day k:

m

_{k}= k + 1/5*(m - k - [Sum(m_{j}) j = 1..k-1])day k+1:

m

_{k+1}= k + 1 + 1/5*(m - k - 1 - [Sum(m_{j}) j = 1..k]) = = k + 1 + 1/5*(m - k - 1 - m

_{k}- [Sum(m_{j}) j = 1..k-1]) = = 1 - 1/5 - m

_{k}/5 + k + 1/5*(m - k - [Sum(m_{j}) j = 1..k-1]) = = 1 - 1/5 - m

_{k}/5 + m_{k}= = 4/5*(m

_{k}+ 1)Therefore, all (m

_{k}+ 1) must be a multiple of 5.If (m

_{1}+ 1) = 5 => m_{1}= 4, m_{2}= 4, and all other m_{k}= 4That works. From

m

_{1}= 1 + 1/5*(m - 1)you can calculate that m = 16. Therefore, n = 4.

If (m

_{1}+ 1) = 10 => m_{1}= 9, m_{2}= 8, but m_{3}= 7.2If (m

_{1}+ 1) = 15 => m_{1}= 14, m_{2}= 12, but m_{3}= 10.4If (m

_{1}+ 1) = 20 => m_{1}= 19, m_{2}= 16, but m_{3}= 13.6If (m

_{1}+ 1) = 25 => m_{1}= 24, m_{2}= 20, but m_{3}= 16.8If (m

_{1}+ 1) = 30 => m_{1}= 29, m_{2}= 24, m_{3}= 20, but m_{4}= 16.8If (m

_{1}+ 1) = 35 => m_{1}= 34, m_{2}= 28, but m_{3}= 23.2Increasing the factor that multiplies 5 sometimes pushes the fractional m

_{k}s to higher values of k, but doesn’t eliminate them.This is not foolproof, but a spreadsheet calculation has verified that there are no other solutions for m

_{1}up to more than 800.
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