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5 Digit Number II (Posted on 2011-01-03) Difficulty: 4 of 5
In continuation of 5 Digit Number, let us define a 5-digit non leading zero base N (N > 3) positive integer x as a split number whenever, 3*x is a perfect square and, when the digits of x are split, the first number is double the second one.

How many split numbers are there whenever 11 ≤ N ≤ 36. What are the respective minimum and maximum values?

(Splitting a base-N 5-digit number into two numbers means 12345 into 1 and 2345 or, 123 and 45.)

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): A start. Comment 4 of 4 |
(In reply to re: A start. by Steve Herman)

Sorry if I wasn't clear.  When I wrote:

x = 10000a + 1000b + 100c + 10d + e

My intent was to interpret these numbers as all being in base N:
10 = N
100 = N^2
1000 = N^3
10000 = N^4 

It was a bit of a shorthand that served until I started considering divisibility.

You can restate x = 2001d + 201e as
x = (2N^3 + 1)d + (2N^2 + 1)e
and the divisibility rules might be more clear.  This seems like a dead end though because it doesn't rule out much. 
There are few enough numbers to check anyway that a program like the one Charlie wrote can just churn through all d and e from 0 to N-1 and not worry about just checking values that will force x to be divisible by 3.

I will point out that the most restrictions occurred when N was divisible by 3.  Most of the bases divisible by 3 have no solutions (all except N=12)

  Posted by Jer on 2011-01-04 11:04:31

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