I'm confused. My solution followed the definition of frugal numbers from https://en.wikipedia.org/wiki/Frugal_number, which includes counting the number of digits in the exponents, and also checks frugality in other bases (not just base 10). But most solutions don't follow this. Can you clarify this?
It seems that the problem is not considering the exponent 1 as a valid digit, For instance, 115248 (6 digits) is called a frugal number in this problem since 115248 (6 digits) > 2^4*3*7^4 (5 digits). If we considered 3^1 as 2 digits instead of the 1-digit 3, 115248 should not be frugal. And the problem should have probably declared this restriction and that the considered base is 10.
PS: The OEIS also imposes this same restriction https://oeis.org/A046759, since 3645 is considered frugal: 3^6*5 instead of 3^6*5^1.
I recommend including the number 414720 to the test suite: 3^4*2^10*5.
Remove the small words from a list of words.
Is my wife right?
Doubling elements in a vector
Connect Four - 03
Combinatorics - 01
Decimal to binary conversion (without using built-in function)
Find the treasures in MATLAB Central and discover how the community can help you!
Choose a web site to get translated content where available and see local events and offers. Based on your location, we recommend that you select: .
You can also select a web site from the following list:
Select the China site (in Chinese or English) for best site performance. Other MathWorks country sites are not optimized for visits from your location.
Contact your local office