Numbers with at least One 7

Expert replies

Bullzi: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Thu May 28, 2015 1:13 am; Thanked: 5 times

by Bullzi » Thu Sep 24, 2015 8:32 am

How many positive 3 digit integers contain at least one 7

While I understand the Restrictive Rule way of solving this question below, I couldn't arrive at the same solution using a straightforward method. Please help!

Restrictive Rule - Ignore the problem restrictions and arrive at
a) 900 ways of forming any 3 digit number (9*10*10 ways for 3 digits)
b) 648 ways of forming a 3 digit number with no 7s (8*9*9 ways for digits)
c) 900 - 648 resulting in 252

This is how I approached the problem and I am obviously missing something,:shock:
Straightforward
a) 3 digits with One 7 - 8*10*1
b) 3 digits with Two 7 - 8*1*1
c) 3 digits with Three 7 - 1*1*1

Thanks
Bullzi

Like this post

Source: Beat The GMAT — Problem Solving | Thu Sep 24, 2015 8:32 am

GMAT/MBA Expert

[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members; Location: Palo Alto, CA; Thanked: 2867 times; GMAT Score:800

by [email protected] » Thu Sep 24, 2015 8:43 am

Hi Bullzi,

I'm going to give you some 'hints' so that you can continue to work on this problem:

The math that you use in your approach is meant to break the total number of integers down into smaller groups (which is fine). However, you have to think about what each calculation actually 'translates' into...

For example, if you're looking for the number of 3-digit numbers with TWO 7s, does it matter WHICH TWO digits are 7s?

In the calculation (8)(1)(1)....you assume that the LAST TWO digits have to be the 7s. However, there are OTHER ways to have two 7s that this calculation does not account for.... Couldn't the first and second digits be the 7s? Or the first and third digits? How can you account for those possibilities?

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

Contact Rich at [email protected]

Like this post

Bullzi: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Thu May 28, 2015 1:13 am; Thanked: 5 times

by Bullzi » Thu Sep 24, 2015 9:13 am

Thanks Rich for the hints!

So, I understand that I haven't accounted for the number of arrangements for each of these 3 digit combinations I have arrived at. For example, I have assumed in a that the lone 7 is at the units position while practically it could be at any of the remaining positions too

So, would that mean that I could arrange the numbers themselves in 3! ways resulting in 8*10*1 * 3! = 480 for the option a?
I still see that my numbers are way off the expected solution..

Thanks
Bullzi

Like this post

Bullzi: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Thu May 28, 2015 1:13 am; Thanked: 5 times

by Bullzi » Thu Sep 24, 2015 9:35 am

Alright, another try at getting it all right..
I figured that b and c are 8+3 = 11 (the two 7s and the third digit can be arranged 3 different ways) and 1
That leaves a as 80 * 3, however, I am unable to explain myself why! Why isn't a 80 * 3!

Thanks
Bullzi

Like this post

Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Thu Sep 24, 2015 9:56 am

Bullzi wrote:Alright, another try at getting it all right..
I figured that b and c are 8+3 = 11 (the two 7s and the third digit can be arranged 3 different ways) and 1
That leaves a as 80 * 3, however, I am unable to explain myself why! Why isn't a 80 * 3!

Thanks
Bullzi

One trick to most counting questions is how to correct for overcounting.

For instance, suppose I just count all the two digit numbers with a 7 in them. If I only consider the units place, I have 17, 27, 37, 47, 57, 67, 77, 87, 97. If I only consider the tens place, I have 70, 71, 72, 73, 74, 75, 76, 77, 78, and 79.

So there should be 10 + 9 such numbers, right? But we've OVERCOUNTED 77: it appeared on both lists, yet we only want to count it once.

Like this post

jervizeloy: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Mon Aug 08, 2016 3:05 pm

by jervizeloy » Tue Aug 09, 2016 9:52 am

Matt@VeritasPrep wrote:
Bullzi wrote:Alright, another try at getting it all right..
I figured that b and c are 8+3 = 11 (the two 7s and the third digit can be arranged 3 different ways) and 1
That leaves a as 80 * 3, however, I am unable to explain myself why! Why isn't a 80 * 3!

Thanks
Bullzi
One trick to most counting questions is how to correct for overcounting.

For instance, suppose I just count all the two digit numbers with a 7 in them. If I only consider the units place, I have 17, 27, 37, 47, 57, 67, 77, 87, 97. If I only consider the tens place, I have 70, 71, 72, 73, 74, 75, 76, 77, 78, and 79.

So there should be 10 + 9 such numbers, right? But we've OVERCOUNTED 77: it appeared on both lists, yet we only want to count it once.

Guys, just to make sure I have clear concepts I'm solving this problem considering the restrictions. For doing so, I'm calculating 3-digit numbers with 1 seven, 2 sevens and 3 sevens. However, I'm not getting the same result. How would you guys do it?

Like this post

GMAT/MBA Expert

[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members; Location: Palo Alto, CA; Thanked: 2867 times; GMAT Score:800

by [email protected] » Tue Aug 09, 2016 7:52 pm

Hi jervizeloy,

Since you have not shown any work, it's unclear what mistake(s) you might be making. As a basis for comparison though, I'm going to start the calculation off for you, and you can finish it.

3-digit numbers with JUST one 7...

First digit is the 7: (1)(9)(9).... since the last two digits can be 0, 1, 2, 3, 4, 5, 6, 8, or 9.... but not 7.
Second digit is the 7: (8)(1)(9).... the sine first digit cannot be 0 or 7
Third digit is the 7: (8)(9)(1)

Now, how would the calculations appear if TWO of the digits were 7s?

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

Contact Rich at [email protected]

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

559 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review