Puzzle for September 15, 2019  ( )

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

eq.1) A + B + C + D + E + F = 25 eq.2) C = A + B eq.3) B + D + E = A – B + C eq.4) D – A = B + E – F eq.5) B + C = D – B eq.6)* A + D + E = BC + F

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  BC is a 2-digit number (not B×C).

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Hint #1


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


  

Hint #2


In eq.3a, replace D with 2×B + C (from eq.5a): 2×B + 2×B + C + E = A + C Subtract C from both sides of the above equation: 2×B + 2×B + C + E – C = A + C – C which becomes eq.3b) 4×B + E = A


  

Hint #3


In eq.2, replace A with 4×B + E (from eq.3b): C = 4×B + E + B which becomes eq.2a) C = 5×B + E


  

Hint #4


In eq.5a, substitute 5×B + E for C (from eq.2a): 2×B + 5×B + E = D which becomes eq.5b) 7×B + E = D


  

Hint #5


Substitute 7×B + E for D (from eq.5b), and (4×B + E) for A (from eq.3b) in eq.4: 7×B + E – (4×B + E) = B + E – F which may be written as 7×B + E – 4×B – E = B + E – F which becomes 3×B = B + E – F In the equation above, add F to both sides, and subtract 3×B from both sides: 3×B + F – 3×B = B + E – F + F – 3×B which becomes eq.4a) F = E – 2×B


  

Hint #6


eq.6 may be written as: A + D + E = 10×B + C + F Substitute 4×B + E for A (from eq.3b), 7×B + E for D (from eq.5b), 5×B + E for C (from eq.2a), and E – 2×B for F (from eq.4a) in the above equation: 4×B + E + 7×B + E + E = 10×B + 5×B + E + E – 2×B which becomes 11×B + 3×E = 13×B + 2×E Subtract both 11×B and 2×E from each side of the above equation: 11×B + 3×E – 11×B – 2×E = 13×B + 2×E – 11×B – 2×E which simplifies to E = 2×B


  

Hint #7


Substitute 2×B for E in eq.3b: 4×B + 2×B = A which makes 6×B = A


  

Hint #8


Substitute 2×B for E in eq.2a: C = 5×B + 2×B which makes C = 7×B


  

Hint #9


Substitute 2×B for E in eq.5b: 7×B + 2×B = D which means 9×B = D


  

Hint #10


Substitute 2×B for E in eq.4a: F = 2×B – 2×B which makes F = 0


  

Solution

Substitute 6×B for A, 7×B for C, 9×B for D, 2×B for E, and 0 for F in eq.1: 6×B + B + 7×B + 9×B + 2×B + 0 = 25 which simplifies to 25×B = 25 Divide each side of the equation above by 25: 25×B ÷ 25 = 25 ÷ 25 which means B = 1 making A = 6×B = 6 × 1 = 6 C = 7×B = 7 × 1 = 7 D = 9×B = 9 × 1 = 9 E = 2×B = 2 × 1 = 2 and ABCDEF = 617920