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A hairy problem (Posted on 2019-05-08) Difficulty: 3 of 5
As we all know, human hair grows at a rate of 0.35 mm per day.

On December 31, 2018 Darren, Karen, and Sharon all shaved their heads. Furthermore, whenever Darren’s hair grew to 17mm he shaved it again. Karen did the same thing after her hair reached 20mm, and Sharon did likewise when hers was 21mm.

One day in 2019, Aaron goes to visit Darren, Karen, and Sharon. He finds that Darren is clean shaven, Karen’s hair is 19mm long, and Sharon’s hair is 14mm long.

What is the date of Aaron’s visit?

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution computer solution with assumptions Comment 1 of 1
The program below finds that the answer is December 6, the 340th day of the year, assuming that:

1. Darren, Karen and Sharon all shaved their heads at the same time on December 31, 2018.

2. They each shaved their respective heads at the exact time of day the respective length was achieved, even if this were in the middle of the night. This is the major sticking point; if they waited to check their hair length at, say, the same time as their original shave and did another shave then if the length first met or exceeded the cutoff, there would be no solution.

3. Hair growth is constant during the day.

4. Aaron's visit was at the same time of day as D, K and S's original shaving.

The fact that we get an answer seems to verify these extra facts were in fact the case. 

 molen = Array(0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31)
 mo = 1
 DoM = 0
 For da = 1 To 365
   DoM = DoM + 1
   If DoM > molen(mo) Then
     DoM = DoM - molen(mo)
     mo = mo + 1
   End If
   d = d + 35: k = k + 35: s = s + 35
   If d >= 1700 Then d = d - 1700
   If k >= 2000 Then k = k - 2000
   If s >= 2100 Then s = s - 2100
   If d = 0 And k = 1900 And s = 1400 Then
     Text1.Text = Text1.Text & da & Str(mo) & Str(DoM) & crlf
   End If

  Posted by Charlie on 2019-05-08 17:22:51
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