Thanks for the explanation Don! It looks like I need better math-foo to
completely understand your explanation...
On Tuesday, February 11, 2014 10:07:05 AM UTC-8, Don wrote:
>
>
> It can be shown mathematically that if you use a multiplier A such that
> A*2^15-1 and A*2^16-1 are both prime, you
I like that Python code, and for three digits numbers it is just fine.
If the number could be in a larger range, factoring out digits starting
with larger digits should give the correct sequence, in reverse. If the
number has a prime factor larger than 9, there is no solution.
// Returns the sm
Since the limit is so small, won't a simple bruteforce do ?
Just run a loop from [100, 1000) and return the minimum.
Here is a quick python code I wrote.
>>> N = 24
>>> min (i for i in range (100, 1001) if int (str (i) [0]) * int (str (i)
[1]) * int (str (i) [2]) == N)
138
>>> N = 36
>>> min (i
Given a number N, find the smallest 3 digits number such that product of
its digits is equal to N.
For Eg:-
for input 24 , output :138 (1*3*8 = 24)
for input 36 , output :149 (1*4*9 = 36)
for input 100 , output : 455 (4*5*5 = 100)
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