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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