Puzzle for December 18, 2020  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 25 eq.2) E = A + C + D + F eq.3) C + F = D eq.4) B = E + F eq.5) A + C + D = B – C

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

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Hint #1


Subtract C and D from both sides of eq.5: A + C + D – C – D = B – C – C – D which becomes eq.5a) A = B – 2×C – D   In eq.2, replace A with B – 2×C – D (from eq.5a): E = B – 2×C – D + C + D + F which becomes eq.2a) E = B – C + F


  

Hint #2


In eq.4, replace E with B – C + F (from eq.2a): B = B – C + F + F which becomes B = B – C + 2×F In the equation above, subtract B from both sides, and add C to both sides: B – B + C = B – C + 2×F – B + C which makes C = 2×F


  

Hint #3


In eq.3, substitute 2×F for C: 2×F + F = D which makes 3×F = D


  

Hint #4


Substitute 2×F for C, and 3×F for D in eq.2: E = A + 2×F + 3×F + F which becomes eq.2b) E = A + 6×F


  

Hint #5


To make eq.2b true, check several possible values for F and A:   If F = 1, then E = A + 6×1 = A + 6 which would make A ≤ 3 (since 1 ≤ E ≤ 9) If F = 2, then E = A + 6×2 = A + 12 which would make E > 12 (since A ≥ 1) If F > 2, then E > 18   Since E must be a one-digit integer, then F = 1 making C = 2×F = 2 × 1 = 2 D = 3×F = 3 × 1 = 3


  

Hint #6


Substitute 1 for F in eq.2b: E = A + 6×1 which makes eq.2c) E = A + 6


  

Hint #7


Substitute A + 6 for E (from eq.2c), and 1 for F in eq.4: B = A + 6 + 1 which makes eq.4a) B = A + 7


  

Solution

Substitute A + 7 for B (from eq.4a), 2 for C, 3 for D, A + 6 for E (from eq.2c), and 1 for F in eq.1: A + A + 7 + 2 + 3 + A + 6 + 1 = 25 which simplifies to 3×A + 19 = 25 Subtract 19 from both sides of the above equation: 3×A + 19 – 19 = 25 – 19 which makes 3×A = 6 Divide both sides by 3: 3×A ÷ 3 = 6 ÷ 3 which means A = 2 making B = A + 7 = 2 + 7 = 9 (from eq.4a) E = A + 6 = 2 + 6 = 8 (from eq.2c) and ABCDEF = 292381