Puzzle for May 24, 2021  ( )

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

eq.1) A + B + C + D + E + F = 20 eq.2) B = A + C eq.3) C + E = B + D eq.4) D + E = A + F eq.5) E – C = B + C eq.6) A + E = B + C + D

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

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Hint #1


eq.6 may be written as: A + E = C + B + D In the above equation, replace B + D with C + E (from eq.3): A + E = C + C + E which becomes A + E = 2×C + E Subtract E from both sides of the equation above: A + E – E = 2×C + E – E which makes A = 2×C


  

Hint #2


In eq.2, replace A with 2×C: B = 2×C + C which makes B = 3×C


  

Hint #3


In eq.5, substitute 3×C for B: E – C = 3×C + C Add C to both sides of the above equation: E – C + C = 4×C + C which makes E = 5×C


  

Hint #4


Substitute 5×C for E, and 3×C for B in eq.3: C + 5×C = 3×C + D which becomes 6×C = 3×C + D Subtract 3×C from each side of the equation above: 6×C – 3×C = 3×C + D – 3×C which makes 3×C = D


  

Hint #5


In eq.4, substitute 3×C for D, 5×C for E, and 2×C for A: 3×C + 5×C = 2×C + F which becomes 8×C = 2×C + F Subtract 2×C from both sides of the above equation: 8×C – 2×C = 2×C + F – 2×C which makes 6×C = F


  

Solution

Substitute 2×C for A, 3×C for B and D, 5×C for E, and 6×C for F in eq.1: 2×C + 3×C + C + 3×C + 5×C + 6×C = 20 which simplifies to 20×C = 20 Divide both sides of the above equation by 20: 20×C ÷ 20 = 20 ÷ 20 which means C = 1 making A = 2×C = 2 × 1 = 2 B = D = 3×C = 3 × 1 = 3 E = 5×C = 5 × 1 = 5 F = 6×C = 6 × 1 = 6 and ABCDEF = 231356