Solve the Positive Integer N Puzzle: 6N = defabc

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The problem involves finding a positive integer N represented as abcdef such that 6N equals defabc, indicating a digital rearrangement rather than a direct equation. Initially, there was confusion about the nature of the equation, with some believing that 6N could only equal N if N were zero, which is not a valid solution. The correct solution was identified as abc = 142 and def = 857. The discussion highlights that this type of puzzle is known as a cryptarithm, where letters represent digits, and emphasizes the need for clearer wording in the problem statement. Overall, the puzzle requires understanding of both number manipulation and the properties of digit rearrangement.
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Homework Statement


Im having problem of even starting this question:

"Determine the positive integer N = abcdef such that 6N = defabc"

any ideas?

Homework Equations





The Attempt at a Solution

 
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6N = defabc = abcdef = N

6N doesn't equal N unless it's 0. The only solution is 0 which is not a positive integer.

Or am I missing something really stupid?
 
actually i ended up solving it
abc = 142 and def = 857

...abc and def are just the digits in the question. but thanks!
 
Actually, anyone who recognizes the decimal expansions of 1/7 and 6/7 might suspect immediately that this is an answer... (Ah, but is it unique?)

Feldoh, this type of puzzle is what is referred to as a cryptarithm. Letters stand in for digits and the prospective solver is supposed to "decode" the arithmetic problem. In classical form, this problem would be posed as

ABCDEF
x 6
______
DEFABC

The question does not say that 6N = N , but that 6N is a digital rearrangment of N.
 
Regardless I think the question could be worded better...
 
It did strike me that the intent of the problem might be obscure to anyone not familiar with this kind of puzzle.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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