Determine the total number of concatenated 3-digit positive integers of the form ABC, such that each of their nonzero digits A, B, and C satisfies this relationship:
. . .
___ 0.A + 0.B + 0.C
0.ABC = ------------------
3Notes:
. ---
0.ABC = 0.ABCABCABC.......
.
0.A = 0.AAAA.......
Computer program/excel solver solutions are welcome- but a semi-analytic (p&p with a hand-calculator) methodology is preferred.
Using a computer program on this problem at all is unnecessary overkill.
First, a little exploration: let A=B=C, then both sides of the equation reduce to 0.A(repeating). Then all the repdigits 111, 222, 333, 444, 555, 666, 777, 888, and 999 are solutions.
So in general, start by rewriting in fractional terms:
ABC/999 = (A/3+B/3+C/3)/3
Then multiply by 999:
ABC = 37*(A+B+C)
This is key. The three digit number is 37 times its sum of digits. So there are at most 25 solutions, for the multiples of 37 from 111 to 999.
We already know the repdigits are solutions. Then for each repdigit I'll check if adding 37 increases the sum of digits by 1, or I'll check if adding 74 increases the sum of digits by 2.
Case 1: 37
Adding 37 to a repdigit will increase the sum of digits by 1 when there is exactly one carry. There clearly will not be any carries when the repdigit is 111 or 222, and there will definitely be two carries when the repdigit is 777-999.
That leaves 333-666 to test. 333+37=370, 444+37=481, 555+37=592, and 666+37=703. The first three are good and the last one 703 must be discarded.
But then KS also asks for nonzero digits, so then the first one 370 also needs to be discarded, leaving 481 and 592 as non repdigit solutions.
Case 2:
Adding 74 to a repdigit will increase the sum of digits by 2 when there is exactly one carry. There clearly will not be any carries when the repdigit is 111 or 222, and there will definitely be two carries when the repdigit is 666-999.
That leaves 333-555 to test. 333+74=407, 444+74=518, and 555+74=629. All of these are good.
But then KS also asks for nonzero digits, so then the first one 407 also needs to be discarded, leaving 518 and 629 as non repdigit solutions.
So then the solution set of integers that satisfy the problem's constraints is {111, 222, 333, 444, 481, 518, 555, 592, 629, 666, 777, 888, and 999}.
Analytic Solution
