Puzzle for November 3, 2019  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 24 eq.2) B + E = C eq.3) C = A + D + F eq.4) D + E = A + B eq.5) E + F = A + C + D

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

Scratchpad

 

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Hint #1


eq.1 may be written as: A + D + F + B + E + C = 24 In the equation above, substitute C for both A + D + F (from eq.3) and for B + E (from eq.2): C + C + C = 24 which means 3×C = 24 Divide both sides of the above equation by 3: 3×C ÷ 3 = 24 ÷ 3 which makes C = 8


  

Hint #2


Subtract F from both sides of eq.3: C – F = A + D + F – F which becomes eq.3a) C – F = A + D   Subtract A, D, and F from both sides of eq.5: E + F – A – D – F = A + C + D – A – D – F which becomes eq.5a) E – A – D = C – F


  

Hint #3


In eq.5a, substitute A + D for C – F (from eq.3a): E – A – D = A + D Add A and D to both sides of the equation above: E – A – D + A + D = A + D + A + D which becomes eq.5b) E = 2×A + 2×D


  

Hint #4


Subtract E from both sides of eq.2: B + E – E = C – E which becomes B = C – E In the above equation, substitute 8 for C, and (2×A + 2×D) for E (from eq.5b): B = 8 – (2×A + 2×D) which is equivalent to eq.2a) B = 8 – 2×A – 2×D


  

Hint #5


Substitute 2×A + 2×D for E (from eq.5b) in eq.4: D + 2×A + 2×D = A + B which becomes 2×A + 3×D = A + B Subtract A from both sides of the equation above: 2×A + 3×D – A = A + B – A which becomes eq.4a) A + 3×D = B


  

Hint #6


Substitute A + 3×D for B (from eq.4a) in eq.2a: A + 3×D = 8 – 2×A – 2×D In the above equation, add 2×D to both sides, and subtract A from each side: A + 3×D + 2×D – A = 8 – 2×A – 2×D + 2×D – A which becomes 5×D = 8 – 3×A Divide both sides by 5: 5×D ÷ 5 = (8 – 3×A) ÷ 5 which becomes eq.2b) D = (8 – 3×A) ÷ 5


  

Hint #7


To make eq.2b true, check several possible values for A and D:   If A = 0, then D = (8 – 3×0) ÷ 5 = (8 – 0) ÷ 5 = 8 ÷ 5 = 1.6 If A = 1, then D = (8 – 3×1) ÷ 5 = (8 – 3) ÷ 5 = 5 ÷ 5 = 1 If A = 2, then D = (8 – 3×2) ÷ 5 = (8 – 6) ÷ 5 = 2 ÷ 5 = 0.4 If A = 3, then D = (8 – 3×3) ÷ 5 = (8 – 9) ÷ 5 = –1 ÷ 5 = –0.2 If A > 3, then D < –0.2   Since D must be a one-digit non-negative integer, then D = 1 which makes A = 1


  

Hint #8


Substitute 1 for A and D in eq.5b: E = 2×1 + 2×1 which becomes E = 2 + 2 which means E = 4


  

Hint #9


Substitute 1 for A and D in eq.4a: 1 + 3×1 = B which becomes 1 + 3 = B which makes 4 = B


  

Solution

Substitute 8 for C, and 1 for A and D in eq.3: 8 = 1 + 1 + F which becomes 8 = 2 + F Subtract 2 from both sides of the equation above: 8 – 2 = 2 + F – 2 which makes 6 = F and ABCDEF = 148146