Puzzle for November 5, 2021 ( )
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.
Scratchpad
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Hint #1
Add B to both sides of eq.2: B – D + B = A – B + B which becomes eq.2a) 2×B – D = A Add D to both sides of eq.3: D + E + D = A + B – D + D which becomes eq.3a) 2×D + E = A + B
Hint #2
In eq.3a, replace A with 2×B – D (from eq.2a): 2×D + E = 2×B – D + B which becomes 2×D + E = 3×B – D Add D to both sides of the above equation: 2×D + E + D = 3×B – D + D which becomes eq.3b) 3×D + E = 3×B
Hint #3
Add B to both sides of eq.3b: 3×D + E + B = 3×B + B which becomes 3×D + E + B = 4×B which may be written as eq.3c) 2×D + D + E + B = 4×B Add D and F to both sides of eq.4: F – D + D + F = B + E – F + D + F which becomes 2×F = B + E + D which may be written as eq.4a) 2×F = D + E + B
Hint #4
In eq.3c, replace D + E + B with 2×F (from eq.4a): 2×D + 2×F = 4×B Divide both sides of the above equation by 2: (2×D + 2×F) ÷ 2 = 4×B ÷ 2 which becomes eq.3d) D + F = 2×B
Hint #5
In eq.2a, substitute D + F for 2×B (from eq.3d): D + F – D = A which makes F = A
Hint #6
eq.6 may be written as: D + E + B – C = A + C + F Substitute 2×F for D + E + B (from eq.4a), and F for A in the equation above: 2×F – C = F + C + F which becomes 2×F – C = 2×F + C Subtract 2×F and C from each side of the equation above: 2×F – C – 2×F – C = 2×F + C – 2×F – C which simplifies to –2×C = 0 which means C = 0
Hint #7
Substitute 0 for C, and A for F in eq.5: A – 0 + E = D – E + A which becomes A + E = D – E + A In the above equation, subtract A from both sides, and add E to both sides: A + E – A + E = D – E + A – A + E which simplifies to 2×E = D
Hint #8
Substitute (2×E) for D in eq.3b: 3×(2×E) + E = 3×B which becomes 6×E + E = 3×B which makes 7×E = 3×B Divide both sides of the above equation by 3: 7×E ÷ 3 = 3×B ÷ 3 which makes 2⅓×E = B
Hint #9
Substitute (2⅓×E) for B, and 2×E for D in eq.2a: 2×(2⅓×E) – 2×E = A which becomes 4⅔×E – 2×E = A which makes 2⅔×E = A and also makes F = A = 2⅔×E
Solution
Substitute 2⅔×E for A and F, 2⅓×E for B, 0 for C, and 2×E for D in eq.1: 2⅔×E + 2⅓×E + 0 + 2×E + E + 2⅔×E = 32 which simplifies to 10⅔×E = 32 Divide both sides of the above equation by 10⅔: 10⅔×E ÷ 10⅔ = 32 ÷ 10⅔ which means E = 3 making A = F = 2⅔×E = 2⅔ × 3 = 8 B = 2⅓×E = 2⅓ × 3 = 7 D = 2×E = 2 × 3 = 6 and ABCDEF = 870638