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:

If x represents the sum of all the positive three-digit [#permalink]

Show Tags

29 Jul 2008, 05:53

1

This post was BOOKMARKED

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

If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?

a) 3 b) 6 c) 11 d) 22 e)222

how on earth can you solve a problem like this efficiently? thanks

If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?

a) 3 b) 6 c) 11 d) 22 e)222 how on earth can you solve a problem like this efficiently? thanks

i dont know my Q skills anymore..but here is how i would do it..

pick any 3 digit..say 123 or 456 or 345

x=sum of all possible ways of say writng say 124..

my approach is to calculate the sum of all possibility is to just calculate the sum for one of the digit..say for example the 3-digit number is 123 then just focus on the sum of the unit digit..here is an example

123 132 231 213 312 321

sum of unit digit=2(3)+2(2)+2(1)=12

so then the sum of all possible combination would be 1200+120+12=1332 now notice the sum is divisible by 3 and its even..which means its gotta be divisible 6..

If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible? a) 3 b) 6 c) 11 d) 22 e)222 how on earth can you solve a problem like this efficiently? thanks

possible 3 digit number = 100a + 10b + c possible combinations abc acb bca bac cab cba

sum = (100a + 10b + c) + (100a + 10c + b) + ........... = 222a + 222b + 222c --> a is multipled twice by 100, twice by 10 and twice by 1.. simalarly b and c = 222(a+b+c)

yeah..i agree i didnt realize that sum of a=222 b=222 and c=222..

durgesh79 wrote:

tarek99 wrote:

If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible? a) 3 b) 6 c) 11 d) 22 e)222 how on earth can you solve a problem like this efficiently? thanks

possible 3 digit number = 100a + 10b + c possible combinations abc acb bca bac cab cba

sum = (100a + 10b + c) + (100a + 10c + b) + ........... = 222a + 222b + 222c --> a is multipled twice by 100, twice by 10 and twice by 1.. simalarly b and c = 222(a+b+c)

If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible? a) 3 b) 6 c) 11 d) 22 e)222 how on earth can you solve a problem like this efficiently? thanks

possible 3 digit number = 100a + 10b + c possible combinations abc acb bca bac cab cba

sum = (100a + 10b + c) + (100a + 10c + b) + ........... = 222a + 222b + 222c --> a is multipled twice by 100, twice by 10 and twice by 1.. simalarly b and c = 222(a+b+c)

Answer E

Too good Qhere do u practice quant from
_________________

If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible? a) 3 b) 6 c) 11 d) 22 e)222 how on earth can you solve a problem like this efficiently? thanks

possible 3 digit number = 100a + 10b + c possible combinations abc acb bca bac cab cba

sum = (100a + 10b + c) + (100a + 10c + b) + ........... = 222a + 222b + 222c --> a is multipled twice by 100, twice by 10 and twice by 1.. simalarly b and c = 222(a+b+c)

Answer E

Although I really love this approach, is there another way to figure out that we have 2 a's in the hundreds place, then another 2 a's in the tens place, and then 2 a's in the units place? because the only reason we managed to list the 6 permutations above is that we only have 6 of them. What if we had a permutation that is much larger??? On the GMAT, expect anything as such to occur. So if we had a permutation of 24, for example, i'm not gonna list them all down, you know?

So is there a more efficient way to figure out how to add up such a thing without manually listing all the permutation? that will be a big time safer for all of us.

Whatever the no. of objects, they will be evenly placed in all permutations at all spots.What I am saying is that even if there were 6! numbers made from 6 distinct numbers, each number would appear in each of the 6 spots 6!/6 times.

Whatever the no. of objects, they will be evenly placed in all permutations at all spots.What I am saying is that even if there were 6! numbers made from 6 distinct numbers, each number would appear in each of the 6 spots 6!/6 times.

ok, let's say we had a four digit integer abcd. So it's permutation would be 4!, which is 24 distinct integers. How many of each digit would there be then?

Whatever the no. of objects, they will be evenly placed in all permutations at all spots.What I am saying is that even if there were 6! numbers made from 6 distinct numbers, each number would appear in each of the 6 spots 6!/6 times.

ok, let's say we had a four digit integer abcd. So it's permutation would be 4!, which is 24 distinct integers. How many of each digit would there be then?

Looking at durgesh79's approach, my clue is that each of the digits has to evenly appear in all the 24 numbers and since there are four place holders, each of the digits will appear 6 times in each of the placeholders....i.e. 6 times in 1000th place, 6 times in 100th place, 6 times in 10th place and 6 times in units place.

Extending the same logic.....if there are n possible numbers that can be formed using k distince digits, then each of the digits will appear n/k times at units place, n/k times at 10th place, n/k times at 100th place and so on.....

Whatever the no. of objects, they will be evenly placed in all permutations at all spots.What I am saying is that even if there were 6! numbers made from 6 distinct numbers, each number would appear in each of the 6 spots 6!/6 times.

ok, let's say we had a four digit integer abcd. So it's permutation would be 4!, which is 24 distinct integers. How many of each digit would there be then?

Looking at durgesh79's approach, my clue is that each of the digits has to evenly appear in all the 24 numbers and since there are four place holders, each of the digits will appear 6 times in each of the placeholders....i.e. 6 times in 1000th place, 6 times in 100th place, 6 times in 10th place and 6 times in units place.

Extending the same logic.....if there are n possible numbers that can be formed using k distince digits, then each of the digits will appear n/k times at units place, n/k times at 10th place, n/k times at 100th place and so on.....

yeah, based on what you have said, I tried experimenting with different numbers and I came to the same realization. So basically, in order to know the number of times that each digit will appear in each placement, the formula is: n!/n. So for example, in our first example regarding the placement of 3!, the number of times each digit will appear in each placement is 3!/3, which is 2. And in the example that I just posted, it will be 4!/4, which is 6. Cool! thanks a lot.

Re: If x represents the sum of all the positive three-digit [#permalink]

Show Tags

04 Sep 2015, 10:31

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.
_________________