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:

Re: What is the sum of all 3 digit positive integers that can be formed [#permalink]

Show Tags

29 Jun 2009, 03:40

1

This post received KUDOS

Is it C .. This is how I did it .. Keeping 1 as Hundred digit .. 158+185=343 -----1 Keeping 5 as Hundred digit .. 518+581=1099-------2 Keeping 8 as Hundred digit .. 815+851=1666-------------3 Adding 1+2+3 = 3108 ..
_________________

Re: What is the sum of all 3 digit positive integers that can be formed [#permalink]

Show Tags

29 Jun 2009, 03:49

20

This post received KUDOS

4

This post was BOOKMARKED

There can be 27 numbers. Every digit has three possibilities because question permits repetition. So it becomes a very good summation problem. If there are 27 digits. There are 27 hundred digits, 27 tens digit and 27 ones digit to be summed. Of 27 hundred digits, 9 of them are 1, 9 of them are 5 and the last 9 are 8. Same is true for the other two digits. Thus the sum is: 9x100+9x500+9x800+9x10+9x50+9x80+9x1+9x5+9x8= 999x(1+5+8)=999x14=13986 E

Re: What is the sum of all 3 digit positive integers that can be formed [#permalink]

Show Tags

29 Jun 2009, 06:01

1

This post received KUDOS

Since problem permits repetition. There are 27 numbers that satisfies. e.g.: Lets say: first digit is 1, then numbers can be: 111 115 118 151 155 158 181 185 188 Same is true when first digits are 5 and 8. As you can see, there are nine 1 in first digit. Nine 5 in first digit. And nine 8 in first digit. Same is true for the other digits.

Re: What is the sum of all 3 digit positive integers that can be formed [#permalink]

Show Tags

29 Jun 2009, 13:08

2

This post received KUDOS

maliyeci wrote:

Since problem permits repetition. There are 27 numbers that satisfies. e.g.: Lets say: first digit is 1, then numbers can be: 111 115 118 151 155 158 181 185 188 Same is true when first digits are 5 and 8. As you can see, there are nine 1 in first digit. Nine 5 in first digit. And nine 8 in first digit. Same is true for the other digits.

One more tip: when we add up all the numbers, we can start with the hundreds. We know each number (1, 5, & 8) will appear in the hundreds place a total of 9 times. So let's see how many hundreds we have. 1X9=9 5X9=45 8X9=72 Add this up we have a total of 126 hundreds, or also expressed as 12,600. We see that only one answer could possibly match the size of this sum, which is (E), so without calculating the exact sum, we already know (E) is the only possible choice. On a real test however, the writers could make life difficult by adding a few answer choices that are close to this sum (i.e. 11,950, or 14,088, etc).

Re: What is the sum of all 3 digit positive integers that can be formed [#permalink]

Show Tags

29 Jun 2009, 21:30

maliyeci wrote:

There can be 27 numbers. Every digit has three possibilities because question permits repetition. So it becomes a very good summation problem. If there are 27 digits. There are 27 hundred digits, 27 tens digit and 27 ones digit to be summed. Of 27 hundred digits, 9 of them are 1, 9 of them are 5 and the last 9 are 8. Same is true for the other two digits. Thus the sum is: 9x100+9x500+9x800+9x10+9x50+9x80+9x1+9x5+9x8= 999x(1+5+8)=999x14=13986 E

Now these are my early days here. I even have some problems to use this site When I saw this squestion, I thought how is it possible to do that, to add twenty seven numbers, but thanks to maliyeci!!

I learned a new approach today to add numbers!! +1
_________________

Re: What is the sum of all 3 digit positive integers that can be formed [#permalink]

Show Tags

29 Jun 2009, 22:29

1

This post received KUDOS

maliyeci wrote:

There can be 27 numbers. Every digit has three possibilities because question permits repetition. So it becomes a very good summation problem. If there are 27 digits. There are 27 hundred digits, 27 tens digit and 27 ones digit to be summed. Of 27 hundred digits, 9 of them are 1, 9 of them are 5 and the last 9 are 8. Same is true for the other two digits. Thus the sum is: 9x100+9x500+9x800+9x10+9x50+9x80+9x1+9x5+9x8= 999x(1+5+8)=999x14=13986 E

Great ! Kudos to you.

Another approach is intelligent guess, based on which I would have opted E. Explanation: Total possibilities = 3*3*3 =27 Now, taking examples of numbers starting with 8. Sum of any four 3-digit numbers starting with 8 > 3200,

We know that there are 9 possible nos starting with 8 (apart form other 18 numbers), so sum would certainly be much much greater then 3200.

All other options, except E is less then 3200. (Btw, one can eliminate A, B and D on the 1st glance itself)
_________________

(1+5+8)/3 - "average" digit. (1+5+8)/3 * 111 - another way to write 3-digit number formed from "average digit": xyz = (1+5+8)/3 (1+5+8)/3 (1+5+8)/3 or (1+5+8)/3 * 111
_________________

Re: What is the sum of all 3 digit positive integers that can be formed [#permalink]

Show Tags

24 Nov 2011, 10:36

maliyeci wrote:

There can be 27 numbers. Every digit has three possibilities because question permits repetition. So it becomes a very good summation problem. If there are 27 digits. There are 27 hundred digits, 27 tens digit and 27 ones digit to be summed. Of 27 hundred digits, 9 of them are 1, 9 of them are 5 and the last 9 are 8. Same is true for the other two digits. Thus the sum is: 9x100+9x500+9x800+9x10+9x50+9x80+9x1+9x5+9x8= 999x(1+5+8)=999x14=13986 E

Great explanation .. +1 kudos.
_________________

If Electricity comes from Electrons , Does Morality come from Morons ??

If you find my post useful ... then please give me kudos ......

Re: What is the sum of all 3 digit positive integers that can be formed [#permalink]

Show Tags

10 Nov 2015, 08:07

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: What is the sum of all 3 digit positive integers that can be formed [#permalink]

Show Tags

26 Nov 2015, 23:06

3

This post received KUDOS

Used POE, No need to solve the whole question to get an exact value Explanation:

since three digit numbers formed by 1, 5, 8 would be : Lets start with numbers starting with 8 : 888, 885, 881, 855, 851, 815, 811 sum of these numbers is greater than Options A , B , C , D Hence Ans: E.

There's a great 'pattern-matching' shortcut built into this question that can help you to avoid much of the 'math work' involved.

We're told to use the digits 1, 5 and 8 to form every possible 3-digit number (including those with duplicate digits) and then take the sum of those numbers.

Since the digits can be repeated, we're dealing with the numbers that fall into the range of 111 to 888, inclusive. There are (3)(3)(3) = 27 total numbers and 1/3 of those numbers will begin with an 8. From THAT deduction, we know that the sum of those 9 numbers will be greater than (9)(800) = 7200. There's only one answer that fits that description...

Re: What is the sum of all 3 digit positive integers that can be formed [#permalink]

Show Tags

28 Dec 2015, 13:26

With 1 in the hundreds place, there are 9 numbers that could be created, since repetition is allowed. So 1 occurs 9 times in hundreds place = 900 Similarly it occurs 9 times in tens place = 90 and 9 times in ones place = 9

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...