Puzzle for February 23, 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, CD, and EF are 2-digit numbers (not A×B, C×D, or E×F).
** ABC, BCD, and DEF are 3-digit numbers (not A×B×C, B×C×D, or D×E×F).
Scratchpad
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Hint #1
Add D and C to both sides of eq.4: F – A + C – D + D + C = E – C + D + C which becomes F – A + 2×C = E + D which may be written as F – A + 2×C = D + E In the above equation, replace D + E with F (from eq.2): F – A + 2×C = F Subtract F from each side, and add A to each side: F – A + 2×C – F + A = F – F + A which simplifies to 2×C = A
Hint #2
eq.5 may be written as: 10×E + F – (10×C + D) = 10×A + B – E which is equivalent to 10×E + F – 10×C – D = 10×A + B – E In the above equation, replace F with D + E (from eq.2), and add E to both sides: 10×E + D + E – 10×C – D + E = 10×A + B – E + E which becomes eq.4a) 12×E – 10×C = 10×A + B
Hint #3
In eq.4a, replace A with (2×C): 12×E – 10×C = 10×(2×C) + B which is the same as 12×E – 10×C = 20×C + B Subtract 20×C from both sides of the equation above: 12×E – 10×C – 20×C = 20×C + B – 20×C which becomes eq.5a) 12×E – 30×C = B
Hint #4
In eq.3, substitute 2×C for A, and 12×E – 30×C for B (from eq.5a): 2×C + 12×E – 30×C = D which becomes eq.3a) 12×E – 28×C = D
Hint #5
eq.6 may be written as: 100×B + 10×C + D + F = 100×D + 10×E + F – (10×C + D) – (100×A + 10×B + C) which is equivalent to 100×B + 10×C + D + F = 100×D + 10×E + F – 10×C – D – 100×A – 10×B – C which becomes 100×B + 10×C + D + F = 99×D + 10×E + F – 11×C – 100×A – 10×B Subtract D and F from both sides of the above equation: 100×B + 10×C + D + F – D – F = 99×D + 10×E + F – 11×C – 100×A – 10×B – D – F which becomes 100×B + 10×C = 98×D + 10×E – 11×C – 100×A – 10×B Add 11×C and 10×B to both sides: 100×B + 10×C + 11×C + 10×B = 98×D + 10×E – 11×C – 100×A – 10×B + 11×C + 10×B which becomes eq.6a) 110×B + 21×C = 98×D + 10×E – 100×A
Hint #6
Substitute (12×E – 28×C) for D (from eq.3a), (2×C) for A, and (12×E – 30×C) for B (from eq.5a) in eq.6a: 98×(12×E – 28×C) + 10×E – 100×(2×C) = 110×(12×E – 30×C) + 21×C which becomes 1176×E – 2744×C + 10×E – 200×C = 1320×E – 3300×C + 21×C which becomes 1186×E – 2944×C = 1320×E – 3279×C In the above equation, add 3279×C to both sides, and subtract 1186×E from both sides: 1186×E – 2944×C + 3279×C – 1186×E = 1320×E – 3279×C + 3279×C – 1186×E which becomes 335×C = 134×E Divide both sides by 134: 335×C ÷ 134 = 134×E ÷ 134 which makes 2½×C = E
Hint #7
Substitute (2½×C) for E in eq.5a: 12×(2½×C) – 30×C = B which becomes 30×C – 30×C = B which means 0 = B
Hint #8
Substitute 2×C for A, and 0 for B in eq.3: 2×C + 0 = D which makes 2×C = D
Hint #9
Substitute 2×C for D, and 2½×C for E in eq.2: 2×C + 2½×C = F which makes 4½×C = F
Solution
Substitute 2×C for A and D, 0 for B, 2½×C for E, and 4½×C for F in eq.1: 2×C + 0 + C + 2×C + 2½×C + 4½×C = 24 which simplifies to 12×C = 24 Divide both sides of the equation above by 12: 12×C ÷ 12 = 24 ÷ 12 which means C = 2 making A = D = 2×C = 2 × 2 = 4 E = 2½×C = 2½ × 2 = 5 F = 4½×C = 4½ × 2 = 9 and ABCDEF = 402459