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

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swerve: Legendary Member; Posts: 2545; Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

by swerve » Tue Oct 31, 2017 9:30 am

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.

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Source: Beat The GMAT — Problem Solving | Tue Oct 31, 2017 9:30 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Thu Nov 02, 2017 9:48 am

...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?

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swerve: Legendary Member; Posts: 2545; Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

by swerve » Fri Nov 03, 2017 6:59 am

Hi, here's the source where I transcribed this question. I don't understand the solution that they explain.

https://gmatclub.com/forum/there-is-a-s ... 34724.html

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Fri Nov 03, 2017 7:38 am

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

Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8154; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Fri Dec 13, 2019 1:13 pm

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

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Dec 13, 2019 2:00 pm

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 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

The correct answer is E.

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