Puzzle for November 8, 2020  ( )

Scratchpad

Find the 6-digit number ABCDEF by solving the following equations:

eq.1) A – E = B – F eq.2) C – E = A + E eq.3) B + E = A + C eq.4) A × E = B eq.5) D = B ÷ C eq.6) log [base C] (F) = D

A, B, C, D, E, and F each represent a one-digit non-negative integer.

Scratchpad

 

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Hint #1


Add E and F to both sides of eq.1: A – E + E + F = B – F + E + F which becomes A + F = B + E In eq.3, replace B + E with A + F: A + F = A + C Subtract A from both sides of the above equation: A + F – A = A + C – A which makes F = C


  

Hint #2


eq.6 may be re-written as an exponential equation: C ^ D = F ("C ^ D" means "C raised to the power of D") In the equation above, replace F with C: eq.6a) C ^ D = C   To make eq.6a true, then either: C = 0 and D ≠ 0 or: C = 1 or: D = 1


  

Hint #3


First, check: C = 0, and D ≠ 0 ...   Substituting 0 for C in eq.5 would yield: D = B ÷ 0 which would make D = undefined Since D must be an integer, then: D ≠ undefined which means C ≠ 0


  

Hint #4


Next, begin checking: C = 1 ...   Substituting 1 for C in eq.2 would yield: 1 – E = A + E Adding E to both sides of the above equation would yield: 1 – E + E = A + E + E which would make 1 = A + 2×E Since A and E must be non-negative integers, the above equation would make: A = 1 and E = 0


  

Hint #5


Finish checking: C = 1 ...   Substituting 1 for A, and 0 for E in eq.4 would yield: 1 × 0 = B which would make 0 = B Substituting 0 for B and E, and 1 for A in eq.3 would yield: 0 + 0 = 1 + C which would make –1 = C which contradicts C = 1 Therefore C ≠ 1 which means D = 1


  

Hint #6


Substitute 1 for D in eq.5: 1 = B ÷ C Multiply both sides of the equation above by C: 1 × C = B ÷ C × C which makes C = B and also makes F = C = B


  

Hint #7


Substitute B for C in eq.3: B + E = A + B Subtract B from both sides of the equation above: B + E – B = A + B – B which makes E = A


  

Hint #8


Substitute A for E in eq.2: C – A = A + A which becomes C – A = 2×A Add A to both sides of the equation above: C – A + A = 2×A + A which makes C = 3×A and also makes eq.2a) F = B = C = 3×A


  

Hint #9


Substitute A for E, and 3×A for B in eq.4: eq.4a) A × A = 3×A   To make eq.4a true, then either: A = 0 which would make E = A = 0 or: A = 3 which would make E = A = 3


  

Hint #10


Check: E = A = 0 ...   Substituting 0 for A and E in eq.2 would yield: C – 0 = 0 + 0 which would make C = 0 Substituting 0 for C in eq.5 would yield: D = B ÷ 0 which would make D = undefined Since D must be an integer, then: D ≠ undefined which means E and A ≠ 0 and therefore means E = A = 3


  

Solution

Substitute 3 for A in eq.2a: F = B = C = 3×3 which makes F = B = C = 9 and makes ABCDEF = 399139