Truncate The Square
Does there exist a positive integer n and digits a 1 = 0 , a 2 , … , a n such that
a 1 a 2 … a n − 1 a n − a 1 a 2 … a n − 1 = a n ?
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2 solutions
Are there any other solutions? / How can we find all solutions?
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Interesting question... At the moment I'm trying to figure out the number of solutions...I couldn't solve it yet. Do you know the number of solutions?
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With the "right" interpretation, it's a pretty simple problem.
I just got back from vacation. Give me some time to catch up on emails.
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@Calvin Lin – You can just convert the given equation into an algebraic expression: sqrt(10x + y) - sqrt(x) = y, and we want to find positive integers to this equation with 0<y<10.
A simple trial and error shows that this is the only solution.
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@Pi Han Goh – Indeed. Introducing the variables a 1 , … a n just complicates the problem, though it has a "nice" structure to it.
The clever switch to 1 0 x + y − x = y makes this into a much more 'standard' problem, esp since there is a unique x for each y .
Otherwise did you get my email?
[This is not a complete solution.]
Hint: a n ∈ { 0 , 1 , 4 , 5 , 6 , 9 } .
A possible solution:
⎩ ⎪ ⎨ ⎪ ⎧ a 1 = 1 a 2 = 6 a 3 = 9
Hence, a 1 a 2 a 3 − a 1 a 2 = 1 6 9 − 1 6 = 1 3 − 4 = 9 = a 3