Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Guys,

If I count numbers between 11 to 99, I can find it as 99-11+1 = 89. So, there are 89 numbers between 11 and 99 (both inclusive).

Is there a way to get the answer through counting methods. What I mean is that to calculate the numbers between 11 and 99, we have :

the unit digit changes from 1 - 9 , so a total of 9 ways I can fill the unit digit. I can also fill the tens digit in 9 ways. So, the total numbers between 11 and 99 = 9*9 = 81. This is not the correct answer. This is because I missed 20,30,40,50,60,70 and 80. How would I include these missed numbers (20,30..80) in my above calculation, so that I can find the answer just by multiplying numbers?

If I count numbers between 11 to 99, I can find it as 99-11+1 = 89. So, there are 89 numbers between 11 and 99 (both inclusive).

Is there a way to get the answer through counting methods. What I mean is that to calculate the numbers between 11 and 99, we have :

the unit digit changes from 1 - 9 , so a total of 9 ways I can fill the unit digit. I can also fill the tens digit in 9 ways. So, the total numbers between 11 and 99 = 9*9 = 81. This is not the correct answer. This is because I missed 20,30,40,50,60,70 and 80. How would I include these missed numbers (20,30..80) in my above calculation, so that I can find the answer just by multiplying numbers?

thanks sanjay

Yes, you can use counting methods:

-there are 9 choices for the tens' digit (1-9) -there are 10 choices for the units' digit (0-9) -we therefore have 90 possible two digit numbers from 10 to 99 -from 11 to 99, we need to subtract one, because we don't want to count the number 10. We thus have 89 numbers between 11 and 99 (inclusive).

It is not straightforward, however, to adapt this method to other ranges of numbers, so the alternative approach you mentioned (largest - smallest + 1) is normally preferable. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

If I count numbers between 11 to 99, I can find it as 99-11+1 = 89. So, there are 89 numbers between 11 and 99 (both inclusive).

Is there a way to get the answer through counting methods. What I mean is that to calculate the numbers between 11 and 99, we have :

the unit digit changes from 1 - 9 , so a total of 9 ways I can fill the unit digit. I can also fill the tens digit in 9 ways. So, the total numbers between 11 and 99 = 9*9 = 81. This is not the correct answer. This is because I missed 20,30,40,50,60,70 and 80. How would I include these missed numbers (20,30..80) in my above calculation, so that I can find the answer just by multiplying numbers?

thanks sanjay

Yes, you can use counting methods:

-there are 9 choices for the tens' digit (1-9) -there are 10 choices for the units' digit (0-9) -we therefore have 90 possible two digit numbers from 10 to 99 -from 11 to 99, we need to subtract one, because we don't want to count the number 10. We thus have 89 numbers between 11 and 99 (inclusive).

It is not straightforward, however, to adapt this method to other ranges of numbers, so the alternative approach you mentioned (largest - smallest + 1) is normally preferable.

Ian, thanks for your inputs. You are right that counting method is not a preferable method in that it can involve lengthy calculations. However, y'day I came across a question on gmatclub's test m01. The question is :

How many integers between 324,700 and 458,600 have a 2 in the tens digit and a 1 in the units digit?

For this question, we need to count numbers from 324721 to 458521, which is basically the same as counting numbers from 3247 to 4586. This should be easy if one spots this pattern.

However, if I just tackle the question without giving too much thought to the pattern, it's going to be a nightmare trying to find the answer. Taking a look at these different digits :

hundredth digit varies from 7 to 6. - 10 ways thousandth digit varies from 4 to 8. - 5 ways ten thousandth digit varies from 2 to 5 - 14 ways hundred thousandth digit varies from 3 to 4 - 2 ways .

total number of ways = 10*5*14*2 = 1400. However, this is not the right answer. Could you please point out the flaw in my calculations?

Tricky problem..I do the following way. Consider all numbers which are between 324,700 and 399999 with the given condition: First digit: 3 = 1 Second: 2-9 = 8 Third: 4-9 = 6 Fourth: 7-9 = 3 Fifth: 2 = 1 Sixth : 1 = 1 Total possibilities = 1*8*6*3*1*1 = 144

Consider all numbers which are between 400000 and 458600 with the given condition: First digit: 4 = 1 Second: 0-5 = 6 Third: 0-8 = 9 Fourth: 0-6 = 7 Fifth: 2 = 1 Sixth : 1 = 1 Total possibilities = 1*6*9*7*1*1 = 378

How many integers between 324,700 and 458,600 have a 2 in the tens digit and a 1 in the units digit?

For this question, we need to count numbers from 324721 to 458521, which is basically the same as counting numbers from 3247 to 4586. This should be easy if one spots this pattern.

However, if I just tackle the question without giving too much thought to the pattern, it's going to be a nightmare trying to find the answer. Taking a look at these different digits :

hundredth digit varies from 7 to 6. - 10 ways thousandth digit varies from 4 to 8. - 5 ways ten thousandth digit varies from 2 to 5 - 14 ways hundred thousandth digit varies from 3 to 4 - 2 ways .

total number of ways = 10*5*14*2 = 1400. However, this is not the right answer. Could you please point out the flaw in my calculations?

It's because of questions like this that I said above: "It is not straightforward, however, to adapt this method to other ranges of numbers, so the alternative approach you mentioned (largest - smallest + 1) is normally preferable."

You can't easily apply the multiplication principle from counting to the problem above, without breaking the problem into several cases. The number of choices, for example, for the 'thousands' digit is not 5; it can be anything from 0 to 9, since 330,721 is, for example, a valid number here, as is 339,721. The problem here is that the number of choices you have, say, for the ten thousands' digit depends on what you choose for the hundred thousands' digit. If you choose a '3' for the hundred thousands' digit, you have eight choices for the ten thousands' digit, and if you choose a '4' for the hundred thousands', you only have six choices for the ten thousands' digit. If you really want to use the product rule to count here, you can only resolve this by considering these two cases separately. You'll then discover at each stage you have to consider different cases, and the problem becomes a bit of a mess. It's not a good approach to solving the problem. That said, you've already noticed a good method here, so I'm not sure why you're looking for another one:

sanjay_gmat wrote:

For this question, we need to count numbers from 324721 to 458521, which is basically the same as counting numbers from 3247 to 4586. This should be easy if one spots this pattern.

_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com