Puzzle for September 30, 2022  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 24 eq.2) B = D + E eq.3) C + F = A + D eq.4) F - B = C + D - F eq.5) B + C - D + F = A - C + D + E eq.6) E × F = B + C + D

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

Scratchpad

 

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Hint #1


Add B and F to both sides of eq.4: F - B + B + F = C + D - F + B + F which becomes 2×F = C + D + B which is the same as eq.4a) 2×F = B + C + D


  

Hint #2


In eq.6, replace B + C + D with 2×F (from eq.4a): E × F = 2×F Divide both sides of the above equation by F (assumes F ≠ 0): E × F ÷ F = 2×F ÷ F which makes E = 2


  

Hint #3


Begin confirming: F ≠ 0 ...   If F = 0, then substituting 0 for F in eq.6 would yield: E × 0 = B + C + D which would make 0 = B + C + D Since B and C and D are all non-negative, the above equation would make: 0 = B = C = D


  

Hint #4


Finish confirming: F ≠ 0 ...   If F and B and C and D = 0, then in eq.1: A + 0 + 0 + 0 + E + 0 = 24 which would become A + E = 24 Since A and E are non-negative one-digit integers, then: A ≤ 9 and E ≤ 9 making A + E ≤ 18 which means A + E ≠ 24 and therefore means F ≠ 0 and makes E = 2


  

Hint #5


In eq.5, replace D + E with B (from eq.2): B + C - D + F = A - C + B In the above equation, subtract B from both sides, and add D and C to both sides: B + C - D + F - B + D + C = A - C + B - B + D + C which simplifies to eq.5a) 2×C + F = A + D


  

Hint #6


In eq.5a, substitute C + F for A + D (from eq.3): 2×C + F = C + F Subtract C and F from each side of the equation above: 2×C + F - C - F = C + F - C - F which makes C = 0


  

Hint #7


Substitute 2 for E in eq.2: eq.2a) B = D + 2


  

Hint #8


Substitute D + 2 for B (from eq.2a), and 0 for C in eq.4a: 2×F = D + 2 + 0 + D which becomes 2×F = 2×D + 2 Divide both sides of the above equation by 2: 2×F ÷ 2 = (2×D + 2) ÷ 2 which makes eq.6a) F = D + 1


  

Hint #9


Substitute 0 for C, and D + 1 for F (from eq.6a) in eq.3: 0 + D + 1 = A + D which becomes D + 1 = A + D Subtract D from each side of the above equation: D + 1 - D = A + D - D which makes 1 = A


  

Hint #10


Substitute 1 for A, D + 2 for B (from eq.2a), 0 for C, 2 for E, and D + 1 for F (from eq.6a) in eq.1: 1 + D + 2 + 0 + D + 2 + D + 1 = 24 which becomes 6 + 3×D = 24 Subtract 6 from both sides of the above equation: 6 + 3×D - 6 = 24 - 6 which makes 3×D = 18 Divide both sides by 3: 3×D ÷ 3 = 18 ÷ 3 which means D = 6


  

Solution

Since D = 6, then: B = D + 2 = 6 + 2 = 8 (from eq.2a) F = D + 1 = 6 + 1 = 7 (from eq.6a) and ABCDEF = 180627