Puzzle for December 27, 2018  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 17 eq.2) A - B = D - E eq.3) E - A = A + F eq.4) A + C + F = D + E eq.5) B = C + E eq.6)* CD = (A + C)×E

A, B, C, D, E, and F each represent a one-digit positive integer.
*  CD is a 2-digit number (not C×D).

Scratchpad

 

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Hint #1


Add A to both sides of eq.3: E - A + A = A + F + A which becomes eq.3a) E = 2×A + F   Subtract the left and right sides of eq.2 from the respective sides of eq.4: A + C + F - (A - B) = D + E - (D - E) which becomes A + C + F - A + B = D + E - D + E which becomes eq.4a) C + F + B = 2×E


  

Hint #2


In eq.4a, replace B with C + E (from eq.5): C + F + C + E = 2×E Subtract E from both sides: C + F + C + E - E = 2×E - E which becomes eq.4b) 2×C + F = E


  

Hint #3


In eq.3a, substitute 2×C + F for E (from eq.4b): 2×C + F = 2×A + F Subtract F from each side: 2×C + F - F = 2×A + F - F which simplifies to 2×C = 2×A Divide both sides by 2: 2×C ÷ 2 = 2×A ÷ 2 which makes C = A


  

Hint #4


In eq.2, Substitute (C + E) for B (from eq.5): A - (C + E) = D - E which is the same as A - C - E = D - E Replace C with A, and add E to both sides of the equation above: A - A - E + E = D - E + E which makes 0 = D


  

Hint #5


eq.6 may be written as: 10×C + D = (A + C)×E Substitute A for C, and 0 for D in the equation above: 10×A + 0 = (A + A)×E which becomes 10×A = (2×A)×E Divide both sides by 2×A: 10×A ÷ 2×A = (2×A)×E ÷ 2×A which makes 5 = E


  

Hint #6


Substitute 5 for E in eq.3a: 5 = 2×A + F Subtract 2×A from each side: 5 - 2×A = 2×A + F - 2×A which becomes eq.3b) 5 - 2×A = F


  

Hint #7


Substitute A for C, and 5 for E in eq.5: eq.5a) B = A + 5


  

Solution

Substitute A + 5 for B (from eq.5a), A for C, 0 for D, 5 for E, and 5 - 2×A for F (from eq.3b) in eq.1: A + A + 5 + A + 0 + 5 + 5 - 2×A = 17 which becomes A + 15 = 17 Subtract 15 from each side: A + 15 - 15 = 17 - 15 which makes A = 2 making B = 2 + 5 = 7 (from eq.5a) C = A = 2 F = 5 - 2×2 = 5 - 4 = 1 = F (from eq.3b) making ABCDEF = 272051