Puzzle for August 4, 2019 ( )
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.
* CD is a 2-digit number (not C×D).
Scratchpad
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Hint #1
Add both E and F to each side of eq.1: F – E + E + F = A – F + E + F which becomes eq.1a) 2×F = A + E Add E to both sides of eq.3: A + B – E + E = C + E + E which becomes eq.3a) A + B = C + 2×E
Hint #2
eq.6 may be written as: B + D = C + E + F – A + F which is the same as B + D = C + E + 2×F – A Add A to both sides of the equation above: B + D + A = C + E + 2×F – A + A which becomes B + D + A = C + E + 2×F which may be written as eq.6a) A + B + D = C + E + 2×F
Hint #3
In eq.6a, replace A + B with C + 2×E (from eq.3a): C + 2×E + D = C + E + 2×F Subtract both C and E from each side: C + 2×E + D – C – E = C + E + 2×F – C – E which simplifies to E + D = 2×F In eq.1a, replace 2×F with E + D: E + D = A + E Subtract E from both sides: E + D – E = A + E – E which makes D = A
Hint #4
Substitute A for D in eq.4: B – F = A ÷ A which becomes B – F = 1 Add F to both sides of the above equation: B – F + F = 1 + F which makes eq.4a) B = 1 + F
Hint #5
eq.2 may be written as: E + F – D + E = A + C which becomes 2×E + F – D = A + C Add D to both sides of the above equation: 2×E + F – D + D = A + C + D which becomes eq.2a) 2×E + F = A + C + D
Hint #6
Add C to both sides of eq.6a: A + B + D + C = C + E + 2×F + C which is equivalent to A + C + D + B = 2×C + E + 2×F Substitute 2×E + F for A + C + D (from eq.2a) in the equation above: 2×E + F + B = 2×C + E + 2×F Subtract E and F from both sides: 2×E + F + B – E – F = 2×C + E + 2×F – E – F which simplifies to eq.6b) E + B = 2×C + F
Hint #7
Substitute 1 + F for B (from eq.4a) in eq.6b: E + 1 + F = 2×C + F Subtract both 1 and F from each side: E + 1 + F – 1 – F = 2×C + F – 1 – F which means eq.6c) E = 2×C – 1
Hint #8
eq.5 may be written as: 10×C + D = E × E Substitute (2×C – 1) for E (from eq.6c) in the above equation: 10×C + D = (2×C – 1) × (2×C – 1) which becomes 10×C + D = 4×C² – 4×C + 1 Subtract 10×C from each side: 10×C + D – 10×C = 4×C² – 4×C + 1 – 10×C which means D = 4×C² – 14×C + 1 and which also makes eq.5a) A = D = 4×C² – 14×C + 1
Hint #9
Substitute 4×C² – 14×C + 1 for A (from eq.5a), and (2×C – 1) for E (from eq.6c) in eq.1a: 2×F = 4×C² – 14×C + 1 + (2×C – 1) which becomes 2×F = 4×C² – 12×C Divide both sides by 2: 2×F ÷ 2 = (4×C² – 12×C) ÷ 2 which means eq.1b) F = 2×C² – 6×C
Hint #10
Substitute 2×C² – 6×C for F (from eq.1b) in eq.4a: eq.4b) B = 1 + 2×C² – 6×C
Hint #11
Substitute 4×C² – 14×C + 1 for A (from eq.5a), 1 + 2×C² – 6×C for B (from eq.4b), and (2×C – 1) for E (from eq.6c) in eq.3a: 4×C² – 14×C + 1 + 1 + 2×C² – 6×C = C + 2×(2×C – 1) which becomes 6×C² – 20×C + 2 = C + 4×C – 2 which becomes 6×C² – 20×C + 2 = 5×C – 2 In the above equation, subtract 5×C from both sides, and add 2 to both sides: 6×C² – 20×C + 2 – 5×C + 2 = 5×C – 2 – 5×C + 2 which becomes eq.3b) 6×C² – 25×C + 4 = 0
Solution
eq.3b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for C in eq.3b yields: C = { (–1)×(–25) ± sq.rt.[(–25)² – (4 × 6 × 4)] } ÷ (2 × 6) which becomes C = {25 ± sq.rt.[625 – 96]} ÷ 12 which becomes C = {25 ± sq.rt.[529]} ÷ 12 which becomes C = {25 ± 23} ÷ 12 In the above equation, either C = {25 + 23} ÷ 12 = 48 ÷ 12 = 4 or C = {25 – 23} ÷ 12 = 2 ÷ 12 = 0.1666666667 Since C must be a non-negative integer, then C ≠ 0.1666666667 Therefore, C = 4 making A = D = 4×C² – 14×C + 1 = 4×4² – 14×4 + 1 = 4×16 – 56 + 1 = 64 – 55 = 9 (from eq.5a) B = 1 + 2×C² – 6×C = 1 + 2×4² – 6×4 = 1 + 2×16 – 24 = 1 + 32 – 24 = 9 (from eq.4b) E = 2×C – 1 = 2×4 – 1 = 8 – 1 = 7 (from eq.6c) F = 2×C² – 6×C = 2×4² – 6×4 = 2×16 – 24 = 32 – 24 = 8 (from eq.1b) and ABCDEF = 994978