Puzzle for November 17, 2023  ( )

Scratchpad

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

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

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

Scratchpad

 

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Hint #1


Add E and A to both sides of eq.4: B - E + E + A = E - A + E + A which becomes B + A = 2×E which may be written as eq.4a) A + B = 2×E


  

Hint #2


In eq.6, replace A + B with 2×E (from eq.4a): C + D + F = 2×E + E which becomes eq.6a) C + D + F = 3×E


  

Hint #3


eq.1 may be written as: A + B + E + C + D + F = 30 In the equation above, substitute 2×E for A + B (from eq.4a), and 3×E for C + D + F (from eq.6a): 2×E + E + 3×E = 30 which becomes 6×E = 30 Divide both sides by 6: 6×E ÷ 6 = 30 ÷ 6 which makes E = 5


  

Hint #4


Substitute 5 for E in eq.2: eq.2a) F = C + 5


  

Hint #5


Substitute C + 5 for F (from eq.2a), and 5 for E into eq.6a: C + D + C + 5 = 3×5 which becomes 2×C + D + 5 = 15 Subtract 2×C and 5 from both sides of the above equation: 2×C + D + 5 - 2×C - 5 = 15 - 2×C - 5 which becomes eq.6b) D = 10 - 2×C


  

Hint #6


eq.1 may be written as: A + D + B + C + E + F = 30 Substitute B for A + D (from eq.3), 5 for E, and C + 5 for F (from eq.2a) into the equation above: B + B + C + 5 + C + 5 = 30 which becomes 2×B + 2×C + 10 = 30 Subtract 10 from both sides: 2×B + 2×C + 10 - 10 = 30 - 10 which becomes 2×B + 2×C = 20 which may be written as eq.1a) 2×(B + C) = 20


  

Hint #7


Divide both sides of eq.1a by 2: 2×(B + C) ÷ 2 = 20 ÷ 2 which becomes B + C = 10 Subtract C from each side of the above equation: B + C - C = 10 - C which becomes eq.1b) B = 10 - C


  

Hint #8


Substitute 10 - C for B (from eq.1b), and 5 for E in eq.4a: A + 10 - C = 2×5 which becomes A + 10 - C = 10 In the equation above, subtract 10 from both sides, and add C to both sides: A + 10 - C - 10 + C = 10 - 10 + C which makes A = C


  

Hint #9


Substitute 5 for E, (C + 5) for F (from eq.2a), 10 - C for B (from eq.1b), and 10 - 2×C for D (from eq.6b) in eq.5: 5 + (C + 5) = 10 - C + 10 - 2×C - (C + 5) which becomes 10 + C = 20 - 3×C - C - 5 which becomes 10 + C = 15 - 4×C In the above equation, subtract 10 from both sides, and add 4×C to both sides: 10 + C - 10 + 4×C = 15 - 4×C - 10 + 4×C which makes 5×C = 5 Divide both sides by 5: 5×C ÷ 5 = 5 ÷ 5 which makes C = 1


  

Solution

Since C = 1, then: A = C = 1 B = 10 - C = 10 - 1 = 9 (from eq.1b) D = 10 - 2×C = 10 - 2×1 = 10 - 2 = 8 (from eq.6b) F = C + 5 = 1 + 5 = 6 (from eq.2a) and ABCDEF = 191856