Two Six-digit Numbers

by nsato, Dec 7, 2010, 7:10 PM

Here is a nice number theory problem to end off the year: Find two six-digit numbers $A$ and $B$ , such that the 12-digit number obtained from concatenating $A$ and $B$ is divisible by the product $AB$ . The answer is unique.

math

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Interesting.

EDIT: Hide tags seem to not be working... we'll just post this outside of them.

Solution with proof of uniqueness: Let $k$ be the ratio of divisibility between $AB$ and the concatenation of $AB$ . Writing out equations and doing simple rearrangements gives $10^6 = (Ak - 1) (B/A)$ . Since $(Ak-1,A) = 1$ , $B/A$ is an integer, and since both numerator and denominator are six digits $B/A$ is a one digit number. It's also a divisor of $10^6$ . Finally, observe that if $k = 1$ , $B > 10^6$ , so $k > 1$ . Casework time.

$B/A = 2$ ? Then $Ak = 500001$ . Notice that $k = 3$ and $A = 166667$ satisfy this, giving $B = 333334$ . This is the solution. Testing all other $1 < k \leq 5$ shows there are no other nontrivial six digit divisors of $500001$ .

$B/A$ some other single digit divisor of $10^6$ ? Then $Ak = 1000001, 250001, 200001,$ or $125001$ . By using the fact that $k > 1$ and $k$ is bounded above by $10$ in the first case and $2$ in the others, we can fairly quickly determine that there are no solutions here.

Seems to me that six is not so special; as long as $10^n + 1$ is not divisible by seven and $n$ is not too small the same argument should give a unique solution.

This post has been edited 4 times. Last edited by MellowMelon, Dec 7, 2010, 8:02 PM

by MellowMelon, Dec 7, 2010, 7:34 PM

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If I knew what concatenating was I'd be glad to try, but....

by bluecarneal, Dec 7, 2010, 8:27 PM

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@bluecarneal

concatenation is just a fancy term for putting two strings together, for example the number formed from the concatenation of 56 and 49 is 5649.

by fortenforge, Dec 7, 2010, 8:32 PM

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Art of Problem Solving Blog

Two Six-digit Numbers

by nsato, Dec 7, 2010, 7:10 PM

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