Question

Find the largest k such that \(\large{3^{11}}\) is the sum of k consecutive positive integers.

Answer 1

Okay, I tried in the following method:

Let the integers be n, n+1, n+2, ..., n+k-1. Then,

\[\large{\text{Sum of k consecutive positive integers} = \cfrac{(n+k-1)(n+k) + (n-1)n}{2}}\]

Answer 2

@ganeshie8

Answer 3

\[\large{\text{Sum of k consecutive positive integers} = \cfrac{(n+k-1)(n+k)}{2} +}\]

\(\large{\cfrac{(n-1)n}{2}}\)

Answer 4

Now I tried to equate it to \(\large{3^{11}}\). But, where I am stuck is how I am going to calculate 2 variables from 1 equation.

Answer 5

\[3^{11 } = nk + \dfrac{k(k-1)}{2}\]

Answer 6

Okay why ?

Answer 7

ive just simplified..

Answer 8

\[2 \times 3^{11 } = 2nk + k(k-1)\]

Answer 9

Ohh okay

Answer 10

\[2 \times 3^{11 } = k(2n + k-1)\]

Answer 11

Okay I got that but how should I proceed after that. I am thinking of something like divisibility

Answer 12

\[\large{k|2\cdot3^{11}}\]

Answer 13

yes, there are (1+1)(11+1) = 24 divosors

Answer 14

Okay. But I am not going to check them one by one.. Its not fair :(

Answer 15

k can be any one of them, but we can reduce the number of cases significantly with few observations

Answer 16

Okay. I think we would have to use the fact that k has to be maximum.

Answer 17

(2n+k-1) > k

Answer 18

But that is already true...

Answer 19

that observation reduces the number of cases to more than half

Answer 20

Okay I would still have to check out 12 options ?

Answer 21

yes, this is a quadratic diaphontine equation 12 options is not too unfair :) i see we can reduce further..

Answer 22

Okay.. I too would try to get something out of it.

Well thanks for reducing the options ;)

Answer 23

thats it, i think we're done! all 12 cases are solutions to the diaphontine equation

out of these solutions,
the largest possible value for k is : \(2*3^5\) since it has to be less than the other factor \(3^6\)

Answer 24

Great !! I found the solution here: http://www.artofproblemsolving.com/Wiki/index.php/1987_AIME_Problems/Problem_11

Thanks a lot @ganeshie8 .

Answer 25

I think I would have to review Diophantine Equations :)

Answer 26

Thanks again

Answer 27

nice :) we worked a similar problem recently @ikram002p

Answer 28

it is about finding the largest sequence of consecutive positive integers whose sum is 2000

Answer 29

Ohh ?

Answer 30

Well that would have lesser number of possible cases then ?

Answer 31

yes so much reduction is possible and can be bit tricky :

\(2000 = a + (a+1) + \cdots (a+k-1)\)

Answer 32

but the method is exact same as earlier one... we get a diaphontine equation and solve it

Answer 33

we can also found it by finding odd factors

Answer 34

how does that work @ikram002p ?

Answer 35

ik how algorithm ,but dnt know the proof lol

Answer 36

i think greatest n will be ( 3^11+1)/2
then we find k from sumation