Problem 45249. Frugal number

Created by Asif Newaz

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Last Solution submitted on Oct 27, 2022

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3 Comments

Karl Ezra Pilario on 22 Apr 2020

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?

Rafael S.T. Vieira on 21 Sep 2020

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.

Rafael S.T. Vieira on 21 Sep 2020

I recommend including the number 414720 to the test suite: 3^4*2^10*5.

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Problem 45249. Frugal number

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Problem 45249. Frugal number

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