Puzzle for April 14, 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.
* CD and EF are 2-digit numbers (not C×D or E×F).
Scratchpad
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Hint #1
Add B + C + E to both sides of eq.4: A - B - C + B + C + E = D - E + B + C + E which becomes eq.4a) A + E = D + B + C Subtract B from both sides of eq.5: C + E - B = B + F - B which becomes eq.5a) C + E - B = F
Hint #2
eq.6 can be written as: 10×C + D = A + 10×E + F which may also be written as 10×C + D = A + E + 9×E + F In the equation above, replace A + E with D + B + C (from eq.4a): 10×C + D = D + B + C + 9×E + F Subtract both C and D from each side: 10×C + D - C - D = D + B + C + 9×E + F - C - D which becomes eq.6a) 9×C = B + 9×E + F
Hint #3
In eq.6a, substitute (C + E - B) for F (from eq.5a): 9×C = B + 9×E + (C + E - B) which becomes 9×C = 10×E + C Subtract C from both sides: 9×C - C = 10×E + C - C which makes 8×C = 10×E Divide both sides by 8: 8×C ÷ 8 = 10×E ÷ 8 which means C = 1¼×E
Hint #4
Add D and F to both sides of eq.2: C - D + D + F = D - F + D + F which becomes C + F = 2×D Subtract C from both sides: C + F - C = 2×D - C which becomes eq.2a) F = 2×D - C
Hint #5
Add A and E to both sides of eq.3: F - A + A + E = C - E + A + E which becomes eq.3a) F + E = C + A
Hint #6
Substitute (2×D - C) for F (from eq.2a) into eq.3a: (2×D - C) + E = C + A Subtract C from both sides of the equation above: (2×D - C) + E - C = C + A - C which becomes 2×D + E - 2×C = A Substitute (1¼×E) for C: 2×D + E - 2×(1¼×E) = A which is equivalent to 2×D + E - 2½×E = A which becomes eq.3b) 2×D - 1½×E = A
Hint #7
Substitute (2×D - C) for F (from eq.2a) in eq.5: B + 2×D - C = C + E Add (C - 2×D) to both sides of the above equation: B + 2×D - C + (C - 2×D) = C + E + (C - 2×D) which becomes B = 2×C + E - 2×D Substitute (1¼×E) for C: B = 2×(1¼×E) + E - 2×D which becomes B = 2½×E + E - 2×D which makes eq.5b) B = 3½×E - 2×D
Hint #8
Substitute (2×D - 1½×E) for A (from eq.3b), (3½×E - 2×D) for B (from eq.5b), and 1¼×E for C in eq.4a: (2×D - 1½×E) + E = D + (3½×E - 2×D) + 1¼×E which becomes 2×D - ½×E = -D + 4¾×E Add D + ½×E to both sides: 2×D - ½×E + D + ½×E = -D + 4¾×E + D + ½×E which simplifies to 3×D = 5¼×E Divide both sides by 3: 3×D ÷ 3 = 5¼×E ÷ 3 which makes D = 1¾×E
Hint #9
In eq.3b, substitute (1¾×E) for D: 2×(1¾×E) - 1½×E = A which becomes 3½×E - 1½×E = A which makes 2×E = A
Hint #10
In eq.5b, substitute (1¾×E) for D: B = 3½×E - 2×(1¾×E) which is the same as B = 3½×E - 3½×E which makes B = 0
Hint #11
In eq.2a, substitute (1¼×E) for C, and 1¾×E for D: F = 2×(1¾×E) - 1¼×E which becomes F = 3½×E - 1¼×E which makes F = 2¼×E
Solution
Substitute 2×E for A, 0 for B, 1¼×E for C, 1¾×E for D, and 2¼×E for F in eq.1: 2×E + 0 + 1¼×E + 1¾×E + E + 2¼×E = 33 which simplifies to 8¼×E = 33 Divide both sides by 8¼: 8¼×E ÷ 8¼ = 33 ÷ 8¼ which means E = 4 making A = 2×E = 2 × 4 = 8 C = 1¼×E = 1¼ × 4 = 5 D = 1¾×E = 1¾ × 4 = 7 F = 2¼×E = 2¼ × 4 = 9 and ABCDEF = 805749