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

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What is the sum of all 4-digit numbers that can be formed [#permalink] New post 25 May 2010, 23:38
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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?

I am sorry I don't have the OA. But I think it is solvable without the OA
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Re: Can someone help? [#permalink] New post 26 May 2010, 02:32
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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.
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Re: Can someone help? [#permalink] New post 26 May 2010, 22:47
merci !! very well done
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Re: Can someone help? [#permalink] New post 26 May 2010, 22:57
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

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Re: Can someone help? [#permalink] New post 27 May 2010, 07:13
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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: Can someone help? [#permalink] New post 27 May 2010, 21:41
Thanks Bunuel.. +1..
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Re: Can someone help? [#permalink] New post 23 Aug 2010, 06:43
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] New post 14 Oct 2013, 12:41
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Re: Can someone help? [#permalink] New post 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.
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Re: Can someone help? [#permalink] New post 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.

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Re: Can someone help? [#permalink] New post 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.

Hope it clears your doubt!!

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Re: Can someone help? [#permalink] New post 19 May 2014, 23:10
Expert's post
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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Can someone help? [#permalink] New post 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.

Does this make sense?[/quote]


Ah yes , got it thanks.
Re: Can someone help?   [#permalink] 20 May 2014, 18:53
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