Puzzle for August 7, 2019  ( )

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

eq.1) A + B + C + D + E + F = 29 eq.2) C = A + D eq.3) F = B + C eq.4) A + B + C = E eq.5) E – B = F – D eq.6) B × D = C + F

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

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Hint #1


In eq.4, replace B + C with F (from eq.3): eq.4a) A + F = E


  

Hint #2


In eq.5, replace E with A + F (from eq.4a): A + F – B = F – D Subtract F from both sides of the above equation: A + F – B – F = F – D – F which becomes A – B = –D Add B and D to both sides: A – B + B + D = –D + B + D which becomes eq.5a) A + D = B


  

Hint #3


In eq.2, replace A + D with B (from eq.5a): C = B


  

Hint #4


In eq.3, substitute C for B: F = C + C which makes F = 2×C


  

Hint #5


Substitute C for B, and 2×C for F in eq.6: C × D = C + 2×C which makes C × D = 3×C Divide both sides of the above equation by C: C × D ÷ C = 3×C ÷ C which makes D = 3


  

Hint #6


Substitute 3 for D in eq.2: C = A + 3 Subtract 3 from each side: C – 3 = A + 3 – 3 which makes eq.2a) C – 3 = A


  

Hint #7


Substitute C – 3 for A (from eq.2a), and C for B in eq.4: C – 3 + C + C = E which becomes eq.4a) 3×C – 3 = E


  

Solution

Substitute C – 3 for A (from eq.2a), C for B, 3 for D, 3×C – 3 for E (from eq.4a), and 2×C for F in eq.1: C – 3 + C + C + 3 + 3×C – 3 + 2×C = 29 which simplifies to 8×C – 3 = 29 Add 3 to both sides of the above equation: 8×C – 3 + 3 = 29 + 3 which makes 8×C = 32 Divide both sides by 8: 8×C ÷ 8 = 32 ÷ 8 which means C = 4 making A = C – 3 = 4 – 3 = 1 (from eq.2a) B = C = 4 E = 3×C – 3 = 3×4 – 3 = 12 – 3 = 9 (from eq.4a) F = 2×C = 2×4 = 8 and ABCDEF = 144398