Find all three digit 'special' numbers which equal the average of all permutations of their digits [closed]

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The number $518$ has a special feature. Lets make the six permutations of this number and add them together ($158+185+518+581+815+851=3108$). This number's average is $\dfrac{3108}{6}=518$.

How can I find all three digit numbers that have this same feature?

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2 Answers

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Hint: Let the three digit number be $100a+10b+c$, then the sum of its permutations is,

$$200(a+b+c)+20(a+b+c)+2(a+b+c)=222(a+b+c).$$

Hence the problem is equivalent to solve the diophantine equation

$$37(a+b+c)=100a+10b+c.$$

Note that $1\le a\le 9$ and $0\le b,c \le 9$.

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As per Jack Frost and Patrick Stevens's contributions:

$100a+10b+c = 37(a+b+c)$

$63a = 27b+36c$

$7a = 3b+4c$

$a = (3b+4c)/7$

for $1\le a\le 9$ and $0\le b,c \le 9$

Note the lowest solution is a=1,b=1,c=1. And there are 9 symmetric solutions: a=b=c for each value $1\le a\le 9$

{(1,1,1), (2,2,2), ... (9,9,9)}Note that (0,0,0) is excluded, but it's a useful startpoint for considering perturbations of solutions.

(Note also by congruences you get $a == b$ (mod-3) and $a == c$ (mod-4))

By setting c=0 we also find a=3,b=7,c=0

By setting b=0 we also find a=4,b=0,c=7

Noting 7*1 = 4*1 + 3*1, we can convert one legal solution tuple to another by adding (+1,+1,+1) as long as the largest digits doesn't exceed 9, hence we find:

a=4,b=8,c=1 and again a=5,b=9,c=2

a=5,b=1,c=8 and again a=6,b=2,c=9

Similarly, other legal perturbations are (+4,0,+7) (+3,+7,0), (0,+4,-3), (0,+3,-4) as well as (+1,+1,+1); and these are the only legal ones, since any other would send some digit > 9

Hence there are no other solutions to the nine symmetric ones and the six asymmetric ones above.

You can also elegantly prove that graphically by drawing the grid of (b,c) and coloring in all legal solutions; (0,0,0) is excluded but special, all legal solutions can be reached from (0,0,0) by repeatedly adding any of the perturbations (+1,+1,+1), (+4,0,+7) (+3,+7,0), (0,+4,-3), (0,+3,-4)

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