Puzzle for April 4, 2020  ( )

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

eq.1) A + B + C + D + E + F = 23 eq.2) E + F = B + D eq.3) F = A – D eq.4) C + E = F eq.5) B + D + E = A eq.6)* AB + C – DE = CD + EF

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

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Hint #1


Add D to both sides of eq.3: F + D = A – D + D which becomes F + D = A In eq.5, replace A with F + D: B + D + E = F + D Subtract D from each side of the above equation: B + D + E – D = F + D – D which becomes eq.5a) B + E = F


  

Hint #2


In eq.4, replace F with B + E (from eq.5a): C + E = B + E Subtract E from both sides of the equation above: C + E – E = B + E – E which makes C = B


  

Hint #3


In eq.2, substitute B + E for F (from eq.5a): E + B + E = B + D Subtract B from each side of the equation above: E + B + E – B = B + D – B which makes 2×E = D


  

Hint #4


eq.6 may be written as: 10×A + B + C – (10×D + E) = 10×C + D + 10×E + F which is the same as 10×A + B + C – 10×D – E = 10×C + D + 10×E + F In the above equation, add E to both sides, and subtract C and D from both sides: 10×A + B + C – 10×D – E + E – C – D = 10×C + D + 10×E + F + E – C – D which becomes eq.6a) 10×A + B – 11×D = 9×C + 11×E + F


  

Hint #5


Substitute (2×E) for D, and B for C in eq.6a: 10×A + B – 11×(2×E) = 9×B + 11×E + F which is equivalent to 10×A + B – 22×E = 9×B + 11×E + F In the above equation, subtract B from each side, and add 22×E to each side: 10×A + B – 22×E – B + 22×E = 9×B + 11×E + F – B + 22×E which becomes 10×A = 8×B + 33×E + F which may be written as eq.6b) 10×A = 8×(B + E) + 25×E + F


  

Hint #6


Substitute F for B + E (from eq.5a) in eq.6b: 10×A = 8×(F) + 25×E + F which becomes eq.6c) 10×A = 25×E + 9×F


  

Hint #7


Substitute 2×E for D in eq.3: eq.3a) F = A – 2×E


  

Hint #8


Substitute (A – 2×E) for F (from eq.3a) in eq.6c: 10×A = 25×E + 9×(A – 2×E) which is equivalent to 10×A = 25×E + 9×A – 18×E which becomes 10×A = 7×E + 9×A Subtract 9×A from each side of the equation above: 10×A – 9×A = 7×E + 9×A – 9×A which makes A = 7×E


  

Hint #9


Substitute 7×E for A in eq.3a: F = 7×E – 2×E which makes F = 5×E


  

Hint #10


Substitute 5×E for F in eq.4: C + E = 5×E Subtract E from each side of the above equation: C + E – E = 5×E – E which makes C = 4×E and also makes B = C = 4×E


  

Solution

Substitute 7×E for A, 4×E for B and C, 2×E for D, and 5×E for F in eq.1: 7×E + 4×E + 4×E + 2×E + E + 5×E = 23 which simplifies to 23×E = 23 Divide both sides of the equation above by 23: 23×E ÷ 23 = 23 ÷ 23 which means E = 1 making A = 7×E = 7 × 1 = 7 B = C = 4×E = 4 × 1 = 4 D = 2×E = 2 × 1 = 2 F = 5×E = 5 × 1 = 5 and ABCDEF = 744215