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Which of the following cannot be the sum of two or more cons

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Which of the following cannot be the sum of two or more cons  [#permalink]

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New post Updated on: 26 Mar 2013, 01:14
4
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

48% (02:01) correct 52% (02:09) wrong based on 367 sessions

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Which of the following cannot be the sum of two or more consecutive positive integers?

(A) 3^7
(B) 4^6
(C) 5^5
(D) 6^4
(E) 7^3

This Week's Challenge Problem from MGMAT : "Consecutive Positive Madness"

https://www.manhattangmat.com/challenge_thisweek.cfm?submitted=1

I got this wrong ... Can anybody please explain?

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Originally posted by TheNona on 25 Mar 2013, 09:00.
Last edited by Bunuel on 26 Mar 2013, 01:14, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Challenge Problem from MGMAT: Consecutive Positive Madness  [#permalink]

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New post 25 Mar 2013, 20:45
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TheNona wrote:
This Week's Challenge Problem from MGMAT : "Consecutive Positive Madness"

https://www.manhattangmat.com/challenge_thisweek.cfm?submitted=1

Which of the following cannot be the sum of two or more consecutive positive integers?

(A) 3^7 (B) 4^6 (C) 5^5 (D) 6^4 (E) 7^3

I got this wrong ... Can anybody please explain?



There are lots of properties of numbers and GMAT does not expect you to know them. So a question that appears of GMAT must be solvable without knowing the properties. So think hard about what you know and what you can apply. Use pattern recognition.

Try some numbers to start off:
1+2 = 3
2+3 = 5
3+4 = 7
ok, so is there a pattern here? We are getting all odd numbers. 3, 5, 7, 9, 11 etc. Every odd number can be written as sum of two consecutive numbers. Why? Say an odd number is N. When you divide it by 2, you get half of it which has a .5. You take the integer above it and below it and they will add up to give N
e.g. N = 11.
11/2 = 5.5 so take numbers 5 and 6 and they will add to give 11. Why? because 5.5 is the arithmetic mean of 2 consecutive numbers:5 and 6.

Takeaway: Every odd number can be written as sum of two consecutive integers.

So rule out (A), (C) and (E).

Now, try to sum up three consecutive numbers.
1+2+3 = 6
2+3 +4 = 9 (ignore odd numbers so we have already dealt with them)
3+4+5 = 12
4+5+6 = 15
5+6+7 = 18
6+7+8 = 21
7+8+9 = 24

You are getting all multiples of 3. The important thing is that no multiple of 3 is getting skipped. You are getting all of them. Hence we can represent all multiples of 3 as sum of 3 numbers. Hence (D) is also out since it is a multiple of 3.
Now think why?
Sum of three consecutive integers is given by (n-1) + n + (n + 1) = 3n

Takeaway: Sum of any three consecutive numbers will be a multiple of 3.

Hence answer must be (B) i.e. 4^6
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Re: Challenge Problem from MGMAT: Consecutive Positive Madness  [#permalink]

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New post 25 Mar 2013, 09:32
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Every natural number not of the form 2^k for some natural number k can be written as the sum of two or more consecutive positive integers.
Hence answer is B.
I solved this question manually but later found on internet that numbers which can be expressed as 2^n, n is a natural number, cannot be represented as sum of 2 or more consecutive numbers.

Please give a kudo if you like my explanation.
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Re: Challenge Problem from MGMAT: Consecutive Positive Madness  [#permalink]

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New post 25 Mar 2013, 21:23
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TheNona wrote:
This Week's Challenge Problem from MGMAT : "Consecutive Positive Madness"

https://www.manhattangmat.com/challenge_thisweek.cfm?submitted=1

Which of the following cannot be the sum of two or more consecutive positive integers?

(A) 3^7 (B) 4^6 (C) 5^5 (D) 6^4 (E) 7^3



We know that the sum of consecutive integers is :S = n/2(f+l) --> Here n is the number of terms and f and l are the first and last terms respectively. Now, we have been told that n>=2.

We can have only the following cases :

f = odd, l = odd, n = (odd-odd)+1 = odd. Thus, S = odd*even/2.

f = odd, l = even, n = (even-odd)+1 = even. S = odd.even/2. Similarly, for the other two combinations also, the sum S = odd*even/2. Now, out of the given options, all the options can be in this pattern except 4^6 = 2^12.

B.
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Re: Which of the following cannot be the sum of two or more cons  [#permalink]

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New post 04 Apr 2013, 07:44
jbisht wrote:
TheNona wrote:
Which of the following cannot be the sum of two or more consecutive positive integers?

(A) 3^7
(B) 4^6
(C) 5^5
(D) 6^4
(E) 7^3

This Week's Challenge Problem from MGMAT : "Consecutive Positive Madness"

https://www.manhattangmat.com/challenge_thisweek.cfm?submitted=1

I got this wrong ... Can anybody please explain?


Is this a 700 level question ?

answer simply depends upon logic of even and odd number



4^(any Number) = will never be odd
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Re: Which of the following cannot be the sum of two or more cons  [#permalink]

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New post 13 Sep 2015, 17:13
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whenever if you sum two consecutive positive integers,
one is odd another one is even, sum always will be odd.
so we can easily take away options A ,C and E beacuse all are odd numbers .
Whenever if we power odd numbers how many times sum is odd.

Now remaining is 4^6and 6^4.
both are even numbers if you power it , its sum always even.

Now if you sum two numbers is odd.if you sum three numbers sum can be even or odd.
But here sum of even power numbers is even.
any sum of three numbers is divisible by 3.

Take sum of 4^6 and 6^4.

first take 6^4 is 36*36.
both numbers are divisible by3.

Now look 4^6. sum is 16*16*16.
all three numbers not divisibly by 3.


So option B is not a sum of consecutive numbers..
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Which of the following cannot be the sum of two or more cons  [#permalink]

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New post 03 Apr 2016, 07:47
Any sum of any number of consecutive integers will always have an odd factor other 1 .

Hence, B is the answer. because B does not have any odd factor other than 1.
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Which of the following cannot be the sum of two or more cons  [#permalink]

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New post 23 Sep 2017, 23:44
OFFICIAL SOLUTIONF FROM MANHATTAN

The problem seeks a quantity that cannot be a sum of the type described. Process of elimination, then, will likely be an efficient solution method. Specifically, if an answer choice can be shown to be the sum of two or more consecutive positive integers, then that answer can be eliminated.

One approach: Test Cases
The problem specifically discusses the sum of two or more consecutive positive integers. Start with the simplest possibility: the sum of two consecutive positive integers.
1 + 2 = 3
2 + 3 = 5
3 + 4 = 7

What’s the pattern? First, any two consecutive positive integers will sum to an add number. Second, any odd sum greater than 1 can be created (the sums will continue to increase in this manner – 3, 5, 7, 9, 11, … – forever).

Therefore, any odd number greater than 1 can be created by adding together two consecutive positive integers. Answers A, C, and E all represent odd numbers; eliminate them.

Try the next simplest case: the sum of three consecutive positive integers.
1 + 2 + 3 = 6
2 + 3 + 4 = 9
3 + 4 + 5 = 12

What’s the pattern here? The sums can be odd or even – no apparent pattern there. Hmm. All three are multiples of 3… is that an actual pattern, though?

Yes, it is! The sum of any set of three consecutive integers can also be calculated by taking the average and multiplying by the number of terms – that is, multiplying by 3. So any sum of three consecutive integers will be a multiple of 3.

As a result, any multiple of 3 (starting with 6) can be created by finding the sum of the appropriate set of three consecutive positive integers. Answer D represents a multiple of 3; eliminate it.

By process of elimination, the only remaining answer is B. The correct answer is B.
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Which of the following cannot be the sum of two or more cons  [#permalink]

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New post 15 Dec 2017, 20:06
TheNona wrote:
Which of the following cannot be the sum of two or more consecutive positive integers?

(A) 3^7
(B) 4^6
(C) 5^5
(D) 6^4
(E) 7^3

This Week's Challenge Problem from MGMAT : "Consecutive Positive Madness"

https://www.manhattangmat.com/challenge_thisweek.cfm?submitted=1

I got this wrong ... Can anybody please explain?


Let, the 2 consecutive integers be a, a +d and 3 consecutive integers be a-d, a , a+d

Now, \(3^7 , 5^5 & 7^3\) can be sum of 2 Consecutive Integers for d =1 in 2a + d equation. The confusion here will be for \(4^6 & 6^4\)as we cant decide with 2a+d equation. Lets take 3 consecutive integers a-d, a, a+d
3a = \(4^6 & 6^4\) .. In this only \(6^4\) is divisible.

Hi Bunuel & chetan2u,

Can you please confirm if the above approach is correct to solve the problem or am I missing something??
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Re: Which of the following cannot be the sum of two or more cons  [#permalink]

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New post 16 Dec 2017, 05:13
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rahul16singh28 wrote:
TheNona wrote:
Which of the following cannot be the sum of two or more consecutive positive integers?

(A) 3^7
(B) 4^6
(C) 5^5
(D) 6^4
(E) 7^3

This Week's Challenge Problem from MGMAT : "Consecutive Positive Madness"

https://www.manhattangmat.com/challenge_thisweek.cfm?submitted=1

I got this wrong ... Can anybody please explain?


Let, the 2 consecutive integers be a, a +d and 3 consecutive integers be a-d, a , a+d

Now, \(3^7 , 5^5 & 7^3\) can be sum of 2 Consecutive Integers for d =1 in 2a + d equation. The confusion here will be for \(4^6 & 6^4\)as we cant decide with 2a+d equation. Lets take 3 consecutive integers a-d, a, a+d
3a = \(4^6 & 6^4\) .. In this only \(6^4\) is divisible.

Hi Bunuel & chetan2u,

Can you please confirm if the above approach is correct to solve the problem or am I missing something??


Hi..

You are absolutely correct in your approach..

Why 4^6 cannot be put as SUM of two consecutive integers..
Because you cannot add 2 or more integers to get any pure power of 2..
check 2, 4, 8 , 16 ...and so on
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Re: Which of the following cannot be the sum of two or more cons &nbs [#permalink] 16 Dec 2017, 05:13
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