Puzzle for March 10, 2019 ( )
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, CD, and DE are 2-digit numbers (not A×B, B×C, C×D, or D×E).
Scratchpad
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Hint #1
Subtract D from both sides of eq.3: C + E + F - D = A + B + D - D which becomes eq.3a) C + E + F - D = A + B
Hint #2
eq.4 may be written as: 10×A + B = 10×D + E + F Subtract 10×D from both sides of the above equation: 10×A + B - 10×D = 10×D + E + F - 10×D which becomes eq.4a) 10×A + B - 10×D = E + F which may also be written as eq.4b) 9×A + A + B - 10×D = E + F
Hint #3
In eq.4b, replace A + B with C + E + F - D (from eq.3a): 9×A + C + E + F - D - 10×D = E + F Subtract both E and F from each side of the above equation: 9×A + C + E + F - D - 10×D - E - F = E + F - E - F which becomes 9×A + C - 11×D = 0 Add (11×D - 9×A) to both sides: 9×A + C - 11×D + (11×D - 9×A) = 0 + (11×D - 9×A) which becomes eq.4c) C = 11×D - 9×A
Hint #4
In eq.2, replace C with 11×D - 9×A (from eq.4c): B + D = A - B + 11×D - 9×A Add (B - D) to each side of the above equation: B + D + (B - D) = A - B + 11×D - 9×A + (B - D) which becomes 2×B = 10×D - 8×A Divide both sides of the above equation by 2: 2×B ÷ 2 = (10×D - 8×A) ÷ 2 which becomes eq.2a) B = 5×D - 4×A
Hint #5
eq.6 may be written as: 10×D + E - (10×B + C) = 10×B + C - F which is equivalent to 10×D + E - 10×B - C = 10×B + C - F Add C + F to both sides of the above equation: 10×D + E - 10×B - C + C + F = 10×B + C - F + C + F which becomes 10×D + E - 10×B + F = 10×B + 2×C Add (10×B - 10×D) to both sides: 10×D + E - 10×B + F + (10×B - 10×D) = 10×B + 2×C + (10×B - 10×D) which becomes eq.6a) E + F = 20×B + 2×C - 10×D
Hint #6
In eq.6a, substitute 10×A + B - 10×D for E + F (from eq.4a): 10×A + B - 10×D = 20×B + 2×C - 10×D Add (10×D - B) to each side of the equation above: 10×A + B - 10×D + (10×D - B) = 20×B + 2×C - 10×D + (10×D - B) which becomes eq.6b) 10×A = 19×B + 2×C
Hint #7
Substitute (5×D - 4×A) for B (from eq.2a), and (11×D - 9×A) for C (from eq.4c) in eq.6b: 10×A = 19×(5×D - 4×A) + 2×(11×D - 9×A) which is equivalent to 10×A = 95×D - 76×A + 22×D - 18×A which becomes 10×A = 117×D - 94×A Add 94×A to each side: 10×A + 94×A = 117×D - 94×A + 94×A which becomes 104×A = 117×D Divide both sides by 104: 104×A ÷ 104 = 117×D ÷ 104 which makes A = 1⅛×D
Hint #8
Substitute (1⅛×D) for A in eq.2a: B = 5×D - 4×(1⅛×D) which becomes B = 5×D - 4½×D which makes B = ½×D
Hint #9
Substitute (1⅛×D) for A in eq.4c: C = 11×D - 9×(1⅛×D) which becomes C = 11×D - 10⅛×D which makes C = ⅞×D
Hint #10
eq.5 may be written as: A + 10×C + D = 10×D + E Subtract 10×D from each side: A + 10×C + D - 10×D = 10×D + E - 10×D which becomes A + 10×C - 9×D = E Substitute 1⅛×D for A, and (⅞×D) for C: 1⅛×D + 10×(⅞×D) - 9×D = E which becomes 1⅛×D + 8¾×D - 9×D = E which makes ⅞×D = E
Hint #11
Substitute (1⅛×D) for A, ½×D for B, and ⅞×D for E in eq.4b: 9×(1⅛×D) + (1⅛×D) + ½×D - 10×D = ⅞×D + F which becomes 10⅛×D + 1⅝×D - 10×D = ⅞×D + F which becomes 1¾×D = ⅞×D + F Subtract ⅞×D from each side of the above equation: 1¾×D - ⅞×D = ⅞×D + F - ⅞×D which becomes ⅞×D = F
Solution
Substitute 1⅛×D for A, ½×D for B, and ⅞×D for C and E and F in eq.1: 1⅛×D + ½×D + ⅞×D + D + ⅞×D + ⅞×D = 42 which simplifies to 5¼×D = 42 Divide both sides by 5¼: 5¼×D ÷ 5¼ = 42 ÷ 5¼ which becomes D = 8 making A = 1⅛×D = 1⅛ × 8 = 9 B = ½×D = ½ × 8 = 4 C = E = F = ⅞×D = ⅞ × 8 = 7 and ABCDEF = 947877