MHB Find all three-digit integers.

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The problem involves finding all three-digit integers \( n \) represented as \( \overline{abc} \) such that the function \( d(n) = a + b + c + ab + ac + bc + abc \) equals \( n \). Participants discuss the complexity of the equation and the need for a systematic approach to identify valid integers. Several examples are analyzed, but no conclusive solutions have been reached. The discussion highlights the mathematical challenges and encourages further exploration of potential methods to solve the equation. The quest for three-digit integers satisfying \( d(n) = n \) remains an open problem in number theory.
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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 $d(n) = n$.
 
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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
 
RLBrown said:
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

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
 
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