Puzzle for September 8, 2019  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 22 eq.2) E = A + D eq.3) C – B = D – A eq.4) B – A + F = A + C eq.5) F = C + E eq.6) A × D = A + B + E

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

Scratchpad

 

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Hint #1


Add A and B to both sides of eq.3: C – B + A + B = D – A + A + B which becomes C + A = D + B which is the same as eq.3a) A + C = D + B   In eq.4, replace A + C with D + B (from eq.3a): eq.4a) B – A + F = D + B


  

Hint #2


In eq.4a, add A to both sides, and subtract B from both sides: B – A + F + A – B = D + B + A – B which simplifies to F = D + A which is the same as F = A + D In eq.2, replace A + D with F (from eq.4a): E = F


  

Hint #3


In eq.5, replace F with E: E = C + E Subtract E from both sides of the equation above: E – E = C + E – E which makes 0 = C


  

Hint #4


In eq.3a, substitute 0 for C: A + 0 = D + B which becomes eq.3b) A = D + B   In eq.2, substitute D + B for A (from eq.3b): E = D + B + D which means E = B + 2×D and also means eq.2a) F = E = B + 2×D


  

Hint #5


Substitute (D + B) for A (from eq.3b), and B + 2×D for E (from eq.2a) in eq.6: (D + B) × D = (D + B) + B + B + 2×D which becomes D×D + B×D = 3×D + 3×B which may be written as eq.6a) D × (D + B) = 3 × (D + B)   To make eq.6a true, then either: D = 3 or: D + B = 0


  

Hint #6


Begin checking: D + B = 0 ...   Since B and D ≥ 0, this would make D = 0 and B = 0 and would make A = 0 + 0 = 0 (from eq.3b) F = E = 0 + 2×0 = 0 (from eq.2a)


  

Hint #7


Finish checking: D + B = 0 ...   Substitute 0 for A, B, C, D, E, and F in eq.1: 0 + 0 + 0 + 0 + 0 + 0 = 22 which would make 0 = 22 Since 0 ≠ 22, then D + B ≠ 0 and, therefore: D = 3


  

Hint #8


Substitute 3 for D in eq.2a: F = E = B + 2×3 which means eq.2b) F = E = B + 6


  

Hint #9


Substitute 3 for D in eq.3b: eq.3c) A = 3 + B


  

Solution

Substitute 3 + B for A (from eq.3c), 0 for C, 3 for D, and B + 6 for E and F (from eq.2b) in eq.1: 3 + B + B + 0 + 3 + B + 6 + B + 6 = 22 which simplifies to 18 + 4×B = 22 Subtract 18 from both sides of the above equation: 18 + 4×B – 18 = 22 – 18 which makes 4×B = 4 Divide both sides by 4: 4×B ÷ 4 = 4 ÷ 4 which means B = 1 making A = 3 + B = 3 + 1 = 4 (from eq.3c) F = E = B + 6 = 1 + 6 = 7 (from eq.2b) and ABCDEF = 410377