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07 Jan 2010, 05:07
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vitaliy
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
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.
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.
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.
22 Mar 2015, 15:01
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Bunuel
vitaliy
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
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]
