Puzzle for March 3, 2021  ( )

Scratchpad

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

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

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

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Hint #1


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


  

Hint #2


eq.5 may be re-written as: B + F + E = A + D In the above equation, replace B + F with A + E (from eq.3), and replace D with 2: A + E + E = A + 2 which becomes A + 2×E = A + 2 Subtract A from each side of the above equation: A + 2×E – A = A + 2 – A which makes 2×E = 2 Divide both sides by 2: 2×E ÷ 2 = 2 ÷ 2 which makes E = 1


  

Hint #3


In eq.1, substitute 2 for D, and 1 for E: 2 + 1 = F which makes 3 = F


  

Hint #4


Substitute 1 for E, 3 for F, and 2 for D in eq.5: B + 1 + 3 = A + 2 which becomes B + 4 = A + 2 Subtract 2 from each side of the above equation: B + 4 – 2 = A + 2 – 2 which makes eq.5a) B + 2 = A


  

Hint #5


In eq.6, substitute B + 2 for A (from eq.5a), 3 for F, and 2 for D: B + 2 + 3 = B×2 which becomes B + 5 = 2×B Subtract B from each side of the equation above: B + 5 – B = 2×B – B which makes 5 = B


  

Hint #6


Substitute 5 for B in eq.5a: 5 + 2 = A which makes 7 = A


  

Solution

Substitute 2 for D, and 5 for B in eq.2: C + 2 = 5 Subtract 2 from each side of the above equation: C + 2 – 2 = 5 – 2 which makes C = 3 and makes ABCDEF = 753213