Puzzle for August 8, 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
Subtract A from both sides of eq.5: C + D – A = A + E + F – A which becomes C + D – A = E + F In eq.3, replace E + F with C + D – A: C + C + D – A = B + D which becomes 2×C + D – A = B + D In the equation above, add A to both sides, and subtract D from both sides: 2×C + D – A + A – D = B + D + A – D which becomes eq.3a) 2×C = B + A
Hint #2
eq.6 may be written as: F = (A + B + C) ÷ 3 Multiply both sides of the above equation by 3: F × 3 = ((A + B + C) ÷ 3) × 3 which becomes 3×F = A + B + C which is the same as eq.6a) 3×F = B + A + C
Hint #3
In eq.6a, replace B + A with 2×C (from eq.3a): 3×F = 2×C + C which makes 3×F = 3×C Divide both sides by 3: 3×F ÷ 3 = 3×C ÷ 3 which makes F = C
Hint #4
In eq.5, substitute C for F: C + D = A + E + C Subtract C from both sides of the equation above: C + D – C = A + E + C – C which becomes eq.5a) D = A + E
Hint #5
eq.4 may be re–written as: B + F = A + E + D In the above equation, substitute C for F, and D for A + E (from eq.5a): B + C = D + D which becomes eq.4a) B + C = 2×D
Hint #6
Substitute C for F, and B + E for A + C (from eq.2) into eq.6a: 3×C = B + B + E which becomes 3×C = 2×B + E Subtract 2×B from each side of the above equation: 3×C – 2×B = 2×B + E – 2×B which becomes eq.6b) 3×C – 2×B = E
Hint #7
Substitute C for F in eq.3: C + E + C = B + D which becomes 2×C + E = B + D Multiply both sides of the above equation by 2: 2 × (2×C + E) = 2 × (B + D) which becomes eq.3b) 4×C + 2×E = 2×B + 2×D
Hint #8
Substitute (3×C – 2×B) for E (from eq.6b), and B + C for 2×D (from eq.4a) into eq.3b: 4×C + 2×(3×C – 2×B) = 2×B + B + C which becomes 4×C + 6×C – 4×B = 3×B + C which becomes 10×C – 4×B = 3×B + C In the above equation, add 4×B to both sides, and subtract C from both sides: 10×C – 4×B + 4×B – C = 3×B + C + 4×B – C which makes eq.3c) 9×C = 7×B
Hint #9
Multiply both sides of eq.6b by 3: 3 × (3×C – 2×B) = 3 × E which becomes 9×C – 6×B = 3×E Substitute 7×B for 9×C (from eq.3c) in the above equation: 7×B – 6×B = 3×E which makes B = 3×E
Hint #10
Substitute (3×E) for B in eq.3c: 9×C = 7×(3×E) which makes 9×C = 21×E Divide both sides of the above equation by 9: 9×C ÷ 9 = 21×E ÷ 9 which makes C = 2⅓×E and also makes F = C = 2⅓×E
Hint #11
Substitute 3×E for B, and 2⅓×E for C in eq.4a: 3×E + 2⅓×E = 2×D which becomes 5⅓×E = 2×D Divide both sides of the equation above by 2: 5⅓×E ÷ 2 = 2×D ÷ 2 which makes 2⅔×E = D
Hint #12
Substitute 2⅔×E for D in eq.5a: 2⅔×E = A + E Subtract E from each side of the above equation: 2⅔×E – E = A + E – E which makes 1⅔×E = A
Solution
Substitute 1⅔×E for A, 3×E for B, and 2⅓×E for C and F, and 2⅔×E for D in eq.1: 1⅔×E + 3×E + 2⅓×E + 2⅔×E + E + 2⅓×E = 39 which simplifies to 13×E = 39 Divide both sides of the equation above by 13: 13×E ÷ 13 = 39 ÷ 13 which means E = 3 making A = 1⅔×E = 1⅔ × 3 = 5 B = 3×E = 3 × 3 = 9 C = F = 2⅓×E = 2⅓ × 3 = 7 D = 2⅔×E = 2⅔ × 3 = 8 and ABCDEF = 597837