Puzzle for December 16, 2019  ( )

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

eq.1) A + B + C + D + E + F = 15 eq.2) B – E = A – C eq.3) F = D + E eq.4) D = A + B eq.5) A + B – C = D – E eq.6) B = C + E

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

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Hint #1


In eq.2, replace B with C + E (from eq.6): C + E – E = A – C which becomes C = A – C Add C to both sides of the equation above: C + C = A – C + C which makes 2×C = A


  

Hint #2


In eq.5, replace A + B with D (from eq.4): D – C = D – E Subtract D from both sides of the above equation: D – C – D = D – E – D which makes –C = –E which means C = E


  

Hint #3


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


  

Hint #4


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


  

Hint #5


Substitute 4×C for D, and C for E in eq.3: F = 4×C + C which makes F = 5×C


  

Solution

Substitute 2×C for A and B, 4×C for D, C for E, and 5×C for F in eq.1: 2×C + 2×C + C + 4×C + C + 5×C = 15 which simplifies to 15×C = 15 Divide both sides of the equation above by 15: 15×C ÷ 15 = 15 ÷ 15 which means C = 1 making A = B = 2×C = 2 × 1 = 2 D = 4×C = 4 × 1 = 4 E = C = 1 F = 5×C = 5 × 1 = 5 and ABCDEF = 221415