Divisibility Pattern
Assume any number AAAB (B= ones digit, AAA= all other remaining digit) and CCCD (D= ones digit, CCC= all other remaining digit). If CCCD divides AAAB then:
Pattern:
(AAA)(C) - (BBB)(D) will always end/approach zero after some repetition.
1st Ex: Let the two numbers be 14992470 and 176382. And we know that 176382 divides 14992470.
Pattern:
(1499247)(2) - (17638)(0) = 2998494 (Not yet zero, so repeat using this difference as the new AAAB)
(299849)(2) - (17638)(4) = 529146 (Repeat pattern)
(52914)(2) - (17638)(6) = 0 (End of Pattern)
2nd Ex: Let the two numbers be 124029 and 13781. And we know that 13781 divides 124029.
Pattern:
(12402)(1) - (1378)(9) = 0 (End of Pattern)
Question: Would this pattern "always" work for all combination of numbers for which one divides the other?
Bonus: If No, provide a counterexample or a proof that it won't always work. Otherwise, If the answer is Yes, Provide the proof, if possible include along with the prediction on how many repetition the pattern would need to approach 0. (State it in the comment section below).
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1 solution
Let 1st number = 10a+b
Let 2nd number = c(10a+b) = 10ac+bc
We are certain that 10a+b divides 10ac+bc and would result to a quotient c.
Pattern:
(10ac)(b) - (10a)(bc) = 10abc - 10abc = 0
It is important to note however that "bc" won't always be a single digit product. And base on observation if bc is a single digit, the pattern only need one step to terminate. Otherwise, it would take more than a step.