Puzzle for July 4, 2021  ( )

Scratchpad

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

eq.1) A + E = B eq.2) B + C = E + F eq.3) C + D + E = A – D + F eq.4) D = average (A, B, E) eq.5)* EF = B × (B – D) eq.6)** F ÷ A = sine (BC)

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  EF is a 2-digit number (not E×F).
**  BC is a 2-digit number (not B×C). BC is an angle expressed in degrees.

Scratchpad

 

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Hint #1


The sine of any angle is a number from –1 thru 1. There are three possible 2-digit integers for BC that make eq.6 true:   BC = 00, making sine (00) = F ÷ A = 0 which makes F = 0 or BC = 30, making sine (30) = F ÷ A = 1/2 which makes A = 2×F or BC = 90, making sine (90) = F ÷ A = 1 which makes A = F   In any of these three cases: C = 0


  

Hint #2


In eq.2, replace C with 0: B + 0 = E + F which makes eq.2a) B = E + F


  

Hint #3


In eq.1, replace B with E + F (from eq.2a): A + E = E + F Subtract E from each side of the equation above: A + E – E = E + F – E which makes A = F


  

Hint #4


In eq.6, substitute A for F: A ÷ A = sine (BC) which becomes 1 = sine (BC) which makes 1 = sine (90) which means BC = 90 which means B = 9


  

Hint #5


eq.5 may be written as: 10×E + F = B × (B – D) which may be written as 9×E + E + F = B × (B – D) Substitute B for E + F (from eq.2a) in the equation above: eq.5a) 9×E + B = B × (B – D)


  

Hint #6


Substitute 9 for B in eq.5a: 9×E + 9 = 9 × (9 – D) Divide both sides of the above equation by 9: (9×E + 9) ÷ 9 = 9 × (9 – D) ÷ 9 which becomes E + 1 = 9 – D Subtract 1 from each side of the above equation: E + 1 – 1 = 9 – D – 1 which makes eq.5b) E = 8 – D


  

Hint #7


Substitute 0 for C, and 8 – D for E (from eq.5b), and A for F into eq.3: 0 + D + 8 – D = A – D + A which becomes 8 = 2×A – D Add D to both sides of the above equation: 8 + D = 2×A – D + D which becomes 8 + D = 2×A Divide both sides by 2: (8 + D) ÷ 2 = 2×A ÷ 2 which makes 4 + ½×D = A and also makes eq.3a) F = A = 4 + ½×D


  

Hint #8


eq.4 may be written as: D = (A + B + E) ÷ 3 Multiply both sides of the equation above by 3: 3 × D = 3 × (A + B + E) ÷ 3 which becomes eq.4a) 3×D = A + B + E


  

Solution

In eq.4a, substitute 4 + ½×D for A (from eq.3a), 9 for B, and 8 – D for E (from eq.5b): 3×D = 4 + ½×D + 9 + 8 – D which becomes 3×D = 21 – ½×D Add ½×D to both sides of the above equation: 3×D + ½×D = 21 – ½×D + ½×D which makes 3½×D = 21 Divide both sides by by 3½: 3½×D ÷ 3½ = 21 ÷ 3½ which makes D = 6 making F = A = 4 + ½×D = 4 + ½×6 = 4 + 3 = 7 (from eq.3a) E = 8 – D = 8 – 6 = 2 (from eq.5b) and ABCDEF = 790627