Today's MAA number of the day, when multiplied by its successor, gives a number concatenated with itself. (Like 455 * 539 = 245245, except 539 isn't the successor of 455.)

Puzzle: find all such three-digit numbers. (The fact that I'm phrasing it this way should indicate to you that there's more than one.)

Meta-puzzle: generalize this. (I have some generalizations in mind.)

## 07 July 2011

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## 7 comments:

The most interesting number I got after generalizing the problem in, what I assume to be, a pretty standard way:

7272727

Also, some other palindromes:

81818 and 336633

I get 2 or 3 answers depending on weather you're willing to write xy as 0xy or not.

SPOLIERS ALERT

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074*075=006006

286*287=082082

363*364=132132

Jonathan: those are some, but not all, of the answers.

Looks like they're rarer than I expected, with only one four digit and one five digit generator.

I was going to find more, but then decided to use the awesome power of OIES:

http://oeis.org/A116285

There are some very interesting patterns in that sequence, like the fact that 363, 336633, and 333666333 are all generators. That begs to be explored...

Besides Jonathan's, I also found 637*638 = 406406. A key to my solutions, and I think all potential solutions, is that the final product has to be a multiple of 1001 in order to have "concatenation". 1001 is conveniently 7*11*13, and so I tried multiples of 77, multiples of 91, and multiples of 143. In general, any concatenations would be multiples of 11, 101, 1001, 10001, 100001, etc. Or repeated concatenations might be multiples of 111, 10101, 1001001, and so on.

I find

77x78 = 006006

286x287 = 082082

363x364 = 132132

637x638 = 406406

714x715 = 510510

923x924 = 852852

all by a program with HP 11.

The results are always going to have 1001 = 7x11x13 as a factor.

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