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

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..

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.

4^(any Number) = will never be odd

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.

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.

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

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??

We know that the sum from 1 to n is given by

n*(n+1) divided by 2, and it follows that for the number n+x the formula is given by

(n+x)*((n+x)+1) / 2 = (n+x)^2+n+x = (n^2+2nx+x^2+n+x)/2

For x>0, it follows that the sum from 1 to (n+x) minus the sum from 1 to n is equal to the sum of (n+1) to (n+x). In other words

(n^2+2nx+x^2+n+x-(n^2+n))/2 = x^2+2nx+x = x(x+2n+1). This is another way to express the sum from (n+1) to (n+x).

We know that x,n>0, and therefor (x)<(x+2n+1). Also, the two numbers have opposite parity (meaning one of them is always even, other one always odd). So in total we know now:

1. Our number is even because we can divide it by two.
2. We can express it with two factors of opposite parity.

1. kicks out all options and leaves us with (B) and (D).

But (B) is equal to 2^16, and we can't construct any odd factors out of two's only. So it must be (B).

For those confused as to why we can use the sum of (n+1) to (n+x) instead of (n) to (n+x): n is a variable, so if for example x=2 and n=3, we are looking for the sum from 3 to 5. My formula would give the sum from (n+1)=4 to 5, but then, one could just choose n=2 and get the right result ...

I hope it makes sense

Here's an easier approach:

Sum 2 consecutive integers could be:
\((n) + (n+1) = 2n + 1\) -- hence it can be all odd numbers, so eliminate options A, C and E

Sum of 3 consecutive integers: \(3n + 3\) -- nothing new here
Sum of 4 consecutive integers: \(4n + 6\) -- when n = 3 it will be a multiple of 6, so eliminate option D­

Only option B remains, so this must be the answer­