There is a sequence where each term is a positive...

...at least one of each digit of the term has 3 in an ascending order

Hmmmm, what's the source of this question?
Did you transcribe it correctly?

Cheers,
Brent

I see.
So the question is telling us that a certain sequence consists of ascending positive integers such that every integer has at least one 3.
We want the 150th term.

In the link you provided, 0akshay0 offers a nice/quick solution.
Do you have any questions about it?

Cheers,
Brent

swerve wrote:There is a sequence where each term is a positive integer and at least one of each digit of the term has 3 in an ascending order. What is the value of 150th term?

A. 326
B. 329
C. 342
D. 382
E. 392

The OA is E.

Please, can any expert explain this PS question? I don't have it clear. Thanks.

If a term has only one digit, then that term is 3 and it will be the first term of the sequence.

If a term has two digits, then either its units digit or its tens digit is 3. There are 9 two-digit numbers with 3 as the units digit (i.e., 13, 23, 33, …, 93) and 10 two-digit numbers with 3 as the tens digit (30, 31, 32, 33, …, 39). However, since 33 is counted twice, there are actually 9 + 10 - 1 = 18 two-digit numbers with 3 as at least one of its digits. These numbers will be the 2nd to the 19th terms of the sequence.

Adding 100 to each of the first 19 terms, we also have a term that contains 3 as one of its digits. So these numbers (in the 100s) will be the 20th to the 38th terms of the sequence (note: the 20th term is 103 and the 38th term is 193).

Similarly, adding 200 to each of the first 19 terms, we also have a term that contains 3 as one of its digits. So these numbers (in the 200s) will be the 39th to the 57th terms of the sequence (note: the 39th term is 203 and the 57th term is 293).

Now the next 100 terms will be all the numbers in the 300s since the hundreds digit of each of these numbers is 3. In other words, the 58th term is 300 and the 157th term is 399. Since we are looking for the 150th term, we can backtrack from the 157th term, which is 399. Since 150 is 7 less than 157, 7 less than 399 is 392. So the 150th term must be 392.

Answer: E