- #1
magneto1
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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 $d(n) = n$.
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
Three-digit integers are numbers that have three digits, including a hundreds digit, a tens digit, and a ones digit. They range from 100 to 999.
To find all three-digit integers, you can start at 100 and count up to 999. Alternatively, you can use a formula such as 100n + 10m + p, where n, m, and p are any digits from 0 to 9, to generate all possible combinations.
Yes, there are several rules and patterns for three-digit integers. For example, every three-digit integer can be divided by 9. Additionally, the sum of the digits in a three-digit integer is always a multiple of 9.
There are 900 three-digit integers in total, ranging from 100 to 999. This can be calculated by subtracting 99 (the number of two-digit integers) from 999.
The smallest three-digit integer is 100 and the largest is 999. These are known as the lower and upper bounds, respectively, for three-digit integers.