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:

Of the three digits greater than 700, how many have two digi [#permalink]
20 Nov 2009, 10:07

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

56% (03:08) correct
44% (01:46) wrong based on 106 sessions

Of the three digits greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two

A) 90 B) 82 C) 80 D) 45 E) 36

Is there a faster method to find out the answer, because i listed down all the numbers and it took me hardly 1.5 mins. but what if there are more numbers ?

Re: Number system (OG 11 ) [#permalink]
20 Nov 2009, 10:28

Expert's post

1

This post was BOOKMARKED

gurpreet07 wrote:

Of the three digits greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two

A) 90 B) 82 C) 80 D) 45 E) 36

Is there a faster method to find out the answer, because i listed down all the numbers and it took me hardly 1.5 mins. but what if there are more numbers ?

Step by step:

The number can have three forms:

XXY meaning that the first digit is the repeated digit; XYX meaning that the first digit is repeated digit; and YXX meaning that the first digit is not repeated digit.

XXY --> the repeated digits can be 7,8 or 9, so 3 possibilities and the third one can take 9 possible values total 3*9=27.

The same with XYX =3*9=27;

YXX --> the non repeated first digit can be 7, 8, 9, so 3 possibilities and the repeated digits can take 9 possible values, total 3*9=27. But this will give us the number 700 as well and we know that the number should be more than 700, hence 27-1=26.

Re: Number system (OG 11 ) [#permalink]
20 Nov 2009, 10:43

Bunuel wrote:

gurpreet07 wrote:

Of the three digits greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two

A) 90 B) 82 C) 80 D) 45 E) 36

Is there a faster method to find out the answer, because i listed down all the numbers and it took me hardly 1.5 mins. but what if there are more numbers ?

Step by step:

The number can have three forms:

XXY meaning that the first digit is the repeated digit; XYX meaning that the first digit is repeated digit; and YXX meaning that the first digit is not repeated digit.

XXY --> the repeated digits can be 7,8 or 9, so 3 possibilities and the third one can take 9 possible values total 3*9=27.

The same with XYX =3*9=27;

YXX --> the non repeated first digit can be 7, 8, 9, so 3 possibilities and the repeated digits can take 9 possible values, total 3*9=27. But this will give us the number 700 as well and we know that the number should be more than 700, hence 27-1=26.

Re: Number system (OG 11 ) [#permalink]
27 Mar 2010, 17:29

Bunuel,

I don't understand how there is only 27 numbers based on your calculation. For example, XXY the first number has to be 7, 8 or 9 so 3 possibilities and the next number would also have to be 7, 8 or 9 but the last number could be 9 values but how is that equal to 3*9? Wouldn't that be 3*3*9? I'm don't understand.

Re: Number system (OG 11 ) [#permalink]
10 Apr 2010, 03:55

adamsmith2010 wrote:

Bunuel,

I don't understand how there is only 27 numbers based on your calculation. For example, XXY the first number has to be 7, 8 or 9 so 3 possibilities and the next number would also have to be 7, 8 or 9 but the last number could be 9 values but how is that equal to 3*9? Wouldn't that be 3*3*9? I'm don't understand.

Adam in case of XXY, first digit is 7,8 or 9 then so 3 ways 2nd digit has be same as 1st so only one way you can also write it as 3*1*9 = 27

Re: Number system (OG 11 ) [#permalink]
14 Apr 2010, 02:10

How have you guys become so confident, predicting :

700-999

XXY setup = t occurrences without a single calculation line etc...

Isnt there any mathematical way of solving it. _________________

------------------------------------------------------------------------- Ros. Nice Post + Some help + Lucid solution = Kudos

The greatest pleasure in life is doing what people say you cannot do | Great minds discuss ideas, average minds discuss events, small minds discuss people. -------------------------------------------------------------------------

Re: Number system (OG 11 ) [#permalink]
14 Apr 2010, 09:24

1

This post received KUDOS

Expert's post

shrivastavarohit wrote:

How have you guys become so confident, predicting :

700-999

XXY setup = t occurrences without a single calculation line etc...

Isnt there any mathematical way of solving it.

This is how I solved it:

Three digit number can have only following 3 patterns: A. all digits are distinct; B. two are alike and third is different; C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299. A. all digits are distinct = 3(first digit can have only three values 7,8, or 9)*9*8=216; C. all three are alike = 3 (777, 888, 999)

So, 299-216-3=80.

Another way to solve this problem is in my first post.

Re: number property [#permalink]
19 Jun 2010, 21:30

I am getting 78 as the answer. So I went with closest answer 80 (C).

This is how I arrived @ 78:

For numbers greater than 700 we have following possibilities:

a=> 7xx .. Here x being any digit 0-9, excluding 7 and 0 {0 is excluded because number should be >700} b=> 7x7 .. Here x can be any digit from 0-9, excluding 7. {no 3 digits should be same} c=> 77x .. Here x can be any digit from 0-9, excluding 7 for same reason as above.

a + b + c = 8 + 9 + 9 = 26

Hence same is true with 8xx , 8x8, 88x and 9xx,9x9, 99x ...

So summing up I got 78 ...

Please help me if I am wrong here. I am unable to figure out where did I miss 2 more numbers. Any pointers ?

Re: Number system (OG 11 ) [#permalink]
20 Jun 2010, 09:45

Bunuel wrote:

gurpreet07 wrote:

Of the three digits greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two

A) 90 B) 82 C) 80 D) 45 E) 36

Is there a faster method to find out the answer, because i listed down all the numbers and it took me hardly 1.5 mins. but what if there are more numbers ?

Step by step:

The number can have three forms:

XXY meaning that the first digit is the repeated digit; XYX meaning that the first digit is repeated digit; and YXX meaning that the first digit is not repeated digit.

XXY --> the repeated digits can be 7,8 or 9, so 3 possibilities and the third one can take 9 possible values total 3*9=27.

The same with XYX =3*9=27;

YXX --> the non repeated first digit can be 7, 8, 9, so 3 possibilities and the repeated digits can take 9 possible values, total 3*9=27. But this will give us the number 700 as well and we know that the number should be more than 700, hence 27-1=26.

TOTAL=27+27+26=80

Answer: C.

Hope it helps.

thanks! this is an efficient way to solve it. is this method feasible for this kind of questions--fundamental counting?

Re: Number system (OG 11 ) [#permalink]
30 Jun 2010, 16:41

Thanks Bunuel for the effort but I dont really understand why you choose 299, and what is the meaning of distinct, and why is there 3, 9 , 8? Pleas explaing that to me thanks again

Re: Number system (OG 11 ) [#permalink]
30 Jun 2010, 18:44

xmagedo wrote:

Thanks Bunuel for the effort but I dont really understand why you choose 299, and what is the meaning of distinct, and why is there 3, 9 , 8? Pleas explaing that to me thanks again

The question asks about three-digit integers greater than 700. So, all the numbers in the range 701 to 999, inclusive. There will be 999 - 701 + 1 = 299 such numbers in this range. (Whenever you want to count the number of objects in a range, subtract the smallest number from the greatest and add 1. For example, the number of integers in the range of 1 to 6 is 6 -1 +1 = 6).

"distinct" just means different. So, for example, in the number "799", the final two digits are NOT distinct. Here: 798, they are. _________________

Re: Of the three digits greater than 700, how many have 2 digit [#permalink]
01 Feb 2014, 11:28

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. _________________

Re: Of the three digits greater than 700, how many have 2 digit [#permalink]
01 Feb 2014, 19:42

We can fix 2 digits in 3 ways ie 7,8,9 and the third digit can be fixed in 9 ways(10 digits-1 repeating digit) Total=3*9=27 Since this is possible in 3 different ways(first digit diff from other two,second digit diff from other two and third digit diff from other two) we get 3* 27=81 but we subtract 1 for 700 Ans 80

Posted from my mobile device

gmatclubot

Re: Of the three digits greater than 700, how many have 2 digit
[#permalink]
01 Feb 2014, 19:42