Puzzle for April 8, 2022  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 23 eq.2) B + F = D + E eq.3) C – B = B – D + E eq.4) D – A = C – E + F eq.5) E – C = A – B – E eq.6) F ÷ A = C – E

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

Scratchpad

 

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Hint #1


Add B and D to both sides of eq.3: C – B + B + D = B – D + E + B + D which becomes eq.3a) C + D = 2×B + E   Add C, B, and E to both sides of eq.5: E – C + C + B + E = A – B – E + C + B + E which becomes 2×E + B = A + C which may be written as eq.5a) A + C = 2×E + B


  

Hint #2


Subtract the left and right sides of eq.5a from the left and right sides of eq.3a, respectively: C + D – (A + C) = 2×B + E – (2×E + B) which becomes C + D – A – C = 2×B + E – 2×E – B which becomes D – A = B – E Add A and E to both sides of the above equation: D – A + A + E = B – E + A + E which becomes eq.3b) D + E = B + A


  

Hint #3


In eq.2, substitute B + A for D + E (from eq.3b): B + F = B + A Subtract B from both sides of the above equation: B + F – B = B + A – B which makes F = A


  

Hint #4


Substitute A for F in eq.4: D – A = C – E + A Add A and E to both sides of the equation above: D – A + A + E = C – E + A + A + E which becomes eq.4a) D + E = C + 2×A


  

Hint #5


Substitute C + 2×A for D + E (from eq.4a) in eq.3b: C + 2×A = B + A Subtract A from each side of the equation above: C + 2×A – A = B + A – A which becomes C + A = B which is the same as eq.3c) A + C = B


  

Hint #6


Substitute B for A + C (from eq.3c) in eq.5a: B = 2×E + B Subtract B from each side of the above equation: B – B = 2×E + B – B which makes 0 = 2×E which means 0 = E


  

Hint #7


Substitute A for F, and 0 for E in eq.6: A ÷ A = C – 0 which makes 1 = C


  

Hint #8


Substitute 1 for C in eq.3c: eq.3d) A + 1 = B


  

Hint #9


Substitute 0 for E, and A + 1 for B (from eq.3d) in eq.3b: D + 0 = A + 1 + A which becomes eq.3e) D = 2×A + 1


  

Solution

Substitute A + 1 for B (from eq.3d), 1 for C, 2×A + 1 for D (from eq.3e), 0 for E, and A for F in eq.1: A + A + 1 + 1 + 2×A + 1 + 0 + A = 23 which simplifies to 5×A + 3 = 23 Subtract 3 from each side of the above equation: 5×A + 3 – 3 = 23 – 3 which makes 5×A = 20 Divide both sides by 5: 5×A ÷ 5 = 20 ÷ 5 which means A = 4 making B = A + 1 = 4 + 1 = 5 (from eq.3d) D = 2×A + 1 = 2×4 + 1 = 8 + 1 = 9 (from eq.3e) F = A = 4 and ABCDEF = 451904