Puzzle for March 17, 2022  ( )

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

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

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Hint #1


Add B to both sides of eq.3: B + C + B = A – B + B which becomes eq.3a) 2×B + C = A   In eq.2, replace A with 2×B + C (from eq.3a): B + D = 2×B + C Subtract B from each side of the equation above: B + D – B = 2×B + C – B which becomes eq.2a) D = B + C


  

Hint #2


In eq.4, replace A with 2×B + C (from eq.3a), and D with B + C (from eq.2a): C + E = 2×B + C – C + B + C which becomes C + E = 3×B + C Subtract C from each side of the above equation: C + E – C = 3×B + C – C which makes E = 3×B


  

Hint #3


In eq.5, replace D with B + C (from eq.2a), A with 2×B + C (from eq.3a), and E with 3×B: B + C + F = 2×B + C + 3×B – F which becomes B + C + F = 5×B + C – F In the above equation, subtract B and C from both sides, and add F to both sides: B + C + F – B – C + F = 5×B + C – F – B – C + F which simplifies to 2×F = 4×B Divide both sides by 2: 2×F ÷ 2 = 4×B ÷ 2 which makes F = 2×B


  

Hint #4


In eq.6, substitute 3×B for E, and 2×B for F: B + 3×B = C – B – 2×B which becomes 4×B = C – 3×B Add 3×B to both sides of the above equation: 4×B + 3×B = C – 3×B + 3×B which makes 7×B = C


  

Hint #5


Substitute 7×B for C in eq.3a: 2×B + 7×B = A which makes 9×B = A


  

Hint #6


Substitute 7×B for C in eq.2a: D = B + 7×B which makes D = 8×B


  

Solution

Substitute 9×B for A, 7×B for C, 8×B for D, 3×B for E, and 2×B for F in eq.1: 9×B + B + 7×B + 8×B + 3×B + 2×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 = 9×B = 9 × 1 = 9 C = 7×B = 7 × 1 = 7 D = 8×B = 8 × 1 = 8 E = 3×B = 3 × 1 = 3 F = 2×B = 2 × 1 = 2 and ABCDEF = 917832