m2+...+1). Here basically will be either 1 or 11 or 111 or 1111, etc., depending on m. or A*(4*10^(nm))/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.
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Re: There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for posit
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22 Dec 2015, 08:55
St1: m < 6 Minimum value of m = 1. Given: n > m When m = 1, we will have 2 terms > 4 * (anything) is even > Average of 2 terms is an integer When m = 2, we will have 3 terms > 4 * (10^ + 10^ + 10^ ) > Average of 3 terms will be divisible by 3 as sum of the digits inside the bracket equals 3 and hence satisfies the divisibility condition for 3 > Average of 3 terms is an integer When m = 3, we will have 4 terms > 4 * (anything) is divisible by 4 since it the result is a multiple of 4 > Average of 4 terms is an integer When m = 4, we will have 5 terms > 4 * (10^ + 10^ + 10^ + 10^ + 10^) > The result obtained ends with a 0 as n > m > Divisibility condition for 5 is satisfied > Average of 5 terms is an integer When m = 5, we will have 5 terms > 4 * (10^ + 10^ + 10^ + 10^ + 10^ + 10^) > The result obtained is even and a multiple of 3 > Divisibility condition for 6 is satisfied > Average of 6 terms is an integer
Statement 1 alone is sufficient.
St2: n = 12 Average of terms is an integer when m = 1 (as shown before) But, when m = 10, we will have 11 terms > 4(10^12 + 10^11 + ....... 10^2) > This does not satisfy the divisibility condition for 11 > Average of 11 terms is not an integer.
Statement 2 alone is not sufficient
Answer: A



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Re: There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for posit
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22 Dec 2015, 11:29
C 4(10^n)+4(10^n1)+.....+4(10^nm) 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)



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Re: There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for posit
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24 Dec 2015, 07:07
There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), 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^(nm), 4 * 10^(nm+1)....... 4 * 10^(n1), 4 * 10^n
(1) m < 6 => m = 1, 2, 3, 4 or 5 The sequence becomes (for m=5) 4*10^(n5), 4*10^(n4), 4*10^(n3), 4*10^(n2), 4*10^(n1), 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^(nm) * 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; n5=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



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Re: There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for posit
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25 Dec 2015, 06:52
4(10^n), 4(10^(n1)), ..., 4(10^(nm)) , 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.



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Re: There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for posit
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25 Dec 2015, 15:18
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,...5if 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 = 12hence m = 1,2,3.....11. from statement1 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. InSufficientOption A is the correct answer.
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Re: There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for posit
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27 Dec 2015, 12:28
Bunuel wrote: Math Revolution and GMAT Club Contest Starts! QUESTION #16:There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for positive integers n, m (n>m). Is the average (arithmetic mean) of the sequence’s terms an integer? (1) m < 6 (2) n = 12 Check conditions below: Math Revolution and GMAT Club ContestThe Contest Starts November 28th in Quant Forum We are happy to announce a Math Revolution and GMAT Club Contest For the following four (!) weekends we'll be publishing 4 FRESH math questions per weekend (2 on Saturday and 2 on Sunday). To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific. Then a week later, the forum moderator will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6months access to GMAT Club Tests. PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize: PS + DS course with 502 videos that is worth $299! All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time. NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.
Thank you! 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^(nm))+…+4(10^n)]/[n(nm)+1]=4(10^(nm))(4^(m+1)1)/3(m+1). If we substitute, we get 4(1012m)[4m+11]/3(m+1). If, M=1, 4(10^(121))(4^(1+1)1)/3(1+1)=2(10^11)5=integer and this is a “yes”. If, M=2, 4(10^(122))(4^(2+1)1)/3(2+1)=4(10^10)7=integer and this is a “yes”. If, M=3, 4(10^(123))(4^(3+1)1)/3(3+1)=(10^9)85=integer and this is a “yes”. If, M=4, 4(10^(124))(4^(4+1)1)/3(4+1)=4(2^8*5^7)341=integer and this is a “yes”. If, M=5, 4(10^(125))(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^(n1))(4^(1+1)1)/3(1+1)=2(10^(n1))5=integer and this is a “yes”. If M=2, 4(10^(n2))(4^(2+1)1)/3(2+1)=4(10^n2))7=integer and this is a “yes”. If M=3, 4(10^(n3))(4^(3+1)1)/3(3+1)=(10^(n3))85=integer and this is a “yes”. If M=4, 4(10^(n4))(4^(4+1)1)/3(4+1)=4(2^(n4)*5^(n5))341=integer and this is a “yes”. If M=5, 4(10^(n5))(4^(5+1)1)/3(5+1)=2(10^(n5))455=integer and this is a “yes” and is sufficient. In case of con 2), If M=1, 4(10^(121))(4^(1+1)1)/3(1+1)=2(10^11)5=integer and this is a yes. If M=6, 4(10^(126))(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.
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Re: There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for posit
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07 Jun 2017, 10:19
Bunuel wrote: Bunuel wrote: Math Revolution and GMAT Club Contest Starts! QUESTION #16:There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for positive integers n, m (n>m). Is the average (arithmetic mean) of the sequence’s terms an integer? (1) m < 6 (2) n = 12 Check conditions below: Math Revolution and GMAT Club ContestThe Contest Starts November 28th in Quant Forum We are happy to announce a Math Revolution and GMAT Club Contest For the following four (!) weekends we'll be publishing 4 FRESH math questions per weekend (2 on Saturday and 2 on Sunday). To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific. Then a week later, the forum moderator will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6months access to GMAT Club Tests. PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize: PS + DS course with 502 videos that is worth $299! All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time. NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.
Thank you! 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^(nm))+…+4(10^n)]/[n(nm)+1]=4(10^(nm))(4^(m+1)1)/3(m+1). If we substitute, we get 4(1012m)[4m+11]/3(m+1). If, M=1, 4(10^(121))(4^(1+1)1)/3(1+1)=2(10^11)5=integer and this is a “yes”. If, M=2, 4(10^(122))(4^(2+1)1)/3(2+1)=4(10^10)7=integer and this is a “yes”. If, M=3, 4(10^(123))(4^(3+1)1)/3(3+1)=(10^9)85=integer and this is a “yes”. If, M=4, 4(10^(124))(4^(4+1)1)/3(4+1)=4(2^8*5^7)341=integer and this is a “yes”. If, M=5, 4(10^(125))(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^(n1))(4^(1+1)1)/3(1+1)=2(10^(n1))5=integer and this is a “yes”. If M=2, 4(10^(n2))(4^(2+1)1)/3(2+1)=4(10^n2))7=integer and this is a “yes”. If M=3, 4(10^(n3))(4^(3+1)1)/3(3+1)=(10^(n3))85=integer and this is a “yes”. If M=4, 4(10^(n4))(4^(4+1)1)/3(4+1)=4(2^(n4)*5^(n5))341=integer and this is a “yes”. If M=5, 4(10^(n5))(4^(5+1)1)/3(5+1)=2(10^(n5))455=integer and this is a “yes” and is sufficient. In case of con 2), If M=1, 4(10^(121))(4^(1+1)1)/3(1+1)=2(10^11)5=integer and this is a yes. If M=6, 4(10^(126))(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^(n1) and so on... But answer is solved for series 10*4^n,10*4^(n1) and so on... Can you please check if my understanding is correct? If series is 4*10^n,4*10^(n1) and so on... then answer will be E Sum is 4*10^(nm) * (10^m1)/9 and average will be 4*10^(nm) * (10^m1)/((m+1)*9) which is an integer for m = 1 but not for m=2 but if it 10*4^n,10*4^n1 and so on... then answer may be B, I have not solved it for this



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Re: There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for posit
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Re: There is a sequence, 4(10^n), 4(10^(n1)), ..., 4(10^(nm)), for posit
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