Puzzle for December 28, 2019  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 33 eq.2) D – F = A eq.3) B = E – F eq.4) E + F = A + C eq.5) B – C = E – D eq.6) A × C = A + C + D + E

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

Scratchpad

 

Help Area

Hint #1


In eq.5, replace B with E – F (from eq.3): E – F – C = E – D Subtract E from both sides of the above equation: E – F – C – E = E – D – E which becomes eq.5a) –F – C = –D


  

Hint #2


Multiply both sides of eq.5a by (–1): (–1) × (–F – C) = (–1) × (–D) which becomes F + C = D In eq.2, replace D with F + C: F + C – F = A which makes C = A


  

Hint #3


Add F to both sides of eq.2: D – F + F = A + F which becomes D = A + F In eq.6, substitute A + F for D: A × C = A + C + A + F + E which may be written as eq.6a) A × C = A + C + A + E + F


  

Hint #4


In eq.6a, substitute A + C for E + F (from eq.4): A × C = A + C + A + A + C Substitute A for C in the above equation: A × A = A + A + A + A + A which becomes A × A = 5×A Divide both sides by A: A × A ÷ A = 5×A ÷ A which makes A = 5 and also makes C = A = 5


  

Hint #5


Substitute 5 for A in eq.2: D – F = 5 Add F to each side of the above equation: D – F + F = 5 + F which makes eq.2a) D = 5 + F


  

Hint #6


Substitute 5 for A and C in eq.4: E + F = 5 + 5 which becomes E + F = 10 Subtract F from both sides of the above equation: E + F – F = 10 – F which makes eq.4a) E = 10 – F


  

Hint #7


Substitute 10 – F for E in eq.3: B = 10 – F – F which makes eq.3a) B = 10 – 2×F


  

Solution

Substitute 5 for A and C, 10 – 2×F for B (from eq.3a), 5 + F for D (from eq.2a), and 10 – F for E (from eq.4a) in eq.1: 5 + 10 – 2×F + 5 + 5 + F + 10 – F + F = 33 which simplifies to 35 – F = 33 In the equation above, add F to each side, and subtract 33 from each side: 35 – F + F – 33 = 33 + F – 33 which makes 2 = F making B = 10 – 2×F = 10 – 2×2 = 10 – 4 = 6 (from eq.3a) D = 5 + F = 5 + 2 = 7 (from eq.2a) E = 10 – F = 10 – 2 = 8 (from eq.4a) and ABCDEF = 565782