What is the average (arithmetic mean) of all 5-digit numbers that can

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

RJ7X0DefiningMyX · Accepted Answer

The sum of the units digits for all numbers would be (1+3+5+7+9) * 4! = 600 (for detailed explanation : when we take 1 at the unit's digit, we have 4! ways to choose the other 4 numbers of the 5-digit number, similarly for 3, 5, 7, and 9)

Similarly, the sum of tens digits for all numbers would be (1+3+5+7+9) * 4! * 10 = 600
And similarly, hundreds, thousands and ten thousands would be 6000, 60000 and 600000

Adding all of these, we get the total sum to be 6666600

Now, to calculate the average, sum / total number of such combinations = 6666600 / 5! = 6666600 / 120 = 55555 (D)

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gracie · Answer

add least and greatest possible numbers: 13579+97531=111110
111110/2=55555 average
D

GMATGuruNY · Answer

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.

D .

MathRevolution · Answer

=>

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  \frac{66666600}{120} = 555555.

Therefore, the answer is D.
Answer: D

NoMatterWhat · Answer

I doubt it is a right way for such problems.
For E.g

consider any 2 numbers {1,3}
maximum 31
minimum 13

as per your formula (31-13)/2 = 18/2 =9, which is incorrect.

Don't worry about KUDOS

Mansi89 · Answer

I still didn’t understand this problem?

Posted from my mobile device

stne · Answer

Hi  gracie ,

How did you ascertain, that all the numbers can be written in the form of AP to use this formula?

anamika03rm@gmail.com · Answer

can someone explain the solution to this problem to me?

gracie · Answer

hi rupeshnverbal,
actually, I added the two numbers.
per your example, 13+31=44
44/2=22 mean
I hope this helps,
gracie

gracie · Answer

hi stne,
if you define an AP as having one common difference, this is not an AP.
however, looking at the intervals, I do see a clear sequential relationship,
albeit one that I'm unable to characterize.
I wish I could be more helpful with this.
gracie

vivek2611 · Answer

Sum of the digits (1+3+5+7+9=25) . This will appear once in each of ( units, 10s, 100s, 1000s,10000s place )

Hence 
(25 +250+2500+25000+250000)/ 5= 
= 25 ( 1+10+100+1000+100000) /5

= 25 *11111/5

= Ans D

pallabgrahi · Answer

Yes, approached the same way. took around 40 sec

Posted from my mobile device

bv8562 · Answer

Sum of all the numbers that can be formed using 'N' digits ( WITHOUT REPETITION ):

Sum = (N-1)! * (Sum of all the digits) * (1111...N times)

N=5
Sum of digits = 1+3+5+7+9 = 25

Sum = (5-1)! * 25 * 11111 = 24*25*11111 = 6666600

Total no. of 5-digit numbers that can be formed using these digits without repetition = 5! = 120

Average =  \frac{6666600}{120}  = 55555 (D)

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What is the average (arithmetic mean) of all 5-digit numbers that can

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