Puzzle for August 14, 2022 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit non-negative integer.
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Hint #1
Add B to both sides of eq.4: D – E + F + B = A + E + B which may be written as eq.4a) B + D – E + F = A + E + B
Hint #2
In eq.4a, replace B + D with E + F (from eq.2): E + F – E + F = A + E + B which becomes eq.4b) 2×F = A + E + B
Hint #3
eq.6 may be written as: F = (A + B + C + D) ÷ 4 Multiply both sides of the above equation by 4: 4 × F = 4 × (A + B + C + D) ÷ 4 which becomes eq.6a) 4×F = A + B + C + D
Hint #4
Multiply both sides of eq.4b by 2: 2 × (2×F) = 2 × (A + E + B) which becomes eq.4c) 4×F = 2×A + 2×E + 2×B
Hint #5
In eq.6a, replace 4×F with 2×A + 2×E + 2×B (from eq.4c): 2×A + 2×E + 2×B = A + B + C + D Subtract A and B from both sides of the equation above: 2×A + 2×E + 2×B – A – B = A + B + C + D – A – B which becomes eq.6b) A + 2×E + B = C + D
Hint #6
Add E to both sides of eq.4: D – E + F + E = A + E + E which becomes D + F = A + 2×E In eq.6b, substitute D + F for A + 2×E: D + F + B = C + D Subtract D from each side of the equation above: D + F + B – D = C + D – D which becomes F + B = C which may be written as eq.6c) B + F = C
Hint #7
In eq.3, substitute C for B + F (from eq.6c): C + E = C Subtract C from each side of the above equation: C + E – C = C – C which makes E = 0
Hint #8
Substitute 4×F for A + B + C + D (from eq.6a), and 0 for E in eq.1: 4×F + 0 + F = 25 which becomes 5×F = 25 Divide both sides of the above equation by 5: 5×F ÷ 5 = 25 ÷ 5 which makes F = 5
Hint #9
Substitute 0 for E, and 5 for F in eq.2: B + D = 0 + 5 which becomes B + D = 5 Subtract D from each side of the above equation: B + D – D = 5 – D which makes eq.2a) B = 5 – D
Hint #10
Substitute 0 for E, and 5 for F in eq.4: D – 0 + 5 = A + 0 which becomes eq.4d) D + 5 = A
Hint #11
eq.5 may be written as: B = (A + D + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × B = 3 × (A + D + E) ÷ 3 which becomes 3×B = A + D + E Substitute 0 for E: 3×B = A + D + 0 which becomes eq.5a) 3×B = A + D
Hint #12
Substitute 5 – D for B (from eq.2a), and D + 5 for A (from eq.4d) into eq.5a: 3×(5 – D) = D + 5 + D which becomes 15 – 3×D = 2×D + 5 In the above equation, add 3×D to both sides, and subtract 5 from both sides: 15 – 3×D + 3×D – 5 = 2×D + 5 + 3×D – 5 which makes 10 = 5×D Divide both sides by 5: 10 ÷ 5 = 5×D ÷ 5 which makes 2 = D
Hint #13
Substitute 2 for D in eq.4d: 2 + 5 = A which makes 7 = A
Hint #14
Substitute 2 for D in eq.2a: B = 5 – 2 which makes B = 3
Solution
Substitute 3 for B, and 5 for F in eq.6c: 3 + 5 = C which makes 8 = C and makes ABCDEF = 738205