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What is the sum of all possible 3-digit numbers that can be

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What is the sum of all possible 3-digit numbers that can be  [#permalink]

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New post 07 Jan 2010, 05:07
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vitaliy wrote:
What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4 and 5, if each digit can be used only once in each number?


My Q.: How we receive 24s in the final equalization (attached). thnx


Three digit number has the form: 100a+10b+c.

# of three digit numbers with digits {3,4,5} is 3!=6.

These 6 numbers will have 6/3=2 times 3 as hundreds digit (a), 2 times 4 as as hundreds digit, 2 times 5 as hundreds digit.

The same with tens and units digits.

100*(2*3+2*4+2*5)+10*(2*3+2*4+2*5)+(2*3+2*4+2*5)=100*24+10*24+24=2664.

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)


In our original question: n=3. sum of digits=3+4+5=12. --> (3-1)!*(12)*(111)=24*111=2664.

Hope it's clear.
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Re: What is the sum of all possible 3-digit numbers that can be  [#permalink]

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New post 07 Jan 2010, 05:40
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@Vitality - Hope you have understood the solution given by Bunuel.
Now coming to your Qs - how we received 24?

Since each of the 3 digits - 3,4,5 appear twice at each position, hence we have multiplied the sum of these 3 digits i.e 12 by 2.

Sum @hundreds position = (3+4+5) * 100 * 2 = 12 * 2 * 100 = 24 *100
Sum @tens position = (3+4+5) * 10 * 2 = 12 * 2 *10 = 24 * 10
Sum @units position = (3+4+5) * 1 * 2 = 12 * 2 = 24
Total = 2400 + 240 + 24 = 2664.
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Re: What is the sum of all possible 3-digit numbers that can be  [#permalink]

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New post 22 Mar 2015, 15:01
Bunuel wrote:
vitaliy wrote:
What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4 and 5, if each digit can be used only once in each number?


My Q.: How we receive 24s in the final equalization (attached). thnx


Three digit number has the form: 100a+10b+c.

# of three digit numbers with digits {3,4,5} is 3!=6.

These 6 numbers will have 6/3=2 times 3 as hundreds digit (a), 2 times 4 as as hundreds digit, 2 times 5 as hundreds digit.

The same with tens and units digits.

100*(2*3+2*4+2*5)+10*(2*3+2*4+2*5)+(2*3+2*4+2*5)=100*24+10*24+24=2664.

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)


In our original question: n=3. sum of digits=3+4+5=12. --> (3-1)!*(12)*(111)=24*111=2664.

Hope it's clear.


Hi Bunuel,

Thanks for the help. For the above equation, is 111 constant, or does that number increase to match n? Example, if n=5, would we use 11,111?

(n-1)!*(sum of the digits)*(11,111…..n times)[/b]
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Re: What is the sum of all possible 3-digit numbers that can be  [#permalink]

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New post 23 Mar 2015, 03:05
metskj127 wrote:
Bunuel wrote:
vitaliy wrote:
What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4 and 5, if each digit can be used only once in each number?


My Q.: How we receive 24s in the final equalization (attached). thnx


Three digit number has the form: 100a+10b+c.

# of three digit numbers with digits {3,4,5} is 3!=6.

These 6 numbers will have 6/3=2 times 3 as hundreds digit (a), 2 times 4 as as hundreds digit, 2 times 5 as hundreds digit.

The same with tens and units digits.

100*(2*3+2*4+2*5)+10*(2*3+2*4+2*5)+(2*3+2*4+2*5)=100*24+10*24+24=2664.

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)


In our original question: n=3. sum of digits=3+4+5=12. --> (3-1)!*(12)*(111)=24*111=2664.

Hope it's clear.


Hi Bunuel,

Thanks for the help. For the above equation, is 111 constant, or does that number increase to match n? Example, if n=5, would we use 11,111?

(n-1)!*(sum of the digits)*(11,111…..n times)[/b]


Yes, there should be n number of 1's. So, if n=5, it should be 11,111.
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What is the sum of all possible 3-digit numbers that can be  [#permalink]

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New post 28 Dec 2018, 12:36
What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4 and 5, if each digit can be used only once in each number?

add least and greatest numbers: 345+543=888
888/2=444 mean
444*3! total possibilities=2664 sum
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Re: What is the sum of all possible 3-digit numbers that can be  [#permalink]

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New post 29 Dec 2018, 13:31
Bunuel wrote:
vitaliy wrote:
What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4 and 5, if each digit can be used only once in each number?


My Q.: How we receive 24s in the final equalization (attached). thnx


Three digit number has the form: 100a+10b+c.

# of three digit numbers with digits {3,4,5} is 3!=6.

These 6 numbers will have 6/3=2 times 3 as hundreds digit (a), 2 times 4 as as hundreds digit, 2 times 5 as hundreds digit.

The same with tens and units digits.

100*(2*3+2*4+2*5)+10*(2*3+2*4+2*5)+(2*3+2*4+2*5)=100*24+10*24+24=2664.

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)


In our original question: n=3. sum of digits=3+4+5=12. --> (3-1)!*(12)*(111)=24*111=2664.

Hope it's clear.


Hi Bunuel. This formula will break if zero is one of the digits to be used for forming 3 digit numbers. Can you provide an idea for such scenarios? Thanks.
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Re: What is the sum of all possible 3-digit numbers that can be  [#permalink]

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New post 06 Jan 2019, 03:57
metskj127 wrote:
Bunuel wrote:
What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4 and 5, if each digit can be used only once in each number?




Three digit number has the form: 100a+10b+c.

# of three digit numbers with digits {3,4,5} is 3!=6.

These 6 numbers will have 6/3=2 times 3 as hundreds digit (a), 2 times 4 as as hundreds digit, 2 times 5 as hundreds digit.

The same with tens and units digits.

100*(2*3+2*4+2*5)+10*(2*3+2*4+2*5)+(2*3+2*4+2*5)=100*24+10*24+24=2664.

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)


In our original question: n=3. sum of digits=3+4+5=12. --> (3-1)!*(12)*(111)=24*111=2664.

Hope it's clear.


Hi Bunuel,

Thanks for the help. For the above equation, is 111 constant, or does that number increase to match n? Example, if n=5, would we use 11,111?

(n-1)!*(sum of the digits)*(11,111…..n times)[/b]


Yes, there should be n number of 1's. So, if n=5, it should be 11,111.[/quote]

Is this applicable only when the digits can't be repeated ?
What id we were to form four digit numbers using 3,4,5 with any one being repeated ?
Are there more formulae for such questions ?
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Re: What is the sum of all possible 3-digit numbers that can be   [#permalink] 06 Jan 2019, 03:57
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