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 350,000 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:

Re: sum of 3 digit #s [#permalink]
29 Apr 2009, 12:26

Expert's post

hi aismirnov, can you elaborated on your explanation a bit. I'm a bit weak with these types of problems. for example, what did you use tens (1+5+8)*9*10, etc.

Re: sum of 3 digit #s [#permalink]
29 Apr 2009, 23:24

10

This post received KUDOS

1

This post was BOOKMARKED

What is the sum of all 3 digit positive integers that can be formed using the digits 1, 5, and 8, if the digits are allowed to repeat within a number?

Imagine, we have got all these possible numbers written down - there are in total 3^3 numbers (each digit can be either 1 or 5 or 8)

there are 3*3 options for having a number XY1 there are 3*3 options for having a number XY5 there are 3*3 options for having a number XY8

there are 3*3 options for having a number X1Z there are 3*3 options for having a number X5Z there are 3*3 options for having a number X8Z

there are 3*3 options for having a number 1YZ there are 3*3 options for having a number 5YZ there are 3*3 options for having a number 8YZ

we can sum units, tens and hundreds independently: summing units gives (1+5+8)*3*3 summing tens gives (1+5+8)*10*3*3 summing hundreds gives (1+5+8)*100*3*3

Re: sum of 3 digit #s [#permalink]
14 Oct 2009, 17:47

2

This post received KUDOS

I don't have your specific method but by POE I still can have E. for 8xy alone we have 888,881,885,818,855,851,858,815,811. The total of them is larger than 7200 so there is only option E left

Re: sum of 3 digit #s [#permalink]
09 Aug 2010, 20:33

this is how i approached

888 588 188 158 518 118 558 These are the number that can have 8 as the units digit. here, 8(7) = 56 so units dig is 6 similarly, for 5 as the units dig - 5(7) = 35 so units dig is 5 for 1 as the units dig - 1(7) = 7 as the unit dig. therfore, in the sum of these dig, the units dig will be 6+5+7 = 8

Re: sum of 3 digit #s [#permalink]
02 Feb 2011, 02:37

5

This post received KUDOS

Expert's post

10

This post was BOOKMARKED

craky wrote:

I sowe really good formula for solving this problem in some notes downloaded from this forum. I just cannot find it, so I appologize to the author.

The formula says:

Repetition allowed: SUM of digits * (n^n-1)*(11111 ...number composed of n 1digits)

Repetition NOT allowed: SUM of digits * (n-1)!*(11111 ...number composed of n 1digits)

Here we have 3 digits. n is 3. Sum of digits 1+5+8=14

Repetition allowed: 14*(3^2)*111=13986

Repetition not allowed: 14*2*111=3108

It should be:

1. Sum of all the numbers which can be formed by using the \(n\) digits without repetition is: \((n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)\).

2. Sum of all the numbers which can be formed by using the \(n\) digits (repetition being allowed) is: \(n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)\).

Re: What is the sum of all 3 digit positive integers that can be [#permalink]
04 Jan 2013, 05:50

I approach this particular problem without the formulae. Can somebody please help me if this is correct --> If you know that the numbers are allowed to repeat then the possible numbers are 3*3*3 = 27 (instead of 3*2*1 when repetition is not allowed), then you know that there will be 9 ones, 9 fives, 9 eights. So for the first position you can have the 9+45+72 = 12600, then all the answer choices will fall except for E. If you calculate further you get 12600 + 01260 + 00126 = 13,986. Bunuel, Karishma or someone else can you please confirm if this is correct?

Re: What is the sum of all 3 digit positive integers that can be [#permalink]
04 Jan 2013, 06:43

4

This post received KUDOS

asimov wrote:

What is the sum of all 3 digit positive integers that can be formed using the digits 1, 5, and 8, if the digits are allowed to repeat within a number?

A. 126 B. 1386 C. 3108 D. 308 E. 13986

One quickest way to answer this question !!

As we are using digits 1,5, 8 and digits are allowed to repeat. Each of the unit, tenth and hundredth digit can be used by each of three digits. So, Total possible numbers with these digits=3 X 3 X 3 =27.

First, As we have 27 three digit number, Sum will be for sure more than 2700.. Eliminate options A,B,D

Second, If you imagine numbers with the given digits 1,5,8. We have numbers like 888,885,855,858,851. Sum is for sure more than 4000. Eliminate option C.

You are left with answer E.

---------------- consider giving a +kudo if this helps

Re: What is the sum of all 3 digit positive integers that can be [#permalink]
05 Jan 2013, 09:40

3

This post received KUDOS

Expert's post

asimov wrote:

What is the sum of all 3 digit positive integers that can be formed using the digits 1, 5, and 8, if the digits are allowed to repeat within a number?

A. 126 B. 1386 C. 3108 D. 308 E. 13986

Answer to this question is easier to guess than to calculate. e.g. if we take 8 at hundreds place we would get at least 9 nos. So 800 * 9 = 7200 which surpasses every option but E.

Re: sum of 3 digit #s [#permalink]
01 Oct 2013, 17:44

aismirnov wrote:

What is the sum of all 3 digit positive integers that can be formed using the digits 1, 5, and 8, if the digits are allowed to repeat within a number?

Imagine, we have got all these possible numbers written down - there are in total 3^3 numbers (each digit can be either 1 or 5 or 8)

there are 3*3 options for having a number XY1 there are 3*3 options for having a number XY5 there are 3*3 options for having a number XY8

there are 3*3 options for having a number X1Z there are 3*3 options for having a number X5Z there are 3*3 options for having a number X8Z

there are 3*3 options for having a number 1YZ there are 3*3 options for having a number 5YZ there are 3*3 options for having a number 8YZ

we can sum units, tens and hundreds independently: summing units gives (1+5+8)*3*3 summing tens gives (1+5+8)*10*3*3 summing hundreds gives (1+5+8)*100*3*3

How did you know how to do this? I mean, how did you learn? I have 4 weeks to go until my GMAT and I haven't gotten any better at these. I have no idea how to even begin approaching these problems, and none of these formulas make any sense to me.

Re: sum of 3 digit #s [#permalink]
02 Oct 2013, 02:24

Expert's post

3

This post was BOOKMARKED

AccipiterQ wrote:

aismirnov wrote:

What is the sum of all 3 digit positive integers that can be formed using the digits 1, 5, and 8, if the digits are allowed to repeat within a number?

Imagine, we have got all these possible numbers written down - there are in total 3^3 numbers (each digit can be either 1 or 5 or 8)

there are 3*3 options for having a number XY1 there are 3*3 options for having a number XY5 there are 3*3 options for having a number XY8

there are 3*3 options for having a number X1Z there are 3*3 options for having a number X5Z there are 3*3 options for having a number X8Z

there are 3*3 options for having a number 1YZ there are 3*3 options for having a number 5YZ there are 3*3 options for having a number 8YZ

we can sum units, tens and hundreds independently: summing units gives (1+5+8)*3*3 summing tens gives (1+5+8)*10*3*3 summing hundreds gives (1+5+8)*100*3*3

How did you know how to do this? I mean, how did you learn? I have 4 weeks to go until my GMAT and I haven't gotten any better at these. I have no idea how to even begin approaching these problems, and none of these formulas make any sense to me.

Re: What is the sum of all 3 digit positive integers that can be [#permalink]
08 Oct 2014, 15:30

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. _________________

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...