Puzzle for July 22, 2020  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 21 eq.2) D + F = B + C eq.3) E – A = B – E eq.4) A + B = D eq.5) B + F = A + C eq.6) C + D – E = B + E

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

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Hint #1


Add A and E to both sides of eq.3: E – A + A + E = B – E + A + E which becomes 2×E = B + A which may be written as eq.3a) 2×E = A + B   In eq.4, replace A + B with 2×E (from eq.3a): eq.3b) 2×E = D


  

Hint #2


In eq.6, replace D with 2×E: C + 2×E – E = B + E which becomes C + E = B + E Subtract E from both sides of the above equation: C + E – E = B + E – E which makes C = B


  

Hint #3


In eq.5, substitute B for C: B + F = A + B Subtract B from each side of the above equation: B + F – B = A + B – B which makes F = A


  

Hint #4


Substitute A for F, and B for C in eq.2: D + A = B + B which becomes eq.2a) D + A = 2×B


  

Hint #5


In eq.2a, substitute A + B for D (from eq.4): A + B + A = 2×B which becomes 2×A + B = 2×B Subtract B from each side of the equation above: 2×A + B – B = 2×B – B which makes 2×A = B and also makes C = B = 2×A


  

Hint #6


Substitute 2×A for B in eq.3a: 2×E = A + 2×A which makes 2×E = 3×A Divide both sides of the equation above by 2: 2×E ÷ 2 = 3×A ÷ 2 which makes E = 1½×A


  

Hint #7


Substitute (1½×A) for E in eq.3b: 2×(1½×A) = D which makes 3×A = D


  

Solution

Substitute 2×A for B and C, 3×A for D, 1½×A for E, and A for F in eq.1: A + 2×A + 2×A + 3×A + 1½×A + A = 21 which simplifies to 10½×A = 21 Divide both sides of the equation above by 10½: 10½×A ÷ 10½ = 21 ÷ 10½ which means A = 2 making B = C = 2×A = 2 × 2 = 4 D = 3×A = 3 × 2 = 6 E = 1½×A = 1½ × 2 = 3 F = A = 2 and ABCDEF = 244632