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.

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:

Re: NS - Integers!! [#permalink]
14 Aug 2011, 11:42

1

This post received KUDOS

Take numbers between 101 and 199. Numbers that meet criteria 101,111,121,131,141,151,161,171,181,191. Total 10 number between 101 and 199 Similarly, 10 numbers between 200 and 300, 300 and 400, 400 and 500, 500 and 600, 600 and 700, 700 and 800 for total of 70 numbers.

OA C _________________

My dad once said to me: Son, nothing succeeds like success.

Re: NS - Integers!! [#permalink]
14 Aug 2011, 11:49

1

This post received KUDOS

+1 for C. This will be a 3 digit number and it remains same when reversing the digit. So it means 1st and 3rd digit are same. Number of options for 1st and 3rd digit = 7 (1 to 7 inclusive) Number of options for 2nd Digit = 10 (0, 1, to 9)

Total 7 x 10 x 1 = 70 (3rd Digit is already selected when the 1st is selected).

Re: NS - Integers!! [#permalink]
14 Aug 2011, 21:31

1

This post received KUDOS

another way to think of it is like this

I made it 100 just to simplify the below

100 < _ _ _ <= 800 which is the same as 101 <= _ _ _ <= 800

so you have a 3 digit number that must be a palindrome.

you have 7 ways to choose the first number (realize that you can't choose the number 8 because there are no palindromes <= 800) , you have 10 ways to choose the second number, but you only have 1 way to choose the 3rd number, because the third number HAS to be the 1st number

7*10*1 = 70

then 5 digit number you can do the following

10000 < _ _ _ _ _ <= 800000

again you have 7 * 10 * 10 = 700

** the other response was for 80008 so that would be another palindrome that's why that answer is 701

Re: NS - Integers!! [#permalink]
15 Aug 2011, 07:57

Spidy001 wrote:

question is asking for palindrome

first digit possibilities - 1 through 7 = 7

8 is not possible here because it would result in a number greater than 8 (i.e 808 , 818..)

second digit possibilities - 0 though 9 = 10

third digit is same as first digit

=>total possible number meeting the given conditions = 7 *10 = 70

Answer is C.

pinchharmonic wrote:

another way to think of it is like this

I made it 100 just to simplify the below

100 < _ _ _ <= 800 which is the same as 101 <= _ _ _ <= 800

so you have a 3 digit number that must be a palindrome.

you have 7 ways to choose the first number (realize that you can't choose the number 8 because there are no palindromes <= 800) , you have 10 ways to choose the second number, but you only have 1 way to choose the 3rd number, because the third number HAS to be the 1st number

7*10*1 = 70

then 5 digit number you can do the following

10000 < _ _ _ _ _ <= 800000

again you have 7 * 10 * 10 = 700

** the other response was for 80008 so that would be another palindrome that's why that answer is 701

Yups, I like the palindrome approach. Let us remove the unnecessary jargon and say this is a simple approach with little logic and common sense

so we have always 7 numbers vertically and 10 numbers horizontally. so 7*10 =70

actually the answ is predictable. from the beginning u know that u get a set of seven 3-digit numbers (111 ;222; 333 ;*** ;777). so u understand that ur answer choice should be divisible by 7. only 70 is divisible by 7 _________________

Happy are those who dream dreams and are ready to pay the price to make them come true

Re: How many integers from 101 to 800, inclusive, remains the [#permalink]
22 Feb 2013, 22:59

This is how I figured it out: The value will remain same only if all the digits are same OR if the first and the last digits are same. Hence between 101 - 800 : Total number of nos with same digits is 7 i.e. 111, 222, 333, 444, 555, 666, 777

Now for the first and the last digits as same.

since the range is between 101 - 800, Out of the 3 digits the first digit can ( or should ) be between 1 - 7 Total Number of ways in which 7 Numbers can be chosen is 7C1 = 7 ways. This chosen number is same as the 3 rd digit hence the number of ways to chose this same digit is 1 The Middle number can be between 0 - 9 ( exclusive of the number chosen as the first digit ) Therefore, total number of nos is 9: hence total number of ways to choose a number from 9 : 9C1 = 9

Therefore total number of digits is : 9 * 7 = 63 Total number of same 3 digits : 7 ==> 63 + 7 = 70

Re: How many integers from 101 to 800, inclusive, remains the [#permalink]
17 Aug 2014, 03:45

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

It’s been a long time, since I posted. A busy schedule at office and the GMAT preparation, fully tied up with all my free hours. Anyways, now I’m back...

Ah yes. Funemployment. The time between when you quit your job and when you start your MBA. The promised land that many MBA applicants seek. The break that every...

It is that time of year again – time for Clear Admit’s annual Best of Blogging voting. Dating way back to the 2004-2005 application season, the Best of Blogging...