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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.

See The Solution Submitted by K Sengupta

Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

"ab" means 'a' concatenated with 'b'

| Comment 3 of 5 |

(In reply to 45 other solutions by Steve Herman)

When I first started to solve this problem, I thought the first equation was:

a * b^2 + c^2,

but then I realized that the digits are being concatenated, so instead it is:

(10*a+b)^2 + c^2

Posted by Larry on 2023-04-22 09:44:17

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