Puzzle for March 29, 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 DE are 2-digit numbers (not A×B, C×D, or D×E).
** ABC and DEF are 3-digit numbers (not A×B×C or D×E×F).
Scratchpad
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Hint #1
Add D to both sides of eq.3: B + C – D + D = D + F + D which becomes B + C = 2×D + F In the equation above, replace B with E + F (from eq.2): E + F + C = 2×D + F Subtract F and C from both sides: E + F + C – F – C = 2×D + F – F – C which becomes eq.3a) E = 2×D – C
Hint #2
eq.4 may be written as: 10×D + E – (10×A + B) – A = A + C – D which becomes 10×D + E – 11×A – B = A + C – D Add 11×A, B, and D to each side of the equation above: 10×D + E – 11×A – B + 11×A + B + D = A + C – D + 11×A + B + D which simplifies to eq.4a) 11×D + E = 12×A + B + C
Hint #3
In eq.4a, replace E with 2×D – C (from eq.3a): 11×D + 2×D – C = 12×A + B + C which becomes 13×D – C = 12×A + B + C Subtract C from each side of the equation above: 13×D – C – C = 12×A + B + C – C which becomes eq.4b) 13×D – 2×C = 12×A + B
Hint #4
eq.5 may be written as: 10×C + D – A = 10×A + B + 10×D + E + A + C which becomes 10×C + D – A = 11×A + B + 10×D + E + C In the above equation, add A to both sides, and subtract D and C from both sides: 10×C + D – A + A – D – C = 11×A + B + 10×D + E + C + A – D – C which simplifies to eq.5a) 9×C = 12×A + B + 9×D + E
Hint #5
Substitute 13×D – 2×C for 12×A + B (from eq.4b), and 2×D – C for E (from eq.3a) in eq.5a: 9×C = 13×D – 2×C + 9×D + 2×D – C which becomes 9×C = 24×D – 3×C Add 3×C to both sides of the above equation: 9×C + 3×C = 24×D – 3×C + 3×C which makes 12×C = 24×D Divide both sides by 12: 12×C ÷ 12 = 24×D ÷ 12 which makes C = 2×D
Hint #6
Substitute 2×D for C in eq.3a: E = 2×D – 2×D which means E = 0
Hint #7
Substitute 0 for E in eq.2: B = 0 + F which makes B = F
Hint #8
eq.6 may be written as: 100×D + 10×E + F – C = 100×A + 10×B + C + 10×C + D which becomes 100×D + 10×E + F – C = 100×A + 10×B + 11×C + D In the equation above, add C to both sides, and subtract D from each side: 100×D + 10×E + F – C + C – D = 100×A + 10×B + 11×C + D + C – D which becomes eq.6a) 99×D + 10×E + F = 100×A + 10×B + 12×C
Hint #9
Substitute 0 for E, B for F, and (2×D) for C in eq.6a: 99×D + 10×0 + B = 100×A + 10×B + 12×(2×D) which becomes 99×D + B = 100×A + 10×B + 24×D Subtract B and 24×D from both sides of the above equation: 99×D + B – B – 24×D = 100×A + 10×B + 24×D – B – 24×D which becomes eq.6b) 75×D = 100×A + 9×B
Hint #10
Substitute (2×D) for C in eq.4b: 13×D – 2×(2×D) = 12×A + B which is equivalent to 13×D – 4×D = 12×A + B which becomes 9×D = 12×A + B Subtract 12×A from each side of the above equation: 9×D – 12×A = 12×A + B – 12×A which becomes eq.5b) 9×D – 12×A = B
Hint #11
Substitute (9×D – 12×A) for B (from eq.5b) in eq.6b: 75×D = 100×A + 9×(9×D – 12×A) which becomes 75×D = 100×A + 81×D – 108×A which becomes 75×D = 81×D – 8×A In the equation above, add 8×A to both sides, and subtract 75×D from both sides: 75×D + 8×A – 75×D = 81×D – 8×A + 8×A – 75×D which becomes 8×A = 6×D Divide both sides by 8: 8×A ÷ 8 = 6×D ÷ 8 which makes A = ¾×D
Hint #12
Substitute (¾×D) for A in eq.5b: 9×D – 12×(¾×D) = B which becomes 9×D – 9×D = B which makes 0 = B and also makes F = B = 0
Solution
Substitute ¾×D for A, 0 for B and E and F, and 2×D for C in eq.1: ¾×D + 0 + 2×D + D + 0 + 0 = 15 which simplifies to 3¾×D = 15 Divide both sides of the equation above by 3¾: 3¾×D ÷ 3¾ = 15 ÷ 3¾ which means D = 4 making A = ¾×D = ¾ × 4 = 3 C = 2×D = 2 × 4 = 8 and ABCDEF = 308400