1987 AIME Problem 11

Question

Find the largest possible

k

for which

3^{11}

is expressible as the sum of

k

consecutive positive integers.

Answer

Write the sum as

\frac{k(2a+k-1)}{2}

The only possible lengths divide

2\cdot3^{11}

; then enforce a positive first term

If the first term is

a,

then

2\cdot3^{11}=k(2a+k-1).

Thus

k

is

3^j

or

2\cdot3^j.

The largest viable even choice is

k=2\cdot3^5=486,

for which

2a+k-1=3^6=729

and

a=122>0.

The next candidates,

3^6

and

2\cdot3^6,

force a nonpositive first term, as do all larger choices. Hence

k=486.

Problem 11 in other years