Hi, I am confused on this question: 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 the sequence?

(Source: 2006 National Target #5)

321.

Step-by-step explanation:

The next term after 81 is 102 .

Between  102 and 198 inclusive we have  (198-102)  / 5  + 1 = 96/3 + 1 = 33 terms - these will all have at least one 1 and be divisible by 3.

So 198 = 6 + 33 = 39th term.

After 198 we have 201, 210, 213, 219, 231, 261, 291, 312, 315, 318, 321

The answer is 321..

b

step-by-step explanation:

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)