Puzzle for January 19, 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.
* AB, BC, and CD are 2-digit numbers (not A×B, B×C, or C×D).
Scratchpad
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Hint #1
eq.6 may be written as: 10×A + B + 10×B + C = 10×C + D + E Subtract C from each side of the above equation: 10×A + B + 10×B + C – C = 10×C + D + E – C which becomes eq.6a) 10×A + 11×B = 9×C + D + E
Hint #2
Add C and E to both sides of eq.3: B – C + C + E = D – E + C + E which becomes B + E = D + C Subtract B from both sides: B + E – B = D + C – B which becomes eq.3a) E = D + C – B
Hint #3
In eq.6a, replace E with D + C – B (from eq.3a): 10×A + 11×B = 9×C + D + D + C – B which becomes 10×A + 11×B = 10×C + 2×D – B Add B to both sides: 10×A + 11×B + B = 10×C + 2×D – B + B which becomes eq.6b) 10×A + 12×B = 10×C + 2×D
Hint #4
In eq.6b, substitute (A + D) for C (from eq.2): 10×A + 12×B = 10×(A + D) + 2×D which is the same as 10×A + 12×B = 10×A + 10×D + 2×D Subtract 10×A from both sides: 10×A + 12×B – 10×A = 10×A + 10×D + 2×D – 10×A which simplifies to 12×B = 12×D Divide both sides by 12: 12×B ÷ 12 = 12×D ÷ 12 which makes B = D
Hint #5
In eq.3a, substitute D for B : E = D + C – D which means E = C
Hint #6
Substitute D for B, and C for E in eq.4: F – C = C – D – (D – A) which is equivalent to F – C = C – D – D + A Add C to both sides of the equation above: F – C + C = C – D – D + A + C which becomes eq.4a) F = 2×C – 2×D + A
Hint #7
In eq.4a, substitute (A + D) for C (from eq.2): F = 2×(A + D) – 2×D + A which is the same as F = 2×A + 2×D – 2×D + A which makes eq.4b) F = 3×A
Hint #8
Substitute C for E, 3×A for F, and D for B in eq.5: C + C – 3×A = 3×A – D Add 3×A to both sides: C + C – 3×A + 3×A = 3×A – D + 3×A which becomes eq.5a) 2×C = 6×A – D
Hint #9
Substitute (A + D) for C (from eq.2a) in eq.5a: 2×(A + D) = 6×A – D which is equivalent to 2×A + 2×D = 6×A – D Add D to both sides, and subtract 2×A from each side: 2×A + 2×D + D – 2×A = 6×A – D + D – 2×A which simplifies to 3×D = 4×A Divide both sides by 4: 3×D ÷ 4 = 4×A ÷ 4 which means ¾×D = A
Hint #10
Substitute (¾×D) for A in eq.4b: eq.4b) F = 3×(¾×D) which makes F = 2¼×D
Hint #11
Substitute (¾×D) for A in eq.5a: 2×C = 6×(¾×D) – D which is equivalent to 2×C = 4½×D – D which becomes 2×C = 3½×D Divide both sides of the above equation by 2: 2×C ÷ 2 = 3½×D ÷ 2 which makes C = 1¾×D and also makes E = C = 1¾×D
Solution
Substitute ¾×D for A, D for B, 1¾×D for C and E, and 2¼×D for F in eq.1: ¾×D + D + 1¾×D + D + 1¾×D + 2¼×D = 34 which simplifies to 8½×D = 34 Divide both sides by 8½: 8½×D ÷ 8½ = 34 ÷ 8½ which means D = 4 making A = ¾×D = ¾ × 4 = 3 B = D = 4 C = E = 1¾×D = 1¾ × 4 = 7 F = 2¼×D = 2¼ × 4 = 9 and ABCDEF = 347479