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Cryptarithm Crossed Integer Constant Illation (Posted on 2023-04-22)

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: Solution	Math Man	2023-05-06 20:35:48
Solution	Brian Smith	2023-04-22 13:59:50
"ab" means 'a' concatenated with 'b'	Larry	2023-04-22 09:44:17
45 other solutions	Steve Herman	2023-04-22 07:34:23
Computer solution	Larry	2023-04-22 06:40:44

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