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There are 27 different threedigit integers that can be formed using [#permalink]
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13 Jan 2011, 19:58
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There are 27 different threedigit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be? A. 2,704 B. 2,990 C. 5,404 D. 5,444 E. 5,994
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Last edited by Bunuel on 16 Aug 2015, 15:30, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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Re: There are 27 different threedigit integers that can be formed using [#permalink]
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14 Jan 2011, 01:00
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here you go: 27 digits are formed by 3 * 3 *3 that is..1st place can take 1,2,3...again 2nd place can take all three digits...3rd place can also take all three digits.
now lets consider the last digit as 1 out of the three that we can take:
3*3*(1)
so with 1 as the last digit we can have 9 combinations.so we have 9 1's in the last place
similary,lets consider the last digit as 2 out of the three that we can take:
3*3*(2)
so with 2 as the last digit we can have 9 combinations.so we have 9 2's in the last place
similary,lets consider the last digit as 3 out of the three that we can take:
3*3*(3)
so with 3 as the last digit we can have 9 combinations.so we have 9 3's in the last place
therefore sum of 9 1's,9 2's,9 3's will be 9+18+27=54...similarly 2nd and 1st place will give u 54...therefore 5994..hope u got it!!



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Re: There are 27 different threedigit integers that can be formed using [#permalink]
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14 Jan 2011, 01:02
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Probably the easiest and fastest way to answer this problem is to think about the average three digit number you can create using just the digits 1,2,3, and to multiply this average number with 27.
The average three digit number is 222 (choosing an average of 1,2,3 for every single digit of the number). So 222*27 = 5994. Hence, solution A is correct.



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Re: There are 27 different threedigit integers that can be formed using [#permalink]
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07 Aug 2015, 03:53
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The simplest way is this formula n^n1 *sum of the numbers (111)= 5994. where n= total number, which is 3 here. so 3^31=9. 9+6(111)=5994



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Re: There are 27 different threedigit integers that can be formed using [#permalink]
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18 Dec 2015, 13:41
swaraj wrote: here you go: 27 digits are formed by 3 * 3 *3 that is..1st place can take 1,2,3...again 2nd place can take all three digits...3rd place can also take all three digits.
now lets consider the last digit as 1 out of the three that we can take:
3*3*(1)
so with 1 as the last digit we can have 9 combinations.so we have 9 1's in the last place
similary,lets consider the last digit as 2 out of the three that we can take:
3*3*(2)
so with 2 as the last digit we can have 9 combinations.so we have 9 2's in the last place
similary,lets consider the last digit as 3 out of the three that we can take:
3*3*(3)
so with 3 as the last digit we can have 9 combinations.so we have 9 3's in the last place
therefore sum of 9 1's,9 2's,9 3's will be 9+18+27=54...similarly 2nd and 1st place will give u 54...therefore 5994..hope u got it!! This is a great explanation! Thanks!



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Re: There are 27 different threedigit integers that can be formed using [#permalink]
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18 Dec 2015, 14:01
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How I did it:
123 is a possible solution, which is divisible by 3. E is the only option that adds up to a number divisible by 3.



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Re: There are 27 different threedigit integers that can be formed using [#permalink]
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21 Feb 2017, 02:45
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rdhookar wrote: There are 27 different threedigit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be?
A. 2,704 B. 2,990 C. 5,404 D. 5,444 E. 5,994 Since every digit has equal likely chances of appearing at any of the three places in a 3 digit number so out of 27 number 9 will have Unit digit 1, 9 will have unit digit 2 and 9 will have unit digit 3 So sum of the unit digits of all 27 numbers = 9(1+2+3) = 54 We write 4 at the unit digit place and 5 goes carry forward so out of 27 number 9 will have Tens digit 1, 9 will have Tens digit 2 and 9 will have Tens digit 3 So sum of the Tens digits of all 27 numbers = 9(1+2+3) = 54 but 5 is carry forward so the sum becomes 54+5 = 59, We write 9 at the tens digit place and 5 goes carry forward so far the sum looks like _ _ 9 4This much calculation gives us the answer out of given options however we can do the same thing two more times to find exact sum which will be 5,994Answer: option E
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Re: There are 27 different threedigit integers that can be formed using [#permalink]
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22 Feb 2017, 11:58
Hi All, This question has a great series of builtin 'logic shortcuts' that don't require a lot of advanced math (although a bit of arithmetic will still be required to get to the correct answer). We're told that there are 27 3digit numbers that can be formed with the digits 1, 2 and 3 (the smallest being 111 and the largest being 333). Without too much effort, you should be able to figure out that... 9 of the numbers begin with a '1' 9 of the numbers begin with a '2' 9 of the numbers begin with a '3' So when we add up JUST the 100s, we have.... 9(100) + 9(200) + 9(300) = 5400 Next, consider ALL of the 10s that we have... 27 of them (and some are 20s and some are 30s). That would add HUNDREDS (well over 270+) to the total sum. Based on how the answer choices are written, there's only one answer that makes sense... Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: There are 27 different threedigit integers that can be formed using [#permalink]
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27 Feb 2018, 10:09
rdhookar wrote: There are 27 different threedigit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be?
A. 2,704 B. 2,990 C. 5,404 D. 5,444 E. 5,994 Some of the integers that have this property are 123, 111, 213, and 322. Of course, it’s possible to list all 27 such integers and then add them up. However, it will be too timeconsuming. Therefore, we will use a shortcut by arguing that, of these 27 integers, the digits 1, 2, and 3 each must appear in the hundreds position 9 times. Using the same logic, they each will also appear in the tens position 9 times and in the units position 9 times. Thus, the sum of these integers is: (100 + 200 + 300) x 9 + (10 + 20 + 30) x 9 + (1 + 2 + 3) x 9 (600) x 9 + (60) x 9 + (6) x 9 (666) x 9 5,994 Answer: E
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Re: There are 27 different threedigit integers that can be formed using [#permalink]
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06 Mar 2018, 04:33
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Sum of the numbers should be divisible by 9 as each number is appearing 9 times. Divisibility rule by 9 conforms(Sum of all the digits should be divisible by 9) only to the answer option E  hence answer should be E.
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Re: There are 27 different threedigit integers that can be formed using
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