Puzzle for June 12, 2020  ( )

Scratchpad

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

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

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

Scratchpad

 

Help Area

Hint #1


In eq.4, replace C + E with D + F (from eq.1): A + D = D + F + F which becomes A + D = D + 2×F Subtract D from both sides of the above equation: A + D – D = D + 2×F – D which simplifies to A = 2×F


  

Hint #2


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


  

Hint #3


Subtract the left and right sides of eq.3a from the left and right sides of eq.1, respectively: D + F – (D – 5×F) = C + E – (E – C) which is equivalent to D + F – D + 5×F = C + E – E + C which becomes 6×F = 2×C Divide both sides of the above equation by 2: 6×F ÷ 2 = 2×C ÷ 2 which makes 3×F = C


  

Hint #4


In eq.2, substitute 3×F for C, and 2×F for A: B + 3×F = 2×F + F which becomes B + 3×F = 3×F Subtract 3×F from each side of the equation above: B + 3×F – 3×F = 3×F – 3×F which makes B = 0


  

Hint #5


Substitute 2×F for A, and 3×F for C in eq.5: E = (2×F + 3×F) ÷ F which becomes E = (5×F) ÷ F which makes E = 5


  

Hint #6


eq.6 may be written as: C = (A + D + E) ÷ 3 Substitute 3×F for C, 2×F for A, and 5 for E in the above equation: 3×F = (2×F + D + 5) ÷ 3 Multiply both sides of the above equation by 3: 3 × 3×F = 3 × (2×F + D + 5) ÷ 3 which becomes 9×F = 2×F + D + 5 Subtract 2×F and 5 from both sides: 9×F – 2×F – 5 = 2×F + D + 5 – 2×F – 5 which makes eq.6a) 7×F – 5 = D


  

Solution

Substitute 7×F – 5 for D (from eq.6a), 3×F for C, and 5 for E in eq.1: 7×F – 5 + F = 3×F + 5 which becomes 8×F – 5 = 3×F + 5 In the equation above, add 5 to both sides, and subtract 3×F from both sides: 8×F – 5 + 5 – 3×F = 3×F + 5 + 5 – 3×F which makes 5×F = 10 Divide both sides by 5: 5×F ÷ 5 = 10 ÷ 5 which means F = 2 making A = 2×F = 2 × 2 = 4 C = 3×F = 3 × 2 = 6 D = 7×F – 5 = 7×2 – 5 = 14 – 5 = 9 (from eq.5a) and ABCDEF = 406952