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What is the sum of all 4-digit numbers that can be formed

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink] New post  02 May 2021, 22:08
Sum will be number of terms times average of all values

number of terms = 256
each digit can take one of 4 integers, hence 4*4*4*4
the average number will be (min+max)/2 = (1111+4444)/2 = 2,777.5

2,777.5 * 256 = 711,040
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Re: What is the sum of all 4-digit numbers that can be formed [#permalink] New post  01 Oct 2021, 05:00
Bunuel wrote:
dimitri92 wrote:
I am sorry I don't have the OA. But I think it is solvable without the OA

What is the sum of all 4-digit numbers that can be formed using the digits 1,2,3,4 where repetition of digits is allowed?


As each digit of four digit number (abcd) can take 4 options (1, 2, 3, 4) total there can be 4^4 numbers.

Obviously each digit (a, b, c, and d) will take the value of 1, 2, 3, 4 equal number of times, so each digit will take the value of 1, 2, 3, 4 - \(\frac{4^4}{4}=4^3=64\) times: units digit will take the values of 1, 2, 3, 4 - 64 times and the same with tens, hundreds, thousands digits.

So the sum would be 64*(1+2+3+4)+64*10*(1+2+3+4)+64*100*(1+2+3+4)+64*1000*(1+2+3+4)=64*10*(1+10+100+1000)=711040.

Actually there is the direct formula for this kind of problems. Of course it's better to understand the concept, then to memorize the formula but in case someone is interested here it is:

1. Sum of all the numbers which can be formed by using the \(n\) digits without repetition is: (n-1)!*(sum of the digits)*(111…..n times).

2. Sum of all the numbers which can be formed by using the \(n\) digits (repetition being allowed) is: \(n^{n-1}\)*(sum of the digits)*(111…..n times).

Hope it helps.

Hi Bunuel, can you explain the logical way like you did for when the repetition is allowed for when the repetition is not allowed?
I am unable to figure it out. Just curious
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Re: What is the sum of all 4-digit numbers that can be formed [#permalink] New post  01 Oct 2021, 14:34
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Hi ak2121,

You can calculate that total in the same general way - it's just that there wouldn't be as many overall numbers to consider. Here's how:

When forming 4-digit numbers with the digits 1, 2, 3 and 4 - but with NO repetitions - we have (4)(3)(2)(1) = 24 possible numbers. Since each digit can be in any of the 4 'spots', this means that we'll have...

6 numbers that begin with a 4
6 numbers that begin with a 3
6 numbers that begin with a 2
6 numbers that begin with a 1

That will hold true for each of the digits in the 4-digit number, so the overall sum of all those thousands, hundreds, tens and units digits would be...

6(4+3+2+1)(1000) + 6(4+3+2+1)(100) + 6(4+3+2+1)(10) + 6(4+3+2+1)(1)

You can actually simplify this in advance if you realize that 4+3+2+1 = 10...

6(10)(1000) + 6(10)(100) + 6(10)(10) + 6(10)(1) = 66,660

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What is the sum of all 4-digit numbers that can be formed [#permalink] New post  09 Jun 2022, 08:16
Bunuel wrote:
gauravsoni wrote:
Hi Bunuel,

Can we say that the sum of the numbers will be 1111 to 4444 ? since the repetition of the digits are allowed. I think this range would cover all the numbers. Then we can simply use the formula for arithmetic mean to calculate the sum ? Please correct my thinking if wrong.


Yes, the numbers range from 1111 to 4444: 1111, 1112, 1113, 1114, 1121, 1121, ..., 4444. But the numbers are NOT evenly spaced so you cannot get the sum by multiplying the mean by the number of terms.

Does this make sense?

Bunuel

I understand that the numbers are not evenly spaced, but confused as to how we are still getting the sum by multiplying the mean by the number of terms- please refer these posts and provide your valuable inputs-
https://gmatclub.com/forum/what-is-the- ... l#p2292555
https://gmatclub.com/forum/what-is-the- ... l#p2173434
https://gmatclub.com/forum/what-is-the- ... l#p2797965

It may be possible that this would work if the numbers are consecutive.
I tested this on the number 1345 and found that this does not work.
Please confirm my understanding. TIA
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What is the sum of all 4-digit numbers that can be formed [#permalink]
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