Puzzle for September 29, 2019  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 22 eq.2) D + F = A + B + C eq.3) E + F = B + C eq.4) B + C = D – E – F eq.5) A – B = D – C

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 C to both sides of eq.5: A – B + B + C = D – C + B + C which becomes eq.5a) A + C = D + B   eq.2 may be written as: D + F = A + C + B In the above equation, replace A + C with D + B (from eq.5a): D + F = D + B + B Subtract D from both sides: D + F – D = D + B + B – D which makes eq.2a) F = 2×B


  

Hint #2


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


  

Hint #3


eq.1 may be written as: eq.1a) A + D + B + C + E + F = 22   Add E and F to each side of eq.4: B + C + E + F = D – E – F + E + F which becomes B + C + E + F = D Substitute D for B + C + E + F in eq.1a: A + D + D = 22 which becomes A + 2×D = 22 Subtract 2×D from both sides of the above equation: A + 2×D – 2×D = 22 – 2×D which means eq.1b) A = 22 – 2×D


  

Hint #4


To make eq.1b true, check several possible values for D and A:   If D = 9, then A = 22 – 2×9 = 22 – 18 = 4 If D = 8, then A = 22 – 2×8 = 22 – 16 = 6 If D = 7, then A = 22 – 2×7 = 22 – 14 = 8 If D = 6, then A = 22 – 2×6 = 22 – 12 = 10 If D > 6, then A > 10   Since A must be one-digit integer, then A = 4 or 6 or 8 making D = 9 or 8 or 7


  

Hint #5


eq.4a is: 2×(B + C) = D   Since B and C are integers, then 2×(B + C) is an even integer, making D an even integer, which makes D = 8 and A = 6


  

Hint #6


Substitute 8 for D in eq.4a: 2×(B + C) = 8 Divide both sides of the above equation by 2: 2×(B + C) ÷ 2 = 8 ÷ 2 which becomes eq.4b) B + C = 4


  

Hint #7


Substitute 6 for A, and 8 for D in eq.5a: 6 + C = 8 + B Subtract 6 from both sides of the above equation: 6 + C – 6 = 8 + B – 6 which becomes eq.5b) C = 2 + B


  

Hint #8


Substitute 2 + B for C (from eq.5b) in eq.4b: B + 2 + B = 4 Subtract 2 from both sides of the above equation: B + 2 + B – 2 = 4 – 2 which becomes 2×B = 2 Divide both sides by 2: 2×B ÷ 2 = 2 ÷ 2 which means B = 1 making F = 2×1 = 2 (from eq.2a)


  

Hint #9


Substitute 1 for B in eq.5b: C = 2 + 1 which makes C = 3


  

Solution

Substitute 2 for F, 1 for B, and 3 for C in eq.3: E + 2 = 1 + 3 Subtract 2 from both sides of the equation above: E + 2 – 2 = 1 + 3 – 2 which makes E = 2 and ABCDEF = 613822