Puzzle for August 3, 2020  ( )

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

eq.1) A + B + C + D + E + F = 30 eq.2) B + C = A + E eq.3) C = D + E eq.4) D + E + F = A + C eq.5) A + C – D = D + F eq.6) B + (C ÷ D) = C

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

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Hint #1


In eq.4, replace D + E with C (from eq.3): C + F = A + C Subtract C from each side of the above equation: C + F – C = A + C – C which makes F = A


  

Hint #2


In eq.5, replace F with A: A + C – D = D + A In the equation above, subtract A from both sides, and add D to both sides: A + C – D – A + D = D + A – A + D which makes C = 2×D


  

Hint #3


In eq.3, substitute 2×D for C: 2×D = D + E Subtract D from both sides of the equation above: 2×D – D = D + E – D which makes D = E


  

Hint #4


Substitute 2×D for C in eq.6: B + (2×D ÷ D) = 2×D which becomes B + 2 = 2×D Subtract 2 from both sides of the above equation: B + 2 – 2 = 2×D – 2 which makes eq.6a) B = 2×D – 2


  

Hint #5


Substitute 2×D – 2 for B (from eq.6a), 2×D for C, and D for E in eq.2: 2×D – 2 + 2×D = A + D which becomes 4×D – 2 = A + D Subtract D from both sides of the above equation: 4×D – 2 – D = A + D – D which makes 3×D – 2 = A and also makes eq.2a) F = A = 3×D – 2


  

Solution

Substitute 3×D – 2 for A and F (from eq.2a), 2×D – 2 for B (from eq.6a), 2×D for C, and D for E in eq.1: 3×D – 2 + 2×D – 2 + 2×D + D + D + 3×D – 2 = 30 which simplifies to 12×D – 6 = 30 Add 6 to both sides of the above equation: 12×D – 6 + 6 = 30 + 6 which makes 12×D = 36 Divide both sides by 12: 12×D ÷ 12 = 36 ÷ 12 which means D = 3 making A = F = 3×D – 2 = 3×3 – 2 = 9 – 2 = 7 (from eq.2a) B = 2×D – 2 = 2×3 – 2 = 6 – 2 = 4 (from eq.6a) C = 2×D = 2 × 3 = 6 E = D = 3 and ABCDEF = 746337