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Re: There are 27 different three-digit integers that can be formed using [#permalink]
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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 three-digit integers that can be formed using [#permalink]
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The simplest way is this formula n^n-1 *sum of the numbers (111)= 5994. where n= total number, which is 3 here. so 3^3-1=9. 9+6(111)=5994
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Re: There are 27 different three-digit integers that can be formed using [#permalink]
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 three-digit integers that can be formed using [#permalink]
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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 three-digit integers that can be formed using [#permalink]
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Hi All,

This question has a great series of built-in '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 3-digit 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:

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Re: There are 27 different three-digit integers that can be formed using [#permalink]
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rdhookar wrote:
There are 27 different three-digit 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 time-consuming. 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 three-digit integers that can be formed using [#permalink]
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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 three-digit integers that can be formed using [#permalink]
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rdhookar wrote:
There are 27 different three-digit 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


SUPER EASY METHOD that always works

STEP 1 always find the total no. of combinations possible = here we have been given 27 (use fundamental method to check/find 3x3x3 =27)
STEP 2 - find the average of lowest and highest number ( 123 + 321 = 444 . 444/2 = average 222)
STEP 3- multiply the average with the possible numbers . = 27 * 222 = 5994 . :)
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Re: There are 27 different three-digit integers that can be formed using [#permalink]
at ones place there will be nine 1 , nine 2 and nine 3 , adding all 27 will give you 54 . silmilarly you will get 54 at tens place and 54 at hundred place . all all we get e as answer
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Re: There are 27 different three-digit integers that can be formed using [#permalink]
rdhookar wrote:
There are 27 different three-digit 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


111+333=444
444/2=222
222*27=5994
E
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Re: There are 27 different three-digit integers that can be formed using [#permalink]
I think there is a simple formula that can be applied here.

My only confusion is regarding how we can figure out whether or not the question implies repetition of numbers.

The formula without the repetition gives the answer of course, but can we determine that for sure from the question?
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Re: There are 27 different three-digit integers that can be formed using [#permalink]
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salphonso wrote:
I think there is a simple formula that can be applied here.

My only confusion is regarding how we can figure out whether or not the question implies repetition of numbers.

The formula without the repetition gives the answer of course, but can we determine that for sure from the question?


Hi salphonso,

The prompt specifically tells us that we're dealing with "27 DIFFERENT three-digit integers...", so we know that the numbers do NOT repeat. GMAT questions are always carefully worded, so you should pay attention to the specific 'math vocabulary' that is used in each prompt (as that can help you to properly 'restrict' your thinking or get you thinking about possibilities that are not immediately obvious to you).

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Re: There are 27 different three-digit integers that can be formed using [#permalink]
Hi Rich,

Can't believe I missed that! Thank you!
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There are 27 different three-digit integers that can be formed using [#permalink]
Can we use this method?

1 number would be 111 and last would be 333.
Their average will be 222.
Using the average method- 222*27 is 5994.

Bunuel

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Re: There are 27 different three-digit integers that can be formed using [#permalink]
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rdhookar wrote:
There are 27 different three-digit 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


This is a great example of how the GMAT often rewards students for thinking outside the box.

Here, we can apply a divisibility rule that says integer N is divisible by 3 if and only if the sum of the digits of N is divisible by 3.
For example, we know that 11112 is divisible by 3, because 1+1+1+1+5=9, and 9 is divisible by 3.

Notice that 1+2+3 = 6, and 6 is divisible by 3
This means any 3-digit integer consisting of a 1, a 2 and a 3 must be divisible by 3
If each of the 27 integers in the sum is divisible by 3, then the sum must also be divisible by 3.
In other words, the correct answer must be divisible by 3, which means the sum of its digits must be divisible by 3.

Let's check each answer choice...
A. 2+7+0+4 = 13, which is not divisible by 3. Eliminate.
B. 2+9+9+0 = 20, which is not divisible by 3. Eliminate.
C. 5+4+0+4 = 13, which is not divisible by 3. Eliminate.
D. 5+4+4+4 = 17, which is not divisible by 3. Eliminate.
E. 5+9+9+4 = 27, which IS divisible by 3.

By the process of elimination, the correct answer is E.
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Re: There are 27 different three-digit integers that can be formed using [#permalink]
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