Find the sum of all the four digit numbers which are formed

The answer choices make the solution easy:

There are 4!=24 four digit numbers which are formed by digits 1, 2, 5, 6.

Obviously 24/4=6 numbers will end with 1; 6 numbers will end with 2, 6 numbers with 5 and 6 numbers with 6.

6*1+6*2+6*5+6*6=6*14=84, which means that the sum of all these 24 numbers must end by 4, only answer choice with 4 at the end is B.

Answer: B.

But if we were not given such an easy answer choices, the solution would be:

We have 24 numbers of the of the form: 1000a+100b+10c+d, where a, b, c and d can take any value from the set {1, 2, 5, 6} and there will be 6 numbers with same digit (a, b, c, d) at the thousands, hundreds, tens and units digits.

(1000*6a+1000*6b+1000*6c+1000*6d)+(100*6a+100*6b+100*6c+100*6d)+(10*6a+10*6b+10*6c+10*6d)+(6a+6b+6c+6d)=(6a+6b+6c+6d)(1111)=
=6(a+b+c+d)(1111)=6*14*1111=93324

Generally the sum of all the numbers which can be formed by using the n distinct digits, is given by the formula:

(n-1)!*(sum of the digits)*(111…..n times)
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Doesn't 'sum of all' suggest that repetition is allowed ?

No it does not. Ideally it should have been clarified more precisely but it's assumed that we are adding the numbers which could be constructed by re-arranging the numbers 1, 2, 5, and 6.
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Yes, this is not a proper GMAT problem not only because of the wording. Notice that it has 4 options not 5.
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1256+6521=7777
7777/2=3888.5
3888.5*4!=93324
B

3!*1111*(14) will have units digit 4

B is the answer

Hope it helps
Cheers!
J

I have the same query. Experts, please clarify.

Given this ambiguity, it's safe to say this is not an OG problem, and I would not see problems worded similarly on the test?

Could also be solved with estimation:

\(4*1000 + 4*2000 + 4*5000 + 6*6000=68000\) Just looking at the answer choices option B is the only possible solution.

I have made similar mistakes in other problems.
Until repetition is not specified in problem, it is safe to proceed considering non-repetition of digits/letters.
And this rule is valid throughout the GMAT.

Hi Bunuel , thanks for the good explanation as always. Regarding the formula below, can you confirm my understanding that it is not applicable if there is a "0" in one of the digits provided?

Based on a similar question from the link below, the calculation for this question can also be done as:
Total possible 4-digit numbers= 4*3*2*1= 24

Sum of digits at unit place= 24/4*(1+2+5+6)=84
Sum of digits at tens place= 24/4*(1+2+5+6)=84
Sum of digits at hundreds place= 24/4*(1+2+5+6)=84
Sum of digits at thousands place= 24/4*(1+2+5+6)=84

the sum of all possible 4 digit numbers that can be formed using all the digits of the number 1256
= 84* (1000+100+10+1)=93,324



https://gmatclub.com/forum/find-the-sum ... s#p2448163

There are 4 ways to select the thousands digit
There are 4 ways to select the hundreds digit
There are 4 ways to select the tens digit
There are 4 ways to select the ones digit
So, the total number of 4-digit numbers possible = (4)(4)(4)(4) = 256

Now let's focus on the thousands digits of our 256 numbers.
64 (aka 1/4) of the 256 numbers will have thousands digit 1. 64 x 1000 = 64,000
64 (aka 1/4) of the 256 numbers will have thousands digit 2. 64 x 2000 = 128,000
64 (aka 1/4) of the 256 numbers will have thousands digit 5. 64 x 5000 = 320,000
64 (aka 1/4) of the 256 numbers will have thousands digit 6. 64 x 6000 = 384,000
64,000 + 128,000 + 320,000 + 384,000 = 896,000

So, if we ignore the hundreds, tens and ones digits, the sum of our 256 numbers is already 384,000, which means the TOTAL sum must be greater than 384,000

Answer: A

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