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

Question

What is the sum of all the 4 digit numbers formed by digits 2, 3, 4, 5 if repetition is not allowed??

A. 103324

B. 93324

C. 87645

D. 77678

E. 65432

gracie · Accepted Answer

add least and greatest: 2345+5432=7777
7777/2=3888.5=average
4! possible combinations=24
24*3888.5=93324
B

CounterSniper · Answer

let us take 2 as the thousands digit 
the following combinations are possible 
2345
2435
2354
2534
2453
2543

let us take sum of units digit
2*5 + 2*4 + 2*3 = 10+8+6 = 24

similarly 
when we take 3 as the thousands digit 
2*5 + 2*2 + 2*4 = 10+4+8=22

when we take 4 as the thousands digit 
2*5 + 2*2 + 2*3 = 10+4+6=20

when we take 5 as the thousands digit
2*2 + 2*3 + 2*4 = 4+6+8=18

sum of units digit = 84

similarly sum of tens digit = 84*10
sum of hundreds digit =84*100
sum of thousands digit = 84*1000

sum total = 84+84*10+84*100+84*1000 = 84(1111) = 93324 
B

fitzpratik · Answer

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

Kem12 · Answer

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

poojasingavi · Answer

carcass   gmatbusters   VeritasKarishma

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

GMATBusters · Answer

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

KarishmaB · Answer

Request to all forum users: I am one of the forum experts on the PS and CR forums. If you would like me to take a look at any question on these forums, please mention it on my thread (link given below): PS - https://gmatclub.com/forum/veritas-prep ... 78042.html CR - https://gmatclub.com/forum/veritas-prep ... 78026.html Since I receive 100s of mails everyday in my mailbox, unfortunately, it becomes very difficult to follow up on every query even if I am tagged on it. I do take a look at both my threads on a daily basis and hence will certainly get to your query if you mention on it. Thank you! 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)

Archit3110 · Answer

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

bpdulog · Answer

Hi,

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

GMATinsight · Answer

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

Archit3110 · Answer

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..

s111 · Answer

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

Posted from my mobile device

GMATinsight · Answer

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.

Rohit23 · Answer

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.

saukrit · Answer

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

GMATGuruNY · Answer

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

B

ArindamLucky · Answer

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

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