Find all three-digit integers.

[magneto1]

Given a three-digit integer $n$ written in its decimal form $\overline{abc}$. Define a function $d(n) := a + b + c + ab + ac + bc + abc$. Find, with proof, all the (three-digit) integers $n$ such that $d(n) = n$.

 

[RLBrown]

Spoiler

d = a + b + c + ab + ac + bc + abc
d = a((b+c+bc)+1) + (b+c+bc)

If (b+c+bc)<100
(b+c+bc)+1 = 100 for "a" to be first digit of n.
So (b+c+bc)=99

 

[kaliprasad]

Spoiler

d = a + b + c + ab + ac + bc + abc
d = a((b+c+bc)+1) + (b+c+bc)

If (b+c+bc)<100
(b+c+bc)+1 = 100 for "a" to be first digit of n.
So (b+c+bc)=99

This is a problem which is not solved since long. I continue from above

Spoiler

as b and c both are < 10 so b+c + bc <100

now b+c + bc + 1 = 100
or (1+b)(1+c) = 100
so (1+b) = (1+c) = 10 as both <= 10
so b = 9, c = 9
d = 100 a + 99

so a = any digit from 1 to 9 and b = 9 c = 9

numbers are 199 , 299, 399, 499, 599, 699, 799, 899,999