Puzzle for January 6, 2019  ( )

Scratchpad

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

eq.1) A = C + D eq.2) B + C = D eq.3) B - E = E + F eq.4) A + E = B + D - E eq.5)* CD - EF = A + D eq.6) B = C × F

A, B, C, D, E, and F each represent a one-digit positive integer.
*  CD and EF are 2-digit numbers (not C×D or E×F).

Scratchpad

 

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Hint #1


Add (E - F) to both sides of eq.3: B - E + (E - F) = E + F + (E - F) which becomes eq.3a) B - F = 2×E


  

Hint #2


Add E to both sides of eq.4: A + E + E = B + D - E + E which becomes A + 2×E = B + D Substitute B - F for 2×E (from eq.3a) in the equation above: A + B - F = B + D Add (F - B) to both sides: A + B - F + (F - B) = B + D + (F - B) which simplifies to eq.4a) A = D + F


  

Hint #3


In eq.1, replace A with D + F (from eq.4a): D + F = C + D Subtract D from both sides: D + F - D = C + D - D which makes F = C


  

Hint #4


In eq.3a, substitute C for F: B - C = 2×E Add C to both sides of the above equation: B - C + C = 2×E + C which becomes eq.3b) B = 2×E + C


  

Hint #5


In eq.2, replace B with 2×E + C (from eq.3b): 2×E + C + C = D which becomes eq.2a) 2×E + 2×C = D In eq.1, replace D with 2×E + 2×C: A = C + 2×E + 2×C which becomes eq.1a) A = 2×E + 3×C


  

Hint #6


eq.5 may be written as: 10×C + D - (10×E + F) = A + D which is equivalent to 10×C + D - 10×E - F = A + D Subtract D from each side of the equation above: 10×C + D - 10×E - F - D = A + D - D which becomes 10×C - 10×E - F = A Substitute 2×E + 3×C for A, and C for F: 10×C - 10×E - C = 2×E + 3×C Add (10×E - 3×C) to each side: 10×C - 10×E - C + (10×E - 3×C) = 2×E + 3×C + (10×E - 3×C) which simplifies to 6×C = 12×E Divide both sides by 6: 6×C ÷ 6 = 12×E ÷ 6 which makes C = 2×E


  

Hint #7


Substitute (2×E) for C in eq.1a: A = 2×E + 3×(2×E) which becomes A = 2×E + 6×E which makes A = 8×E


  

Hint #8


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


  

Hint #9


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


  

Solution

Substitute 4×E for B, and 2×E for C and F in eq.6: 4×E = 2×E × 2×E which means 4×E = 4×E² Divide each side by 4×E: 4×E ÷ 4×E = 4×E² ÷ 4×E which makes 1 = E making A = 8×E = 8 × 1 = 8 B = 4×E = 4 × 1 = 4 C = F = 2×E = 2 × 1 = 2 D = 6×E = 6 × 1 = 6 and ABCDEF = 842612