Puzzle for September 24, 2021  ( )

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

eq.1) A + B + C + D + E + F = 26 eq.2) F – D = E eq.3) B + D = A – B + F eq.4) A – C + F = B – E eq.5) E + F – A = A + B – D eq.6) A + C – E = E + F – A

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

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Hint #1


Add E and A to both sides of eq.6: A + C – E + E + A = E + F – A + E + A which becomes eq.6a) 2×A + C = 2×E + F   Subtract F from both sides of eq.4: A – C + F – F = B – E – F which becomes eq.4a) A – C = B – E – F


  

Hint #2


Add the left and right sides of eq.4a to the left and right sides of eq.6a, respectively: 2×A + C + A – C = 2×E + F + B – E – F which becomes eq.6b) 3×A = E + B


  

Hint #3


Add B to both sides of eq.3: B + D + B = A – B + F + B which becomes eq.3a) 2×B + D = A + F   Add D to both sides of eq.2: F – D + D = E + D which becomes eq.2a) F = E + D


  

Hint #4


In eq.3a, replace F with E + D (from eq.2a): 2×B + D = A + E + D Subtract D and A from each side of the equation above: 2×B + D – D – A = A + E + D – D – A which becomes eq.3b) 2×B – A = E


  

Hint #5


In eq.6b, substitute 2×B – A for E (from eq.3b): 3×A = 2×B – A + B which becomes 3×A = 3×B – A Add A to both sides of the above equation: 3×A + A = 3×B – A + A which becomes 4×A = 3×B Divide both sides by 4: 4×A ÷ 4 = 3×B ÷ 4 which makes A = ¾×B


  

Hint #6


Substitute ¾×B for A into eq.3b: 2×B – ¾×B = E which makes 1¼×B = E


  

Hint #7


Substitute 1¼×B for E in eq.2a: eq.2b) F = 1¼×B + D


  

Hint #8


Substitute 1¼×B for E, 1¼×B + D for F (from eq.2b), and ¾×B for A in eq.5: 1¼×B + 1¼×B + D – ¾×B = ¾×B + B – D which becomes 1¾×B + D = 1¾×B – D Subtract 1¾×B from each side of the equation above: 1¾×B + D – 1¾×B = 1¾×B – D – 1¾×B which makes D = –D which means D = 0


  

Hint #9


Substitute 0 for D in eq.2b: F = 1¼×B + 0 which means F = 1¼×B


  

Hint #10


Substitute ¾×B for A, and 1¼×B for E and F in eq.6a: 2×(¾×B) + C = 2×(1¼×B) + 1¼×B which becomes 1½×B + C = 2½×B + 1¼×B which becomes 1½×B + C = 3¾×B Subtract 1½×B from both sides of the equation above: 1½×B + C – 1½×B = 3¾×B – 1½×B which makes C = 2¼×B


  

Solution

Substitute ¾×B for A, 2¼×B for C, 0 for D, and 1¼×B for E and F in eq.1: ¾×B + B + 2¼×B + 0 + 1¼×B + 1¼×B = 26 which simplifies to 6½×B = 26 Divide both sides of the above equation by 6½: 6½×B ÷ 6½ = 26 ÷ 6½ which means B = 4 making A = ¾×B = ¾ × 4 = 3 C = 2¼×B = 2¼ × 4 = 9 E = F = 1¼×B = 1¼ × 4 = 5 and ABCDEF = 349055