Puzzle for December 6, 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.
* CD and DE are 2-digit numbers (not C×D or D×E).
Scratchpad
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Hint #1
In eq.5, replace F with A + C (from eq.3): E + A + C – D = A + D In the above equation, subtract A from both sides, and add D to both sides: E + A + C – D – A + D = A + D – A + D which becomes E + C = 2×D which may be written as eq.5a) C + E = 2×D Subtract E from both sides of eq.5a: C + E – E = 2×D – E which becomes eq.5b) C = 2×D – E
Hint #2
Add E to both sides of eq.2: B + C + E = D + E + E In the equation above, replace C + E with 2×D (from eq.5a): B + 2×D = D + E + E Subtract 2×D from both sides: B + 2×D – 2×D = D + E + E – 2×D which becomes eq.2a) B = 2×E – D
Hint #3
In eq.4, substitute 2×D – E for C (from eq.5b), and 2×E – D for B (from eq.2a): 2×D – E + D = A + 2×E – D which becomes 3×D – E = A + 2×E – D In the equation above, subtract 2×E from both sides, and add D to both sides: 3×D – E – 2×E + D = A + 2×E – D – 2×E + D which becomes eq.4a) 4×D – 3×E = A
Hint #4
Substitute 4×D – 3×E for A (from eq.4a), and 2×D – E for C (from eq.5b) in eq.3: F = 4×D – 3×E + 2×D – E which becomes eq.3a) F = 6×D – 4×E
Hint #5
eq.6 may be written as: 10×D + E – F = A + 10×C + D Substitute (6×D – 4×E) for F (from eq.3a), 4×D – 3×E for A (from eq.4a), and (2×D – E) for C (from eq.5b) in the above equation: 10×D + E – (6×D – 4×E) = 4×D – 3×E + 10×(2×D – E) + D which is equivalent to 10×D + E – 6×D + 4×E = 4×D – 3×E + 20×D – 10×E + D which becomes 4×D + 5×E = 25×D – 13×E Subtract 4×D from both sides, and add 13×E to both sides: 4×D + 5×E – 4×D + 13×E = 25×D – 13×E – 4×D + 13×E which becomes 18×E = 21×D Divide both sides by 18: 18×E ÷ 18 = 21×D ÷ 18 which becomes E = 1⅙×D
Hint #6
Substitute (1⅙×D) for E in eq.3a: F = 6×D – 4×(1⅙×D) which becomes F = 6×D – 4⅔×D which makes F = 1⅓×D
Hint #7
Substitute (1⅙×D) for E in eq.4a: 4×D – 3×(1⅙×D) = A which becomes 4×D – 3½×D = A which makes ½×D = A
Hint #8
Substitute (1⅙×D) for E in eq.2a: B = 2×(1⅙×D) – D which becomes B = 2⅓×D – D which makes B = 1⅓×D
Hint #9
Substitute 1⅙×D for E in eq.5b: C = 2×D – 1⅙×D which makes C = ⅚×D
Solution
Substitute ½×D for A, 1⅓×D for B and F, ⅚×D for C, and 1⅙×D for E in eq.1: ½×D + 1⅓×D + ⅚×D + D + 1⅙×D + 1⅓×D = 37 which simplifies to 6⅙×D = 37 Divide both sides of the above equation by 6⅙: 6⅙×D ÷ 6⅙ = 37 ÷ 6⅙ which means D = 6 making A = ½×D = ½ × 6 = 3 B = F = 1⅓×D = 1⅓ × 6 = 8 C = ⅚×D = ⅚ × 6 = 5 E = 1⅙×D = 1⅙ × 6 = 7 and ABCDEF = 385678