Max@Math Revolution wrote:[GMAT math practice question]

What is the average (arithmetic mean) of all 5-digit numbers that can be formed using each of the digits 1, 3, 5, 7 and 9 exactly once?

A. 48000
B. 50000
C. 54000
D. 55555
E. 56000

For any set that is SYMMETRICAL ABOUT THE MEDIAN:
average = median

Number of integers that can be composed from the 5 given digits = 5! = 120.
Since the number of integers is EVEN, the median will be equal to the average of the two middle values:
...53791, 53917, 53971, 57139, 57193, 57319...
The values in green constitute the two middle values.
Notice that the set is symmetrical about these two values and thus is SYMMETRICAL ABOUT THE MEDIAN.
In accordance with the rule in blue, we get:
average = median = (53971 + 57139)/2 = 111,110 = 55,555.

The correct answer is D.

=>

There are 5! = 120 different numbers of this form.


Next, we find the sum of all numbers of this form.

Let's start by considering the numbers with 1 in each position. There are 4! = 24 numbers with 1 in the ten-thousands digit (these contribute 24 x 1 x 10000 to the sum), 24 numbers with 1 in the thousands digit (these contribute 24 x 1 x 1000 to the sum) 24 numbers with 1 in the hundreds digit (these contribute 24 x 1 x 100 to the sum), 24 numbers with 1 in the tens digit (these contribute 24 x 1 x 10 to the sum), and 24 numbers with 1 in the units digit (these contribute 24 x 1 x 1 to the sum).

Thus, the digit 1 contributes a total of 24 x 1 x (10000 + 1000 + 100 + 10 + 1) = 24 x 1 x 11111 to the sum. Similarly, the 3s contribute 24 x 3 x 11111 to the sum, and so on. Thus, the sum of all numbers of this form is
24 * ( 1 + 3 + 5 + 7 + 9 ) * 11111 = 24*25*11111 = 66666600.

The average (arithmetic mean) of these numbers is 66666600 / 120 = 555555.

Therefore, the answer is D.
Answer: D