(Contest). Try to solve it without Google.

[tsh73]

It is high-school test problem my kid got yesterday
I jumped in - but after getting something working... I Googled and it happened that I solved quite anoter task (Doh).

So here is it (Google-translated text, original is in Russian)

Let's call a nontrivial divisor of a natural number its divisor, which is not equal to one and the number itself. For example, the number 6 has two nontrivial divisors: 2 and 3. Find all natural numbers belonging to the segment [123456789; 223456789] and having exactly three nontrivial divisors. For each found number, write down its largest nontrivial divisor in the answer. Arrange the answers in ascending order.

Could you solve it without Google?

[bplus]

' What kind of number has exactly 3 divisors???

row = 1
For i = 2 To 2500
   scan
   Locate 1, row: Print i;"  "; ' <<< progress report OMG! This is going to take far too long for 123456789+ numbers
   test = ThreeDivisors(i)
   If test Then
       Locate 1, row
       For j = 2 To i / 2 'show divisors
           scan
           If i Mod j = 0 Then 'j divides i
               Print " ";j;
           End If
       Next
       Print " divides "; i
       row = row + 1
   End If
Next
Locate 1, row +1
Print "End of run."

Function ThreeDivisors (n)
   For i = 2 To n / 2 ' count divisors
       scan
       If n Mod i = 0 Then 'i divides n
           count = count + 1: lastDivisor = i
           If count > 3 Then Exit Function ' too  many divisors
       End If
   Next
   If count = 3 Then ThreeDivisors = lastDivisor
End Function


p, p^2, p^3 divides p^4 where p is prime.

OK so what integers to power of 4 are between 123456789 and 223456789
' what is the first integer ^ 4 >= 123456789
i = 2
While i ^ 4 < 123456789
   i = i + 1
Wend
a = i ^ 4
Print i, a

' what is the last integer i st i ^ 4 <= 223456789
While i ^ 4 <= 223456789
   i = i + 1
Wend
'Ok backup i by 1
a = (i - 1) ^ 4
Print i - 1, a


OK so here is the solution:

For n = 106 To 122 ' n^4 > 123456789 and n^4 < 223456789
   'if n is prime print the divisors
   Prime = -1
   For j = 2 To Sqr(n)
       If n Mod j = 0 Then Prime = 0: Exit For
   Next
   If Prime Then Print n; ", "; n ^ 2; ", "; n ^ 3; " divides "; n ^ 4
Next
Print



[tsh73]

Yes that's it.
Interesting that if you think before, you could solve it with calculator.
But if you don't - 100 000 000 numbers is very big chank to test...
(and if you do not comprehend the task you could get to know some useless things.
Like, there are some 19 000 000 numbers in that region having exactly 3 primes in factorisation.
Or using // instead if / speeds up Python.
Or using Python 2 instead of Python 3 speeds it up about 2x.
Or using PyPy you could indeed check all these numbers overnight.
But it does not get task solved.
)