Puzzle for July 17, 2022  ( )

Scratchpad

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

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

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

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Hint #1


In eq.5, replace A with D + F (from eq.3), and C with B + F (from eq.2): B + D + E = D + F + B + F – E which becomes B + D + E = D + 2×F + B – E In the above equation, subtract B and D from both sides, and add E to both sides: B + D + E – B – D + E = D + 2×F + B – E – B – D + E which simplifies to 2×E = 2×F Divide both sides by 2: 2×E ÷ 2 = 2×F ÷ 2 which makes E = F


  

Hint #2


Add E and D to both sides of eq.4: C – E + E + D = F – D + E + D which becomes C + D = F + E In the equation above, replace F with E: C + D = E + E which becomes eq.4a) C + D = 2×E


  

Hint #3


Subtract D from each side of eq.3: A – D = D + F – D which becomes A – D = F In eq.2, substitute A – D for F: C = B + A – D Add D to both sides of the above equation: C + D = B + A – D + D which becomes C + D = B + A which is equivalent to eq.3a) C + D = A + B


  

Hint #4


In eq.3a, substitute 2×E for C + D (from eq.4a): eq.3b) 2×E = A + B


  

Hint #5


Substitute 2×E for A + B (from eq.3b) and for C + D (from eq.4a), and E for F in eq.1: 2×E + 2×E + E + E = 30 which makes 6×E = 30 Divide both sides of the above equation by 6: 6×E ÷ 6 = 30 ÷ 6 which makes E = 5 and also makes F = E = 5


  

Hint #6


Substitute D + F for A (from eq.3) in eq.3b: eq.3c) 2×E = D + F + B


  

Hint #7


Substitute 5 for E and F in eq.3c: 2×5 = D + 5 + B which becomes 10 = D + 5 + B Subtract 5 from each side of the equation above: 10 – 5 = D + 5 + B – 5 which becomes eq.3d) 5 = D + B


  

Hint #8


eq.1 may be written as: A + C + D + B + E + F = 30 Substitute 5 for D + B (from eq.3d), and 5 for E and F in the equation above: A + C + 5 + 5 + 5 = 30 which becomes A + C + 15 = 30 Subtract 15 from each side of the equation above: A + C + 15 – 15 = 30 – 15 which becomes eq.3e) A + C = 15


  

Hint #9


Substitute 5 for E, and 15 for A + C (from eq.3e) in eq.6: D × 5 = 15 Divide both sides of the equation above by 5: (D × 5) ÷ 5 = 15 ÷ 5 which makes D = 3


  

Hint #10


Substitute 3 for D in eq.3d: 5 = 3 + B Subtract 3 from each side of the equation above: 5 – 3 = 3 + B – 3 which makes 2 = B


  

Hint #11


Substitute 3 for D, and 5 for F in eq.3: A = 3 + 5 which makes A = 8


  

Solution

Substitute 2 for B, and 5 for F in eq.2: C = 2 + 5 which makes C = 7 and makes ABCDEF = 827355