Puzzle for June 2, 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, DE, and EF are 2-digit numbers (not A×B, D×E, or E×F).
Scratchpad
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Hint #1
eq.5 may be written as: eq.5a) 10×D + E = C + F Subtract both E and F from each sides of eq.5a: 10×D + E - E - F = C + F - E - F which becomes 10×D - F = C - E In eq.4, replace C - E with 10×D - F: 10×D - F = A + B - D Add D and F to both sides: 10×D - F + D + F = A + B - D + D + F which becomes eq.5b) 11×D = A + B + F
Hint #2
In eq.5b, replace F with A + D (from eq.3): 11×D = A + B + A + D Subtract D from both sides of the above equation: 11×D - D = A + B + A + D - D which becomes eq.4a) 10×D = 2×A + B
Hint #3
eq.6 may be written as: 10×E + F - D = 10×A + B + D - (10×E + F) which is equivalent to 10×E + F - D = 10×A + B + D - 10×E - F Add D, 10×E, and F to both sides of the above equation: 10×E + F - D + D + 10×E + F = 10×A + B + D - 10×E - F + D + 10×E + F which becomes 20×E + 2×F = 10×A + B + 2×D Substitute (A + D) for F (from eq.3): 20×E + 2×(A + D) = 10×A + B + 2×D which is equivalent to 20×E + 2×A + 2×D = 10×A + B + 2×D Subtract 10×A and 2×D from both sides: 20×E + 2×A + 2×D - 10×A - 2×D = 10×A + B + 2×D - 10×A - 2×D which becomes eq.6a) 20×E - 8×A = B
Hint #4
Substitute A + B - D for C - E (from eq.4), and A + D for F (from eq.3) in eq.2: A + B + A + B - D = D + E + A + D which is the same as 2×A + 2×B - D = 2×D + E + A In the above equation, subtract A from both sides, and add D to both sides: 2×A + 2×B - D - A + D = 2×D + E + A - A + D which becomes eq.2a) A + 2×B = 3×D + E
Hint #5
Substitute (20×E - 8×A) for B (from eq.6a) in eq.2a: A + 2×(20×E - 8×A) = 3×D + E which is the same as A + 40×E - 16×A = 3×D + E which becomes 40×E - 15×A = 3×D + E Subtract E from both sides of the equation above: 40×E - 15×A - E = 3×D + E - E which becomes 39×E - 15×A = 3×D Divide both sides by 3: (39×E - 15×A) ÷ 3 = 3×D ÷ 3 which becomes eq.2b) 13×E - 5×A = D
Hint #6
Substitute (13×E - 5×A) for D (from eq.2b), and 20×E - 8×A for B (from eq.6a) in eq.4a: 10×(13×E - 5×A) = 2×A + 20×E - 8×A which becomes 130×E - 50×A = 20×E - 6×A In the above equation, add 50×A to both sides, and subtract 20×E from both sides: 130×E - 50×A + 50×A - 20×E = 20×E - 6×A + 50×A - 20×E which simplifies to 110×E = 44×A Divide both sides by 44: 110×E ÷ 44 = 44×A ÷ 44 which makes 2½×E = A
Hint #7
Substitute (2½×E) for A in eq.2b: 13×E - 5×(2½×E) = D which becomes 13×E - 12½×E = D which makes ½×E = D
Hint #8
Substitute (2½×E) for A in eq.6a: 20×E - 8×(2½×E) = B which becomes 20×E - 20×E = B which means 0 = B
Hint #9
Substitute 2½×E for A, and ½×E for D in eq.3: F = 2½×E + ½×E which makes F = 3×E
Hint #10
Substitute (½×E) for D, and 3×E for F in eq.5a: 10×(½×E) + E = C + 3×E which becomes 5×E + E = C + 3×E Subtract 3×E from both sides: 5×E + E - 3×E = C + 3×E - 3×E which makes 3×E = C
Solution
Substitute 2½×E for A, 0 for B, 3×E for C and F, and ½×E for D in eq.1: 2½×E + 0 + 3×E + ½×E + E + 3×E = 20 which simplifies to 10×E = 20 Divide both sides of the equation above by 10: 10×E ÷ 10 = 20 ÷ 10 E = 2 making A = 2½×E = 2½ × 2 = 5 C = F = 3×E = 3 × 2 = 6 D = ½×E = ½ × 2 = 1 and ABCDEF = 506126