Puzzle for November 9, 2019  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 43 eq.2) D + E = A + C eq.3) F – C = A – F eq.4) A – B – C = C – D eq.5) C – B = A + B – E eq.6) C + D – A = A – B

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

Scratchpad

 

Help Area

Hint #1


In eq.4, replace A – B with C + D – A (from eq.6): C + D – A – C = C – D which becomes D – A = C – D Add A and D to both sides of the equation above: D – A + A + D = C – D + A + D which becomes 2×D = C + A which may be written as eq.4a) 2×D = A + C


  

Hint #2


In eq.2, replace A + C with 2×D (from eq.4a): D + E = 2×D Subtract D from both sides of the above equation: D + E – D = 2×D – D which makes E = D


  

Hint #3


In eq.5, substitute D for E: C – B = A + B – D In the equation above, add B to both sides, and subtract A from each side: C – B + B – A = A + B – D + B – A which becomes eq.5a) C – A = 2×B – D   Subtract D from both sides of eq.6: C + D – A + D = A – B + D which becomes eq.6a) C – A = A – B – D


  

Hint #4


Substitute A – B – D for C – A (from eq.6a) into eq.5a: A – B – D = 2×B – D Add B and D to both sides of the above equation: A – B – D + B + D = 2×B – D + B + D which simplifies to eq.5b) A = 3×B


  

Hint #5


Substitute 3×B for A in eq.4: 3×B – B – C = C – D which becomes 2×B – C = C – D Subtract C from each side of the above equation: 2×B – C – C = C – D – C which becomes eq.4a) 2×B – 2×C = –D


  

Hint #6


Substitute 3×B for A in eq.5a: C – 3×B = 2×B – D Add 3×B to each side of the equation above: C – 3×B + 3×B = 2×B – D + 3×B which becomes eq.5c) C = 5×B – D


  

Hint #7


Substitute (5×B – D) for C (from eq.5c) into eq.4a: 2×B – 2×(5×B – D) = –D which is equivalent to 2×B – 10×B + 2×D = –D which becomes –8×B + 2×D = –D Add 8×B and D to each side of the above equation: –8×B + 2×D + 8×B + D = –D + 8×B + D which means 3×D = 8×B Divide both sides by 8: 3×D ÷ 8 = 8×B ÷ 8 which makes ⅜×D = B


  

Hint #8


Substitute (⅜×D) for B in eq.5c: C = 5×(⅜×D) – D which becomes C = 1⅞×D – D which makes C = ⅞×D


  

Hint #9


Substitute (⅜×D) for B in eq.5b: A = 3×(⅜×D) which makes A = 1⅛×D


  

Hint #10


Substitute ⅞×D for C, and 1⅛×D for A in eq.3: F – ⅞×D = 1⅛×D – F Add ⅞×D and F to both sides of the above equation: F – ⅞×D + ⅞×D + F = 1⅛×D – F + ⅞×D + F which becomes 2×F = 2×D Divide both sides by 2: 2×F ÷ 2 = 2×D ÷ 2 which makes F = D


  

Solution

Substitute 1⅛×D for A, ⅜×D for B, ⅞×D for C, and D for E and F in eq.2: 1⅛×D + ⅜×D + ⅞×D + D + D + D = 43 which simplifies to 5⅜×D = 43 Divide both sides of the above equation by 5⅜: 5⅜×D ÷ 5⅜ = 43 ÷ 5⅜ which means D = 8 making A = 1⅛×D = 1⅛ × 8 = 9 B = ⅜×D = ⅜ × 8 = 3 C = ⅞×D = ⅞ × 8 = 7 E = F = D = 8 and ABCDEF = 937888