Puzzle for January 24, 2021  ( )

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

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

A, B, C, D, E, and F each represent a one-digit positive integer.
*  DE is a 2-digit number (not D×E).
**  "C ^ E" means "C raised to the power of E".

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Hint #1


In eq.1, add F to each side, and subtract E from each side: D – F + F – E = A + E + F – E which becomes D – E = A + F In eq.3, replace D – E with A + F: B – A = A + F Add A to both sides of the equation above: B – A + A = A + F + A which becomes eq.3a) B = 2×A + F


  

Hint #2


In eq.4, replace B with 2×A + F (from eq.3a): A + C + F = 2×A + F – C Subtract A and F from both sides of the above equation: A + C + F – A – F = 2×A + F – C – A – F which simplifies to C = A – C Add C to both sides: C + C = A – C + C which makes 2×C = A


  

Hint #3


In eq.4, substitute 2×C for A: 2×C + C + F = B – C which becomes 3×C + F = B – C Add C to both sides of the above equation: 3×C + F + C = B – C + C which becomes eq.4a) 4×C + F = B


  

Hint #4


Substitute 4×C + F for B (from eq.4a) in eq.2: 4×C + F – C = D which becomes eq.2a) 3×C + F = D


  

Hint #5


Substitute 3×C + F for D (from eq.2a), and 2×C for A in eq.1: 3×C + F – F = 2×C + E which becomes 3×C = 2×C + E Subtract 2×C from each side of the equation above: 3×C – 2×C = 2×C + E – 2×C which makes C = E


  

Hint #6


Substitute C for E, and 2×C for A in eq.6: C ^ C = 2×C Divide both sides of the equation above by C: (C ^ C) ÷ C = 2×C ÷ C which means eq.6a) C ^ (C–1) = 2


  

Hint #7


To make eq.6a true, check several possible values for C:   If C = 1, then 2 = 1 ^ (1–1) = 1 ^ 0 = 1 If C = 2, then 2 = 2 ^ (2–1) = 2 ^ 1 = 2 If C = 3, then 2 = 3 ^ (3–1) = 3 ^ 2 = 9 If C = 4, then 2 = 4 ^ (4–1) = 4 ^ 3 = 64 If C > 4, then C ^ (C–1) > 64   Since C ^ (C–1) must equal 2, then C = 2 making A = 2×C = 2 × 2 = 4 E = C = 2


  

Hint #8


eq.5 may be written as: A × B × C = 10×D + E Substitute 4 for A, and 2 for C and E in the above equation: 4 × B × 2 = 10×D + 2 which becomes eq.5a) 8×B = 10×D + 2


  

Hint #9


In eq.5a, substitute (B – C) for D (from eq.2): 8×B = 10×(B – C) + 2 which is equivalent to 8×B = 10×B – 10×C + 2 In the above equation, add 10×C to both sides, and subtract 8×B from both sides: 8×B + 10×C – 8×B = 10×B – 10×C + 2 + 10×C – 8×B which becomes eq.5b) 10×C = 2×B + 2


  

Hint #10


In eq.5b, substitute 2 for C, and subtract 2 from each side: 10×2 – 2 = 2×B + 2 – 2 which becomes 20 – 2 = 2×B which makes 18 = 2×B Divide both sides of the equation above by 2: 18 ÷ 2 = 2×B ÷ 2 which makes 9 = B


  

Hint #11


Substitute 9 for B, and 2 for C in eq.2: 9 – 2 = D which makes 7 = D


  

Solution

Substitute 4 for A, 2 for C, and 9 for B in eq.4: 4 + 2 + F = 9 – 2 which becomes 6 + F = 7 Subtract 6 from each side of the equation above: 6 + F – 6 = 7 – 6 which makes F = 1 and makes ABCDEF = 492721