Puzzle for March 17, 2021  ( )

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

eq.1) A + B + C + D + E + F = 30 eq.2) B + D = C eq.3) F = C + D eq.4) E = B + C + D eq.5) C + F = A – C + E eq.6) A – B + D = B + E

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

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Hint #1


eq.4 may be written as: E = B + D + C In the above equation, replace B + D with C (from eq.2): E = C + C which makes E = 2×C


  

Hint #2


In eq.5, replace E with 2×C: C + F = A – C + 2×C which becomes C + F = A + C Subtract C from each side of the above equation: C + F – C = A + C – C which makes F = A


  

Hint #3


In eq.3, substitute A for F, and B + D for C (from eq.2): A = B + D + D which becomes eq.3a) A = B + 2×D


  

Hint #4


Substitute B + D for C (from eq.2) in eq.4: E = B + B + D + D which becomes eq.4a) E = 2×B + 2×D


  

Hint #5


Substitute B + 2×D for A (from eq.3a), and 2×B + 2×D for E (from eq.4a) in eq.6: B + 2×D – B + D = B + 2×B + 2×D which becomes 3×D = 3×B + 2×D Subtract 2×D from each side of the equation above: 3×D – 2×D = 3×B + 2×D – 2×D which makes D = 3×B


  

Hint #6


Substitute (3×B) for D in eq.3a: A = B + 2×(3×B) which becomes A = B + 6×B which makes A = 7×B which also makes F = A = 7×B


  

Hint #7


Substitute 3×B for D in eq.2: B + 3×B = C which makes 4×B = C


  

Hint #8


Substitute (3×B) for D in eq.4a: E = 2×B + 2×(3×B) which becomes E = 2×B + 6×B which makes E = 8×B


  

Solution

Substitute 7×B for A and F, 4×B for C, 3×B for D, and 8×B for E in eq.1: 7×B + B + 4×B + 3×B + 8×B + 7×B = 30 which simplifies to 30×B = 30 Divide both sides of the above equation by 30: 30×B ÷ 30 = 30 ÷ 30 which means B = 1 making A = F = 7×B = 7 × 1 = 7 C = 4×B = 4 × 1 = 4 D = 3×B = 3 × 1 = 3 E = 8×B = 8 × 1 = 8 and ABCDEF = 714387