the sequence 12,15, 18, 21, 51, 81,... consists of all positive multiples of 3 that contain at least one digit that is a 1. what is the 50th term of this sequence?

what i tried:

I listed out all the multiples of 3 and circled ones with the number 1. but this method has many sources of error. can someones please help. thanks in advance.

mathlete Jun 28, 2020

#1**+1 **

We know that the rule for divisibility by 3 is that the digits of the number must add up to a multiple of 3. So, it's clear that there are no other such two-digit numbers beyond the ones listed in the problem. Every number divisible by 3 between 100 and 99 is in the sequence, so that gets us through the 39th term of the sequence. Using the rule for divisibility by 3 , it is fairly simple to list out the remaining 11 terms of the sequence: 201, 210. 213, 216, 219, 231, 261, 291, 312, 315, 318. Thus, the 50th term is 318.

mathlete Jun 28, 2020