There are 27 different three-digit integers that can be

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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

OA E.

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by fskilnik@GMATH » Tue Nov 06, 2018 6:17 pm
AAPL wrote:GMAT Prep

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
\[? = 111 + 112 + 113 + 121 + 122 + 123 + \ldots + 331 + 332 + 333\]
\[1 \,\, \, \underline {} \, \, \, \underline {} :\,111 + 112 + 113 + 121 + 122 + 123 + 131 + 132 + 133\]
\[\left\{ \begin{gathered}
\,1 \to {\text{9}}\,\,{\text{times}}\,\,{\text{in}}\,\,{\text{hundreds}}\,\,{\text{digit}} \hfill \\
\,1,2,3 \to {\text{3}}\,\,{\text{times}}\,\,{\text{each}}\,\,{\text{in}}\,\,{\text{tens}}\,\,{\text{digit}} \hfill \\
\,1,2,3 \to {\text{3}}\,\,{\text{times}}\,\,{\text{each}}\,\,{\text{in}}\,\,{\text{units}}\,\,{\text{digit}} \hfill \\
\end{gathered} \right.\]
\[{\text{Same}}\,\,{\text{occurs}}\,\,{\text{with}}\,\,2 \,\,\, \underline {} \,\,\, \underline {} \, \,\,\, {\text{and}} \,\,\, \,\,3 \,\,\, \underline {} \,\,\, \underline {} \,\,,\,\,{\text{hence:}}\]
\[{\text{?}}\,\,\,{\text{:}}\,\,\,\left\{ \begin{gathered}
\,1,2,3 \to {\text{9}}\,\,{\text{times}}\,\,{\text{in}}\,\,{\text{hundreds}}\,\,{\text{digit}} \hfill \\
\,1,2,3 \to {\text{3 + 3 + 3}}\,\,{\text{times}}\,\,{\text{each}}\,\,{\text{in}}\,\,{\text{tens}}\,\,{\text{digit}} \hfill \\
\,1,2,3 \to {\text{3 + 3 + 3}}\,\,{\text{times}}\,\,{\text{each}}\,\,{\text{in}}\,\,{\text{units}}\,\,{\text{digit}} \hfill \\
\end{gathered} \right.\]
\[? = 9\left( {1 + 2 + 3} \right) \cdot 100 + 9\left( {1 + 2 + 3} \right) \cdot 10 + 9\left( {1 + 2 + 3} \right)\]
\[? = 9 \cdot 6 \cdot \left( {100+10+1} \right) = 5994\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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by GMATGuruNY » Wed Nov 07, 2018 3:52 am
AAPL wrote:GMAT Prep

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
Given any set that is symmetrical about the median:
Sum = (quantity)(median)

The set of 3-digit integers that can be formed from the digits 1,2 and 3 is symmetrical about the median (222):
...212, 213, 221, 222, 223, 231, 232...

Thus:
Sum of the 27 integers = (quantity)(median) = 27*222 = 5994.

The correct answer is E.

Another approach:

There are 3 positions in each integer: hundreds place, tens place, units place.
Each digit will appear in each position 27/3 = 9 times.
Thus, in each position, there will be nine 1's, nine 2's, and nine 3's.
Sum of the digits in each position = 9(1+2+3) = 54.

Sum for the hundreds place = 54*100 = 5400.
Sum for the tens place = 54*10 = 540.
Sum for the units place = 54*1 = 54.

Sum of all the integers = 5400+540+54 = 5994.

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by fskilnik@GMATH » Wed Nov 07, 2018 7:10 am
GMATGuruNY wrote:
Given any set that is symmetrical about the median:
Sum = (quantity)(median)

The set of 3-digit integers that can be formed from the digits 1,2 and 3 is symmetrical about the median (222):
...212, 213, 221, 222, 223, 231, 232...

Thus:
Sum of the 27 integers = (quantity)(median) = 27*222 = 5994.
Outstanding, Mitch. Congrats!

Regards,
Fabio.
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by Brent@GMATPrepNow » Wed Nov 07, 2018 7:27 am
AAPL wrote:GMAT Prep

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

OA E.
Yes, great work, Mitch!
I thought I'd mention that, once we know the correct answer is a multiple of 222 (which is also a multiple of 3), then the correct answer will be a multiple of 3.
Useful property: If a number is a multiple of 3, then the sum of its digits is also a multiple of 3.
When we use this property to check the answer choices, we see that only E (5994) is a multiple of 3 (since 5 + 9 + 9 + 4 = 27, and 27 is a multiple of 3)

Cheers,
Brent
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by Scott@TargetTestPrep » Wed Nov 07, 2018 6:13 pm
AAPL wrote:GMAT Prep

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 must appear in the hundreds position 9 times each. Using the same logic, they each will also appear in the tens position 9 times each and in the units position 9 times each. 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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