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If x represents the sum of all the positive three-digit [#permalink]
29 Jul 2008, 05:53

1

This post was BOOKMARKED

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)

Re: GMATprep question [#permalink]
03 Aug 2008, 05:04

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 _________________

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.

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