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Cryptarithm Crossed Integer Constant Illation (Posted on 2023-04-22) Difficulty: 3 of 5
Each of a, b, and c is a different base ten digit and n is a positive integer such that:

ab2+c2=ac2+cb2=n
For example:
272+12=212+172=730

Determine the total number of values of the positive integer constant n less than 2023, such that we will have valid values of a,b, and c satisfying the abovementioned condition.

***** No number can contain any leading zero.

  Submitted by K Sengupta    
Rating: 4.0000 (2 votes)
Solution: (Hide)
(a,b,c) = (2,7,1), (4,3,1),(3,4,1) are the only other values of the triplet, that satisfy the given conditions.

The total number of values of the positive integer constant n less than 2023 is three (730, 1156, and 1850).

For an explanation, refer to the analytic solution submitted by Brian Smith in this location.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: SolutionMath Man2023-05-06 20:35:48
SolutionSolutionBrian Smith2023-04-22 13:59:50
Hints/Tips"ab" means 'a' concatenated with 'b'Larry2023-04-22 09:44:17
Some Thoughts45 other solutionsSteve Herman2023-04-22 07:34:23
SolutionComputer solutionLarry2023-04-22 06:40:44
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