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If x represents the sum of all the positive three-digit

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VP
Joined: 21 Jul 2006
Posts: 1390
If x represents the sum of all the positive three-digit  [#permalink]

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29 Jul 2008, 05:53
1
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

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Current Student
Joined: 28 Dec 2004
Posts: 3222
Location: New York City
Schools: Wharton'11 HBS'12
Re: GMATprep question  [#permalink]

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29 Jul 2008, 06:45
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

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..

I am going to say the ans is 6
Director
Joined: 27 May 2008
Posts: 523
Re: GMATprep question  [#permalink]

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29 Jul 2008, 07:00
2
1
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)

Answer E
Current Student
Joined: 28 Dec 2004
Posts: 3222
Location: New York City
Schools: Wharton'11 HBS'12
Re: GMATprep question  [#permalink]

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29 Jul 2008, 07:05
1
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)

Answer E
Current Student
Joined: 11 May 2008
Posts: 545
Re: GMATprep question  [#permalink]

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29 Jul 2008, 09:17
DURGESH ...GOOD ONE AS USUAL!!!
Senior Manager
Joined: 23 May 2006
Posts: 286
Re: GMATprep question  [#permalink]

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29 Jul 2008, 16:44
good explanation durgesh. Kudos for you.
VP
Joined: 21 Jul 2006
Posts: 1390
Re: GMATprep question  [#permalink]

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30 Jul 2008, 05:26
OA is E. thanks guys
VP
Joined: 17 Jun 2008
Posts: 1203
Re: GMATprep question  [#permalink]

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03 Aug 2008, 05:04
1
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)

Answer E

Too good
Qhere do u practice quant from
_________________

cheers
Its Now Or Never

VP
Joined: 21 Jul 2006
Posts: 1390
Re: GMATprep question  [#permalink]

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16 Sep 2008, 06:19
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)

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.

Thanks!
Manager
Joined: 22 Jul 2008
Posts: 134
Re: GMATprep question  [#permalink]

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16 Sep 2008, 12:57
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.
VP
Joined: 21 Jul 2006
Posts: 1390
Re: GMATprep question  [#permalink]

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16 Sep 2008, 13:16
KASSALMD wrote:
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?
VP
Joined: 17 Jun 2008
Posts: 1413
Re: GMATprep question  [#permalink]

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18 Sep 2008, 01:06
tarek99 wrote:
KASSALMD wrote:
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.....
VP
Joined: 21 Jul 2006
Posts: 1390
Re: GMATprep question  [#permalink]

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18 Sep 2008, 04:26
scthakur wrote:
tarek99 wrote:
KASSALMD wrote:
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.
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Re: If x represents the sum of all the positive three-digit  [#permalink]

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26 Nov 2018, 07:02
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Re: If x represents the sum of all the positive three-digit &nbs [#permalink] 26 Nov 2018, 07:02
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