Puzzle for June 23, 2022  ( )

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

eq.1) A + B + C + D + E + F = 31 eq.2) F = C + E eq.3) A = D + E eq.4) D = average (A, C, F) eq.5)* AB = B + CD – A eq.6) E + F – B = B + (A ÷ C)

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  AB and CD are 2-digit numbers (not A×B or C×D).

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Hint #1


eq.4 may be written as: D = (A + C + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × D = 3 × (A + C + F) ÷ 3 which becomes eq.4a) 3×D = A + C + F


  

Hint #2


In eq.4a, replace A with D + E (from eq.3), and F with C + E (from eq.2): 3×D = D + E + C + C + E which becomes 3×D = D + 2×E + 2×C Subtract D from each side of the equation above: 3×D – D = D + 2×E + 2×C – D which becomes 2×D = 2×E + 2×C Divide both sides by 2: 2×D ÷ 2 = (2×E + 2×C) ÷ 2 which becomes D = E + C which may be written as eq.4b) D = C + E


  

Hint #3


In eq.2, replace C + E with D (from eq.4b): F = D


  

Hint #4


eq.5 may be written as: 10×A + B = B + 10×C + D – A In the above equation, subtract B from both sides, and add A to both sides: 10×A + B – B + A = B + 10×C + D – A – B + A which becomes eq.5a) 11×A = 10×C + D


  

Hint #5


Subtract E from each side of eq.4b: D – E = C + E – E which becomes D – E = C In eq.5a, substitute (D + E) for A (from eq.3), and (D – E) for C: 11×(D + E) = 10×(D – E) + D which becomes 11×D + 11×E = 10×D – 10×E + D which becomes 11×D + 11×E = 11×D – 10×E In the above equation, subtract 11×D from both sides, and add 10×E to both sides: 11×D + 11×E – 11×D + 10×E = 11×D – 10×E – 11×D + 10×E which simplifies to 21×E = 0 which means E = 0


  

Hint #6


Substitute 0 for E in eq.2: F = C + 0 which makes F = C


  

Hint #7


Substitute 0 for E in eq.3: A = D + 0 which makes A = D and also makes A = D = F = C


  

Hint #8


Substitute 0 for E, and A for F and C in eq.6: 0 + A – B = B + (A ÷ A) which becomes A – B = B + 1 Add B to both sides of the equation above: A – B + B = B + 1 + B which makes A = 2×B + 1 and also makes eq.6a) A = D = F = C = 2×B + 1


  

Solution

Substitute 2×B + 1 for A and C and D and F (from eq.6a), and 0 for E in eq.1: 2×B + 1 + B + 2×B + 1 + 2×B + 1 + 0 + 2×B + 1 = 31 which simplifies to 9×B + 4 = 31 Subtract 4 from both sides of the above equation: 9×B + 4 – 4 = 31 – 4 which makes 9×B = 27 Divide both sides by 9: 9×B ÷ 9 = 27 ÷ 9 which means B = 3 making A = C = D = F = 2×B + 1 = 2×3 + 1 = 6 + 1 = 7 (from eq.6a) and ABCDEF = 737707