# Thread: Sequence Problem

1. ## Sequence Problem

Hello everyone!!

The following sequence continues indefinitely:

27 = 3 x 3 x 3
207 = 3 x 3 x 23
2,007 = 3 x 3 x 223
20,007 = 3 x 3 x 2223.......

Which of the following integers is a multiple of 81?

A: 200,007
B:20,000,007
C: 2,000,000,007
D: 200,000,000,007
E: 20,000,000,000,007

I know what the answer is, can someone show me how it is done? And how similar problems are done. Thanks.

2. Do you remember your divisibility rule for 9...

if the sum of the digits is divisibile by 9 then so is the number...

ex...2223 2+2+2+3=9 and 9 is divisibile by nine so

so 2223 can be factored to equal $\displaystyle 9*247$

so then

$\displaystyle 20,007=3*3*9*247$

since $\displaystyle 20,007 = 81*247$ it is divisible by 81.

so the next 22...3 will be divisible by nine is when the sum of the digits is divisivle by 9.
2222222222223 would be the next one...

since there are twelve 2's in the list it will be the number with twelve zero's

20,000,000,000,007

I hope this helps

3. That was very clear!! Thank you so much.