Puzzle for February 4, 2020  ( )

Scratchpad

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

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

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

Scratchpad

 

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Hint #1


Add E and C to each side of eq.3: B – E + E + C = D – C + E + C which becomes eq.3a) B + C = D + E   In eq.5, replace D + E with B + C (from eq.3a): B + C + F = A + B + C Subtract B and C from both sides of the equation above: B + C + F – B – C = A + B + C – B – C which simplifies to F = A


  

Hint #2


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


  

Hint #3


In eq.3, substitute C + E for D (from eq.4a): B – E = C + E – C which becomes B – E = E Add E to each side of the equation above: B – E + E = E + E which makes B = 2×E


  

Hint #4


Add B to both sides of eq.6: A + C – B + B = B + D + E + B which becomes A + C = 2×B + D + E Substitute B + C for D + E (from eq.3a) in the above equation: A + C = 2×B + B + C Subtract C from each side: A + C – C = 2×B + B + C – C which makes eq.6a) A = 3×B


  

Hint #5


Substitute (2×E) for B in eq.6a: A = 3×(2×E) which makes A = 6×E and also makes F = A = 6×E


  

Hint #6


Substitute 6×E for F in eq.2: E + 6×E = C which makes 7×E = C


  

Hint #7


Substitute 2×E for B, 7×E for C in eq.3: 2×E – E = D – 7×E which becomes E = D – 7×E Add 7×E to both sides of the equation above: E + 7×E = D – 7×E + 7×E which makes 8×E = D


  

Solution

Substitute 6×E for A and F, 2×E for B, 7×E for C, and 8×E for D in eq.1: 6×E + 2×E + 7×E + 8×E + E + 6×E = 30 which simplifies to 30×E = 30 Divide both sides of the equation above by 30: 30×E ÷ 30 = 30 ÷ 30 which means E = 1 making A = F = 6×E = 6 × 1 = 6 B = 2×E = 2 × 1 = 2 C = 7×E = 7 × 1 = 7 D = 8×E = 8 × 1 = 8 and ABCDEF = 627816