Puzzle for October 29, 2021 ( )
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
Add C to both sides of eq.2: D – C + C = C – E + C which becomes eq.2a) D = 2×C – E In eq.5, replace D with 2×C – E (from eq.2a): B + 2×C – E – E = E – C which becomes B + 2×C – 2×E = E – C In the above equation, subtract 2×C from both sides, and add 2×E to both sides: B + 2×C – 2×E – 2×C + 2×E = E – C – 2×C + 2×E which becomes eq.5a) B = 3×E – 3×C
Hint #2
eq.6 may be written as: 10×C + D – A = 10×D + E + A In the equation above, subtract D from both sides, and add A to both sides: 10×C + D – A – D + A = 10×D + E + A – D + A which becomes eq.6a) 10×C = 9×D + E + 2×A
Hint #3
In eq.6a, substitute (2×C – E) for D (from eq.2a): 10×C = 9×(2×C – E) + E + 2×A which becomes 10×C = 18×C – 9×E + E + 2×A which becomes 10×C = 18×C – 8×E + 2×A In the equation above, subtract 18×C from both sides, and add 8×E to both sides: 10×C – 18×C + 8×E = 18×C – 8×E + 2×A – 18×C + 8×E which becomes –8×C + 8×E = 2×A which may be written as 8×E – 8×C = 2×A Divide both sides by 2: (8×E – 8×C) ÷ 2 = 2×A ÷ 2 which becomes eq.6b) 4×E – 4×C = A
Hint #4
Subtract F from both sides of eq.4: E + F – A – F = A + C – F which becomes E – A = A + C – F Substitute A – B for C – F (from eq.3) in the equation above: E – A = A + A – B which becomes E – A = 2×A – B Add A to both sides: E – A + A = 2×A – B + A which becomes eq.4a) E = 3×A – B
Hint #5
Substitute (4×E – 4×C) for A (from eq.6b), and (3×E – 3×C) for B (from eq.5a) in eq.4a: E = 3×(4×E – 4×C) – (3×E – 3×C) which becomes E = 12×E – 12×C – 3×E + 3×C which becomes E = 9×E – 9×C In the above equation, subtract E from both sides, and add 9×C to both sides: E – E + 9×C = 9×E – 9×C – E + 9×C which makes 9×C = 8×E Divide both sides by 8: 9×C ÷ 8 = 8×E ÷ 8 which makes 1⅛×C = E
Hint #6
Substitute (1⅛×C) for E in eq.6b: 4×(1⅛×C) – 4×C = A which becomes 4½×C – 4×C = A which makes ½×C = A
Hint #7
Substitute (1⅛×C) for E in eq.5a: B = 3×(1⅛×C) – 3×C which becomes B = 3⅜×C – 3×C which makes B = ⅜×C
Hint #8
Substitute 1⅛×C for E in eq.2a: D = 2×C – 1⅛×C which makes D = ⅞×C
Hint #9
Substitute 1⅛×C for E, and ½×C for A in eq.4: 1⅛×C + F – ½×C = ½×C + C ⅝×C + F = 1½×C Subtract ⅝×C from each side of the above equation: ⅝×C + F – ⅝×C = 1½×C – ⅝×C F = ⅞×C
Solution
Substitute ½×C for A, ⅜×C for B, ⅞×C for D and F, and 1⅛×C for E in eq.1: ½×C + ⅜×C + C + ⅞×C + 1⅛×C + ⅞×C = 38 which simplifies to 4¾×C = 38 Divide both sides of the above equation by 4¾: 4¾×C ÷ 4¾ = 38 ÷ 4¾ which makes C = 8 making A = ½×C = ½ × 8 = 4 B = ⅜×C = ⅜ × 8 = 3 D = F = ⅞×C = ⅞ × 8 = 7 E = 1⅛×C = 1⅛ × 8 = 9 and ABCDEF = 438797