Giza IQ-Test - Solution 11

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 (m₁ + 1) = 5 => m₁ = 4, m₂ = 4, and all other m_k = 4

That works. From

     m₁ = 1 + 1/5*(m - 1)

you can calculate that m = 16. Therefore, n = 4.

If (m₁ + 1) = 10 => m₁ = 9, m₂ = 8, but m₃ = 7.2

If (m₁ + 1) = 15 => m₁ = 14, m₂ = 12, but m₃ = 10.4

If (m₁ + 1) = 20 => m₁ = 19, m₂ = 16, but m₃ = 13.6

If (m₁ + 1) = 25 => m₁ = 24, m₂ = 20, but m₃ = 16.8

If (m₁ + 1) = 30 => m₁ = 29, m₂ = 24, m₃ = 20, but m₄ = 16.8

If (m₁ + 1) = 35 => m₁ = 34, m₂ = 28, but m₃ = 23.2

Increasing the factor that multiplies 5 sometimes pushes the fractional m_ks 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 m₁ up to more than 800.