Puzzle for October 31, 2020  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 38 eq.2) A + B + C = D + E eq.3) F – B = C – D eq.4) C + D – A = A + B eq.5) D – F = E – A eq.6) B – C + D – F = C – E + F

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

Scratchpad

 

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Hint #1


Add B and D to both sides of eq.3: F – B + B + D = C – D + B + D which becomes eq.3a) F + D = C + B   In eq.2, replace B + C with D + F: A + B + C = A + D + F = D + E which becomes eq.2a) A + F = E


  

Hint #2


In eq.5 replace E with A + F (from eq.2a): D – F = A + F – A which becomes D – F = F Add F to both sides of the equation above: D – F + F = F + F which makes D = 2×F


  

Hint #3


Add C, F, and E to each side of eq.6: B – C + D – F + C + F + E = C – E + F + C + F + E which becomes B + D + E = 2×C + 2×F Substitute D for 2×F in the above equation: B + D + E = 2×C + D Subtract D from each side: B + D + E – D = 2×C + D – D which becomes eq.6a) B + E = 2×C


  

Hint #4


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


  

Hint #5


Substitute A + E for 2×C (from eq.2b) in eq.6a: B + E = A + E Subtract E from both sides of the above equation: B + E – E = A + E – E which makes B = A


  

Hint #6


Substitute A for B, and 2×F for D in eq.3: F – A = C – 2×F Add 2×F to each side of the above equation: F – A + 2×F = C – 2×F + 2×F which becomes eq.3b) 3×F – A = C


  

Hint #7


Substitute 3×F – A for C (from eq.3b), 2×F for D, and A for B in eq.4: 3×F – A + 2×F – A = A + A which becomes 5×F – 2×A = 2×A Add 2×A to both sides of the equation above: 5×F – 2×A + 2×A = 2×A + 2×A which makes 5×F = 4×A Divide both sides by 4: 5×F ÷ 4 = 4×A ÷ 4 which makes 1¼×F = A and also makes B = A = 1¼×F


  

Hint #8


Substitute 1¼×F for A in eq.3b: 3×F – 1¼×F = C which makes 1¾×F = C


  

Hint #9


Substitute 1¼×F for A in eq.2a: 1¼×F + F = E which makes 2¼×F = E


  

Solution

Substitute 1¼×F for A and B, 1¾×F for C, 2×F for D, and 2¼×F for E in eq.1: 1¼×F + 1¼×F + 1¾×F + 2×F + 2¼×F + F = 38 which simplifies to 9½×F = 38 Divide both sides of the above equation by 9½: 9½×F ÷ 9½ = 38 ÷ 9½ which means F = 4 making A = B = 1¼×F = 1¼ × 4 = 5 C = 1¾×F = 1¾ × 4 = 7 D = 2×F = 2 × 4 = 8 E = 2¼×F = 2¼ × 4 = 9 and ABCDEF = 557894