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There is a sequence where each term is a positive integer an

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by hazelnut01 » Thu May 04, 2017 6:26 pm
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

Source : Math Revolution
OA=E
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Source: — Problem Solving |

by [email protected] » Fri May 05, 2017 9:24 am
Hi ziyuenlau,

This is a really poorly-worded question, and it's not the only poorly-worded question to come from this source. If you're serious about studying for the GMAT, then you might want to focus your studies on more realistic GMAT study materials.

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)
From 100 to 199, there are 19 terms that contain at least one '3' (you can list them out)
From 200 to 299, there are 19 terms that contain at least one '3' (you can list them out)

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

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Rich
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