Puzzle for March 28, 2020 ( )
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 B, D, and E from both sides of eq.5: B + C + D – B – D – E = A + E + F – B – D – E which becomes eq.5a) C – E = A + F – B – D In eq.6, add C to both sides, and subtract D from both sides: A – C – D + E + C – D = D – E + C – D which becomes A – 2×D + E = –E + C which may be written as eq.6a) A – 2×D + E = C – E
Hint #2
In eq.6a, replace C – E with A + F – B – D (from eq.5a): A – 2×D + E = A – B – D + F Add 2×D and B to both sides of the above equation: A – 2×D + E + 2×D + B = A – B – D + F + 2×D + B which becomes A + E + B = A + F + D Subtract A from each side: A + E + B – A = A + F + D – A which becomes E + B = F + D which may be written as eq.6b) B + E = F + D
Hint #3
In eq.6b, replace B + E with C + F (from eq.4): C + F = F + D Subtract F from both sides of the equation above: C + F – F = F + D – F which makes C = D
Hint #4
In eq.3, substitute C for D: C + F = A + C Subtract C from both sides of the above equation: C + F – C = A + C – C which becomes F = A
Hint #5
In eq.6a, substitute C for D: A – 2×C + E = C – E In the equation above, add 2×C to both sides, and subtract E from each side: A – 2×C + E + 2×C – E = C – E + 2×C – E which makes A = 3×C – 2×E and also makes eq.6c) F = A = 3×C – 2×E
Hint #6
Substitute C for D, and 3×C – 2×E for A and F (from eq.6c) in eq.5: B + C + C = 3×C – 2×E + E + 3×C – 2×E which becomes B + 2×C = 6×C – 3×E Subtract 2×C from both sides of the equation above: B + 2×C – 2×C = 6×C – 3×E – 2×C which becomes eq.5b) B = 4×C – 3×E
Hint #7
Substitute 3×C – 2×E for A and F (from eq.6c), and 4×C – 3×E for B (from eq.5b) in eq.2: C + E = 3×C – 2×E + 4×C – 3×E + 3×C – 2×E which becomes C + E = 10×C – 7×E In the above equation, add 7×E to both sides, and subtract C from both sides: C + E + 7×E – C = 10×C – 7×E + 7×E – C which becomes 8×E = 9×C Divide both sides by 8: 8×E ÷ 8 = 9×C ÷ 8 which makes E = 1⅛×C
Hint #8
Substitute (1⅛×C) for E in eq.5b: B = 4×C – 3×(1⅛×C) which becomes B = 4×C – 3⅜×C which makes B = ⅝×C
Hint #9
Substitute (1⅛×C) for E in eq.6c: F = A = 3×C – 2×(1⅛×C) which becomes F = A = 3×C – 2¼×C which makes F = A = ¾×C
Solution
Substitute ¾×C for A and F, ⅝×C for B, C for D, and 1⅛×C for E in eq.1: ¾×C + ⅝×C + C + C + 1⅛×C + ¾×C = 42 which simplifies to 5¼×C = 42 Divide both sides of the equation above by 5¼: 5¼×C ÷ 5¼ = 42 ÷ 5¼ which means C = 8 making A = F = ¾×C = ¾ × 8 = 6 B = ⅝×C = ⅝ × 8 = 5 D = C = 8 E = 1⅛×C = 1⅛ × 8 = 9 and ABCDEF = 658896