Unique number Part 1

Probability Level 3

Define subfactorial ( $!n$ ) as $!n=\displaystyle \sum_{k=0}^n k!(-1)^{n-k}\binom n k$

Find the six-digit number $\overline{\text{abcdef}}$ such that $!\text{a}+!\text{b}+!\text{c}+!\text{d}+!\text{e}+!\text{f}=\overline{\text{abcdef}}$ .

The answer is 148349.

1 solution

Kyle T
Apr 1, 2019

First lets start by defining a couple things on the board

k! means k factorial, where 0!=1, 1!=1, 2!=2, 3!=6, etc.
the parenthesis (n, k) means n choose k, also written as n! / ( k! * (n-k)!)

once we have those defined we can calculate the subfactorial for each number 0-9 as follows:
!0 = 1
!1 = 0
!2 = 1
!3 = 2
!4 = 9
!5 = 44
!6 = 265
!7 = 1854
!8 = 14833
!9 = 133496

Now the brute forcing comes in, we can loop through every 6 digit number 000000-999999 and convert the digits into a sum using the conversion chart above. The first number to come back where the converted total equals the original number is our answer, 148349.

Unique number Part 1

The answer is 148349.

1 solution

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