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How do we find how many values for a Smooth square root?

Mathematica Asked on April 26, 2021

this question is from the Eulers project
The positive integer is called square root smooth if all of the prime factors it has are less than its square root.
Including the number 1, there are 29 square rooot smooth numbers not exceeding 100.

this is what I have so far. but i don’t understand how to check for each value. How can I prove in Mathematica that there are exactly 29 square root numbers that don’t exceed 100?

 g = (n^(1/2)) /. n -> Range[1, 100];
 j = Simplify[g]
 k = Primes[n^(1/2)) /. n -> Range[1, 100], g <= n^(1/2))]

One Answer

Use Select

srs = Select[Range[100], 
  And @@ Thread[FactorInteger[#][[All, 1]] < Sqrt[#]] &]

{8, 12, 16, 18, 24, 27, 30, 32, 36, 40, 45, 48, 50, 54, 56, 60, 63, 64, 70, 
72, 75, 80, 81, 84, 90, 96, 98, 100}

Length@srs

(* 28 *)

With 1 the count is 29

Correct answer by Bob Hanlon on April 26, 2021

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