What is the sum of all 4-digit numbers that can be formed : Problem Solving (PS)

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dimitri92

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]

26 May 2010, 23:47

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merci !! very well done

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]

26 May 2010, 23:57

Could you plz explain this :

so each digit will take the value of 1, 2, 3, 4 - \frac{4^4}{4}=4^3=64 times

Thanks & Regards

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]

27 May 2010, 08:13

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sag wrote:

Could you plz explain this :

so each digit will take the value of 1, 2, 3, 4 - \frac{4^4}{4}=4^3=64 times

Thanks & Regards

Total such numbers = 4^4 = 256.

1/4 of these numbers, or 64 numbers, will have units digit of 1; another 1/4 will have the units digit of 2; another 1/4 will have the units digit of 3; and the last 1/4 will have the units digit of 4.

The same with tens, hundreds, thousands digits.

Hope it's clear.

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]

27 May 2010, 22:41

Thanks Bunuel.. +1..

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]

23 Aug 2010, 07:43

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Agreed. These are powerful formulas but the concept behind it should be understood before roting them.

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]

19 May 2014, 12:11

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 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.

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]

19 May 2014, 13:16

gauravsoni wrote:

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 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.

The range of the numbers do vary from 1111 to 4444 inclusive, and there are only 264 different numbers altogether formed.
however, what do you mean by the sum of the numbers will be 1111 to 4444 is not clear.

The formula as given when repetition is allowed is pretty simple :
n^{n-1}*(sum of the digits)*(111…..n times).

Here, n^{n-1} : the no of times each digit appear at each place
needs to be multiplied by sum (of the digits) as all the digits take that place
multiplied by 111... upto n times - to finally find the value at each place

You could see the manner in which the total sum could be arrived at with the formula.
Hope it helps.

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]

19 May 2014, 20:01

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gauravsoni wrote:

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 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.

All the numbers which are formed lies between 1111 and 4444 but it does not include all numbers from 1111 to 4444. For example, 1235 will not be formed as we have only 1,2,3,4 to choose from. Thus, we can't use the formula of arithmetic mean.

Hope it clears your doubt!!

Kudos if it does!!!

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]

20 May 2014, 00:10

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gauravsoni wrote:

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 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?

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]

20 May 2014, 19:53

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 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?[/quote]

Ah yes , got it thanks.

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]

26 Jul 2015, 16:13

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.

Did you mean "1111"? (highlighted in red)

L

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]

28 Jul 2015, 20:59

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Hi All,

While much of this discussion is over a year old, it's important to note how important it is to include the 5 answer choices to any PS question. The GMAT only rarely offers questions that can only be solved by 'doing math in one specific way', which means that there are normally several different ways to approach each question. By having the answer choices to work with, we can sometimes avoid doing math altogether (since can use estimation or logic to determine that certain answers are 'too small' or 'too big' to be correct).

Here, by not including the 5 answer choices, the original poster forces us to do math, when a more elegant, simpler or faster approach might have been possible.

GMAT assassins aren't born, they're made,
Rich

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]

20 Aug 2015, 18:15

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Here is an alternative solution that you can compute easily on the GMAT if given this question:

First, note that the smallest number you can make is 1111, the largest is 4444. It is reasonable to conclude that the average of all the combinations of digits (1,2,3,4) very close to the average of 1111+4444, which is 5555/2.

Next, compute the possibilities: 4*4*4*4 = 16*16 = 256 (memorize this). 4*4*4*4.

Now all you have to do is compute 256 * 5555 / 2.

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]

24 Sep 2017, 03:30

At 1000nds place 4,3 2,1 can all come likewise at 100th ,10th and unit place also all of it can come
1000(4+3+2+1)+100(4+3+2+1)+10(4+3+2+1)+1(4+3+2+1)=11110 ans

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]

12 Jun 2019, 02:23

Bunuel wrote:

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?

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.

Bunuel: Has there ever been such a question on official GMAT...If so please share some reference link.

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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]

12 Jun 2019, 03:06

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saukrit wrote:

Bunuel wrote:

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?

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.

Bunuel: Has there ever been such a question on official GMAT...If so please share some reference link.

Check here: Constructing Numbers, Codes and Passwords.

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What is the sum of all 4-digit numbers that can be formed [#permalink]

12 Jun 2019, 20:49

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dimitri92 wrote:

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?

least possible value=1111
greatest possible value=4444
1111+4444=5555
5555/2=2777.5=mean
because each of 4 digits can take any of 4 options,
sum=4^4*2777.5=711040

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What is the sum of all 4-digit numbers that can be formed [#permalink]

12 Jun 2019, 20:49

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