There is a sequence where each term is a positive integer and at least

Question

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  150^{th}  term?

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

0akshay0 · Accepted Answer

19 terms have digit 3 from 1 to 99
19 terms have digit 3 from 100 to 199
19 terms have digit 3 from 200 to 299
90 terms have digit 3 from 300 to 389
last 3 terms - 390, 391 and 392

so  150^{th}  is 392

Hence option E is correct

EMPOWERgmatRichC · Answer

Hi All,

This is a really poorly-worded question, but I think the 'intent' of this question is to describe an increasing sequence of positive integers in which every integer that includes at least one 3 among its digits is in the sequence (so the numbers 3, 33, 153, 337, etc. would all be in the sequence). Assuming that the first term in the sequence is "3", we're asked to find the 150th term in the sequence.

From 1 to 99, there are 19 terms that contain at least one '3' (you can list them out - it's not hard to find them) 
From 100 to 199, there are 19 terms that contain at least one '3' (you can list them out if needed) 
From 200 to 299, there are 19 terms that contain at least one '3' (you can list them out if needed)

So far, there are 3(19) = 57 terms accounted for. Once we hit 300, EVERY term in the next 100 terms will contain at least one '3', so we need to go 93 terms 'in' to this set of 100 terms to find that 150th term... Starting at 300, 93 terms in would be 392.

Final Answer:  E

GMAT assassins aren't born, they're made, 
Rich

gvij2017 · Answer

"at least one of each digit of the term has 3 in an ascending order" is perplexing phrase, isn't it?
I understood sequence 3,13,23, 113,....

GMATGuruNY · Answer

The answer choices indicate that the 150th term is less than 400.

Integers between 000 and 399, inclusive, that do NOT include at least one 3:
Number of options for the hundreds place = 3. (0, 1 or 2.)
Number of options for the tens place = 9. (Any digit but 3.)
Number of options for the units place = 9. (Any digit but 3.)
To combine these options, we multiply:
3*9*9 = 243

Of the 400 integers between 0 and 399, inclusive, 243 include no 3's.
Implication:
The number of integers between 0 and 399, inclusive, that include at least one 3 = 400-243 = 157

The greatest integer less than 400 that includes at least one 3 is 399.
Thus:
157th term = 399
156th term = 398
155th term = 397
154th term = 396
153rd term = 395
152nd term = 394
151st term = 393
150th term = 392

E

ScottTargetTestPrep · Answer

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

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There is a sequence where each term is a positive integer and at least

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