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Re: Can someone help? [#permalink]
26 May 2010, 02:32

12

This post received KUDOS

Expert's post

9

This post was BOOKMARKED

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).

Re: Can someone help? [#permalink]
27 May 2010, 07:13

3

This post received KUDOS

Expert's post

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.

Re: What is the sum of all 4-digit numbers that can be formed [#permalink]
14 Oct 2013, 12:41

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Re: Can someone help? [#permalink]
19 May 2014, 11:11

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.

Re: Can someone help? [#permalink]
19 May 2014, 12:16

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.

Re: Can someone help? [#permalink]
19 May 2014, 19:01

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.

Re: Can someone help? [#permalink]
19 May 2014, 23:10

Expert's post

1

This post was BOOKMARKED

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.

Re: Can someone help? [#permalink]
20 May 2014, 18:53

Bunuel 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.

Re: What is the sum of all 4-digit numbers that can be formed [#permalink]
27 Mar 2015, 01:35

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.[/quote]

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.[/quote]

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]

I am no Math Expert. Can you please correct my below approach? We know the smallest number we can make is 1111 and the largest number we can make is 4444. We also know that our numbers will be evenly distributed in the middle (i. e. 1112 is balanced by 4443; 1113 is balanced by 4442). So, we can solve using the average formula. Finally, we know that there are 4*4*4*4 = 256 numbers in our set. Average = sum of terms/# of terms sum of terms = average * # of terms sum of terms = (1111+4444)/2 * 256 = 5555/2 * 256 = 5555*128 = 711040 _________________

Re: What is the sum of all 4-digit numbers that can be formed [#permalink]
26 Jul 2015, 15:13

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).

Re: What is the sum of all 4-digit numbers that can be formed [#permalink]
28 Jul 2015, 19:59

Expert's post

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.

Interested in applying for an MBA? In the fourth and final part of our live QA series with guest expert Chioma Isiadinso, co-founder of consultancy Expartus and former admissions...