What is the sum of all the 4 digit numbers formed by digits 2,3,4,5?

The way I solved it was using the formula suggested by bunuel

on the following link:
https://gmatclub.com/forum/what-is-the- ... 94836.html

I solved this question using another approach,
Since there are 4! Or 24 Ways of arranging the digits without repetition, the sum of all the numbers formed should be a multiple of 24. Using POE I got OA.Although I got OA Bunuel is this reasoning applicable to questions of this sort? Thanks.

Posted from my mobile device

carcass gmatbusters VeritasKarishma

Dear Experts,
Kindly help with a solution on this one please. Thank you.

Four digit numbers =4⋅3⋅2⋅1=24 ways we can form a four digit number. Since it's a 4 digit number, each digit will appear 6=24/4 times in each of units, tens, hundreds, and thousands place. Therefore, the sum of digits in the units place is 6(2+3+4+5)=85

Why they are adding 2+3+4+5?= Because each digit will come at those places.

Similarly sum of digits at tens,hundreds,thousands place is 84

Required sum of all numbers =84+84⋅10+84⋅100+84⋅1000=93324

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Coming to the question:

What is the sum of all the 4 digit numbers formed by digits 2,3,4,5?

A few things to note here:
The numbers will be of the form 2354, 3425 ... etc
There will be a total of 4*3*2*1 = 24 numbers
These will be made symmetrically. What this means is that in the 24 numbers, 2 will appear 6 times, 3 will appear 6 times, 4 will appear 6 times and 5 will appear 6 times (since all digits are equivalent)

2345
3245
...
2354
3254
...

etc

This will be true of digits in tens place, hundreds place and thousands place too.

The sum of digits in units place = 6*2 + 6*3 + 6*4 + 6*5 = 6*(2 + 3 + 4 + 5) = 84
The sum of digits in tens places, hundreds places and thousands places will also be 84 each.

The sum of all numbers then will be = 84 + 84*10 + 84*100 + 84*1000 = 93324

Answer (B)

GMATinsight : sir is there any direct formula to solve such questions?

Hi,

If we are not told can we assume that the numbers can repeat?

YES,
No mentioned on repetition of digit means that repetition of digits is allowed. and therefore all above mentioned solutions are incorrect as they assumed that repetition of digits is not allowed.

Here the solution should be smalled such number = 2222
Largest number = 5555

Sum of the pair of largest and smallest numbers = 2222+5555 = 7777

Total Such numbers = 4*4*4*4 = 256

Total suc pairs = 256/2 = 128

Sum of all pairs = 7777*128 = 995456

GMATinsight : but sir your answer is not one of the given options? also what would be the solution in case repetition of digits is not allowed..

If numbers cnt be repeated then in that case what will be the sum?

Posted from my mobile device

That means that either question couldn't explain properly that repetition of digits is not allowed or Options are incoorect.

Probably the prior assumption is correct.

No of ways of arranging 4- digits = No of possible 4- digit numbers = 4! = 24.
Each digit will appear for 6 times at each of thousandth, hundredth, tens and ones position.
The sum of the digits at ones place = 6(2 + 3 + 4 + 5) = 84.
Similarly, sum of digits at tens , hundredth, thousand place = 6(2 + 3 + 4 + 5) = 84.
Sum of all possible numbers = 1000 x 84 + 100 x 84 + 10 x 84 +84 = 93324.

Bunuel: How do we know that in this question digits can be repeated or not

Given the answer choices, the question stem should include the red portion below:



Total number of integers = number of ways to arrange the four digits = 4! = 24

The middle values in the set are as follows:
...3425, 3452, 3524, 3542, 4235, 4253, 4325, 4352...

Notice that every colored pair sums to 7777:
3542+4235 = 7777
3524+4253 = 7777
3452+4325 = 7777
3425+4352 = 7777

Since the 24 integers are composed of 12 such pairs, we get:
Sum of the set = 12*7777 = 93324

IMO B

There's a formula for such questions

n= total number of digits given =4
H= highest number formed by the digits
L=lowest number formed by the digits

SUM =(H+L)*n! /2

put those values Sum=24/2*(2345+5432) =93324

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