The addition problem shown above shows five of the 27 different three-

Question

111
112
113
121
...
+333
________

The addition problem shown above shows five of the 27 different three-digit integers which can be formed using only the integers 1, 2, and 3. What is the sum of these 27 integers?

A. 1,998
B. 2,700
C. 3,996
D. 5,994
E. 6,660

ankitgoswami · Answer

Out of these 27 different three-digit integers, 9 each will have 1,2 & 3 as their unit's digit. So the sum of the unit's digits will be (9*3) + (9*2) + (9*1) = 54

Similarly, the sum of ten's & hundred's digits will also be 54.

So, adding all these three together, we'll get the sum of these 27 integers i.e. 5994.

Ans is  D

bhima22 · Answer

The information "of the 27 different three-digit integers" helps a lot in navigating this questions.

That means, there would be 9 3-digit integers ending with "1", 9 3-digit integers ending with "2", and 9 3-digit integers integers ending with "3"

Therefore, the number digit will be (1*9) + (2*9) + (3*9) = 54 (we only need to know the number digit)

The only answers with a number digit of 4 is D.

RiyaR · Answer

why do we not use the n-1! formula for this question?

whatsarc · Answer

Smallest of them is 111
Largest of them is 333
Mean is 222

The distance of 333 from 222 and 111 from 222 will be the same. Mean is also the median. This is an AP

There are 27 terms, 13 on either side of the median. 
222 x 27 = 5994. 
Ans D

chetan2u · Answer

This is not an AP as in any AP, the common difference between any two consecutive terms is the same.

Here, the series would be 111, 112, 113, 121...... Thus not an AP.

You can not calulate mean the way you have calculated.
Actually, you have taken the property of average of extremes as mean, (111+333)/2, a property of any AP, so you start with the basis of the sequence being AP at the very beggining.

whatsarc · Answer

OH! I realised my mistake. Thank you for pointing it out.

Dheeraj_079 · Answer

Each of the three digits 1,2,3 appears 9 times in the units, tens and hundres place respectives.
So, the total would be 9 * ( 1+2+3) + 9* (1+2+3)*10 + 9*(1+2+3)*100 = 5994
Option D

The addition problem shown above shows five of the 27 different three-

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The addition problem shown above shows five of the 27 different three-

Prep Toolkit

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