Puzzle for August 7, 2020  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 27 eq.2) B = C – F eq.3) A + D + F = B – F eq.4) C – A = A – D eq.5) E + F = A + B + C + D – E eq.6) average (A, B) = average (C, E, F)

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

Scratchpad

 

Help Area

Hint #1


eq.6 may be written as: (A + B) ÷ 2 = (C + E + F) ÷ 3 Multiply both sides of the equation above by 6: 6 × (A + B) ÷ 2 = 6 × (C + E + F) ÷ 3 which becomes 3 × (A + B) = 2 × (C + E + F) which may be written as eq.6a) 3×A + 3×B = 2×C + 2×E + 2×F


  

Hint #2


Add A and D to both sides of eq.4: C – A + A + D = A – D + A + D which becomes C + D = 2×A In eq.5, replace C + D with 2×A: E + F = A + B + 2×A – E which becomes E + F = 3×A + B – E In the above equation, subtract F from each side, and add E to each side: E + F – F + E = 3×A + B – E – F + E which becomes eq.5a) 2×E = 3×A + B – F


  

Hint #3


In eq.6a, replace 2×E with 3×A + B – F (from eq.5a): 3×A + 3×B = 2×C + 3×A + B – F + 2×F which becomes 3×A + 3×B = 2×C + 3×A + B + F Subtract 3×A and B from both sides of the above equation: 3×A + 3×B – 3×A – B = 2×C + 3×A + B + F – 3×A – B which becomes eq.6b) 2×B = 2×C + F


  

Hint #4


In eq.6b, substitute (C – F) for B (from eq.2): 2×(C – F) = 2×C + F which is equivalent to 2×C – 2×F = 2×C + F In the equation above, subtract 2×C from both sides, and add 2×F to both sides: 2×C – 2×F – 2×C + 2×F = 2×C + F – 2×C + 2×F which makes 0 = 3×F which means 0 = F


  

Hint #5


Substitute 0 for F in eq.2: B = C – 0 which makes B = C


  

Hint #6


Substitute B for C in eq.4: B – A = A – D Add A to both sides of the equation above: B – A + A = A – D + A which becomeq eq.4a) B = 2×A – D


  

Hint #7


Substitute 0 for F, and 2×A – D for B (from eq.4a) in eq.3: A + D + 0 = 2×A – D – 0 which becomes A + D = 2×A – D In the above equation, subtract A from both sides, and add D to both sides: A + D – A + D = 2×A – D – A + D which makes 2×D = A


  

Hint #8


Substitute (2×D) for A in eq.4a: B = 2×(2×D) – D which becomes B = 4×D – D which makes B = 3×D and also makes C = B = 3×D


  

Hint #9


Substitute 2×D for A, 3×D for B, and 0 for F in eq.5a: 2×E = 3×(2×D) + 3×D – 0 which becomes 2×E = 6×D + 3×D which makes 2×E = 9×D Divide both sides of the above equation by 2: 2×E ÷ 2 = 9×D ÷ 2 which makes E = 4½×D


  

Solution

Substitute 2×D for A, 3×D for B and C, 4½×D for E, and 0 for F in eq.1: 2×D + 3×D + 3×D + D + 4½×D + 0 = 27 which simplifies to 13½×D = 27 Divide both sides of the above equation by 13½: 13½×D ÷ 13½ = 27 ÷ 13½ which means D = 2 making A = 2×D = 2 × 2 = 4 B = C = 3×D = 3 × 2 = 6 E = 4½×D = 4½ × 2 = 9 and ABCDEF = 466290