Problem Solving

1,234
1,243
1,324
.....
....
+4,321

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?

A. 24,000
B. 26,664
C. 40,440
D. 60,000
E. 66,660

Using the symmetry in the numbers involved (All formed using all possible combinations of 1,2,3,4), and we know there are 24 of them. We know there will be 6 each with the units digits as 1, as 2, as 3 and as 4. And the same holds true of the tens, hundreds and thousands digit.

The sum is therefore = (1 + 10 + 100 + 1000) * (1*6 +2*6 +3*6 +4*6) = 1111 * 6 * 10 = 66660

Answer: e

OR

Formulas for such kind of problems (just in case):

1. Sum of all the numbers which can be formed by using the nn digits without repetition is: (n−1)!∗(sum of the digits)∗(111... n times)(n−1)!∗(sum of the digits)∗(111... n times)

2. Sum of all the numbers which can be formed by using the nn digits (repetition being allowed) is: nn−1∗(sum of the digits)∗(111... n times)nn−1∗(sum of the digits)∗(111... n times).

For any set that is SYMMETRICAL ABOUT THE MEDIAN:
sum = (count)(median)

rajeet123 wrote:1,234
1,243
1,324
.....
....
+4,321

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exactly once in each integer. What is the sum of these 24 integers?

A. 24,000
B. 26,664
C. 40,440
D. 60,000
E. 66,660

The set is composed of integers in the following ranges:
1234...1432
2134...2431
3124...3421
4123...4321.

Each range contains the same number of integers.
Thus, the median of the set is equal to the average of the two integers in red:
(2431 + 3124)/2 = 5555/2.

Notice that the set is SYMMETRICAL ABOUT THE MEDIAN:
...2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241...
Thus:
sum = (count)(median) = 24 * 5555/2 = 12 * 5555 = 66,660.

The correct answer is E.

For a similar problem that can be solved with the same line of reasoning, check my post here:
https://www.beatthegmat.com/on-consecuti ... 85395.html

Alternate solution:

Each digit will appear in each position 24/4 = 6 times.
Thus, in each position, there will be six 1's, six 2's, six 3's, and six 4's.
Sum of the digits in each position = 6(1+2+3+4) = 60.

Sum of the thousands place = 60*1000 = 60,000.
Sum of the hundreds place =60*100 =6,000.
Sum of the tens place = 60*10 = 600.
Sum of the units place = 60*1 = 60.

Sum of all the integers = 60000 + 6000 + 600 + 60 = 66,660.