Puzzle for July 15, 2023 ( )
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, DE, and EF are 2-digit numbers (not A×B, C×D, D×E, or E×F).
Scratchpad
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Hint #1
eq.5 may be written as: 10×C + D - (10×D + E) = C + D which becomes 10×C + D - 10×D - E = C + D which becomes 10×C - 9×D - E = C + D In the above equation, add 9×D to both sides, and subtract C from both sides: 10×C - 9×D - E + 9×D - C = C + D + 9×D - C which becomes eq.5a) 9×C - E = 10×D
Hint #2
In eq.5a, substitute (C - D) for E (from eq.2): 9×C - (C - D) = 10×D which becomes 9×C - C + D = 10×D which becomes 8×C + D = 10×D Subtract D from each side of the equation above: 8×C + D - D = 10×D - D which makes 8×C = 9×D Divide both sides by 8: 8×C ÷ 8 = 9×D ÷ 8 which makes eq.5b) C = 1⅛×D
Hint #3
In eq.2, replace C with 1⅛×D: E = 1⅛×D - D which makes E = ⅛×D Multiply both sides of the above equation by 8: 8 × E = 8 × (⅛×D) which makes 8×E = D
Hint #4
In eq.5b, substitute (8×E) for D: C = 1⅛×(8×E) which makes C = 9×E
Hint #5
Add C to both sides of eq.4: AB - C + C = E + C which becomes eq.4a) AB = E + C
Hint #6
Substitute (E + C) for AB (from eq.4a) into eq.6: EF - (E + C) = (E + C) - D which may be written as 10×E + F - E - C = E + C - D which becomes 9×E + F - C = E + C - D In the above equation, add C to both sides, and subtract E from both sides: 9×E + F - C + C - E = E + C - D + C - E which becomes eq.6a) 8×E + F = 2×C - D
Hint #7
Substitute (9×E) for C, and 8×E for D in eq.6a: 8×E + F = 2×(9×E) - 8×E which becomes 8×E + F = 18×E - 8×E which becomes 8×E + F = 10×E Subtract 8×E from each side of the equation above: 8×E + F - 8×E = 10×E - 8×E which makes F = 2×E
Hint #8
eq.4a may be written as: 10×A + B = E + C Substitute 9×E for C in the above equation: 10×A + B = E + 9×E which becomes 10×A + B = 10×E Subtract 10×A from both sides of the above equation: 10×A + B - 10×A = 10×E - 10×A which becomes eq.4b) B = 10×E - 10×A
Hint #9
Substitute 2×E for F, and (10×E - 10×A) for B (from eq.4b) in eq.3: 2×E - A = A - (10×E - 10×A) which becomes 2×E - A = A - 10×E + 10×A which becomes 2×E - A = 11×A - 10×E Add A and 10×E to both sides of the equation above: 2×E - A + A + 10×E = 11×A - 10×E + A + 10×E which makes 12×E = 12×A Divide both sides by 12: 12×E ÷ 12 = 12×A ÷ 12 which makes E = A
Hint #10
Substitute E for A in eq.4b: B = 10×E - 10×E which makes B = 0
Solution
Substitute E for A, 0 for B, 9×E for C, 8×E for D, and 2×E for F in eq.1: E + 0 + 9×E + 8×E + E + 2×E = 21 which simplifies to 21×E = 21 Divide both sides of the above equation by 21: 21×E ÷ 21 = 21 ÷ 21 which means E = 1 making A = E = 1 C = 9×E = 9 × 1 = 9 D = 8×E = 8 × 1 = 8 F = 2×E = 2 × 1 = 2 and ABCDEF = 109812