Puzzle for July 15, 2022  ( )

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

eq.1) A + B + C + D + E + F = 32 eq.2) E – D = D + F eq.3) D + E = B + C + F eq.4) E – B = A + C – E eq.5) C + D + F = A + B – F eq.6) B – D + E = A + D + F

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

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Hint #1


Add D to both sides of eq.2: E – D + D = D + F + D which becomes eq.2a) E = 2×D + F   Add D to both sides of eq.6: B – D + E + D = A + D + F + D which becomes eq.6a) B + E = A + 2×D + F


  

Hint #2


In eq.6a, replace 2×D + F with E (from eq.2a): B + E = A + E Subtract E from each side of the equation above: B + E – E = A + E – E which makes B = A


  

Hint #3


In eq.5, replace B with A: C + D + F = A + A – F which becomes C + D + F = 2×A – F Add F to both sides of the above equation: C + D + F + F = 2×A – F + F which becomes eq.5a) C + D + 2×F = 2×A


  

Hint #4


In eq.4, replace B with A: E – A = A + C – E Add A and E to both sides of the above equation: E – A + A + E = A + C – E + A + E which becomes eq.4a) 2×E = 2×A + C


  

Hint #5


In eq.4a, substitute (2×D + F) for E (from eq.2a), and C + D + 2×F for 2×A (from eq.5a): 2×(2×D + F) = C + D + 2×F + C which becomes 4×D + 2×F = 2×C + D + 2×F Subtract 2×F and D from each side of the equation above: 4×D + 2×F – 2×F – D = 2×C + D + 2×F – 2×F – D which makes 3×D = 2×C Divide both sides by 2: 3×D ÷ 2 = 2×C ÷ 2 which makes 1½×D = C


  

Hint #6


Substitute 2×D + F for E (from eq.2a), and 1½×D for C in eq.3: D + 2×D + F = B + 1½×D + F which becomes 3×D + F = B + 1½×D + F Subtract F and 1½×D from each side of the above equation: 3×D + F – F – 1½×D = B + 1½×D + F – F – 1½×D which makes 1½×D = B and also makes 1½×D = B = A


  

Hint #7


Substitute (1½×D) for C and A in eq.5a: (1½×D) + D + 2×F = 2×(1½×D) which becomes 2½×D + 2×F = 3×D Subtract 2½×D from both sides of the above equation: 2½×D + 2×F – 2½×D = 3×D – 2½×D which makes 2×F = ½×D Divide both sides by 2: 2×F ÷ 2 = ½×D ÷ 2 which makes F = ¼×D


  

Hint #8


Substitute ¼×D for F in eq.2a: E = 2×D + ¼×D which makes E = 2¼×D


  

Solution

Substitute 1½×D for A and B and C, 2¼×D for E, and ¼×D for F in eq.1: 1½×D + 1½×D + 1½×D + D + 2¼×D + ¼×D = 32 which simplifies to 8×D = 32 Divide both sides of the above equation by 8: 8×D ÷ 8 = 32 ÷ 8 which means D = 4 making A = B = C = 1½×D = 1½ × 4 = 6 E = 2¼×D = 2¼×D = 2¼ × 4 = 9 F = ¼×D = ¼ × 4 = 1 and ABCDEF = 666491