Puzzle for February 3, 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 F from both sides of eq.3: B + C + F - F = D + E - F which becomes B + C = D + E - F In eq.2, replace B + C with D + E - F: E + F = D + E - F Subtract E from, and add F to each side of the equation above: E + F - E + F = D + E - F - E + F which simplifies to D = 2×F
Hint #2
Add D + E to each side of eq.4: C - D + D + E = A - E + D + E which becomes C + E = A + D In eq.5, replace A + D with C + E: C + E = B + E + F Subtract E from each side of the above equation: C + E - E = B + E + F - E which becomes eq.5a) C = B + F
Hint #3
In eq.3, substitute B + F for C (from eq.5a), and 2×F for D: B + B + F + F = 2×F + E which makes 2×B + 2×F = 2×F + E Subtract 2×F from both sides of the above equation: 2×B + 2×F - 2×F = 2×F + E - 2×F which makes 2×B = E
Hint #4
In eq.5, substitute 2×B for E, and 2×F for D: A + 2×F = B + 2×B + F Subtract both A and F from each side of the equation above: A + 2×F - A - F = B + 2×B + F - A - F which simplifies to eq.5b) F = 3×B - A
Hint #5
Substitute 2×B for E, and 3×B - A for F (from eq.5b) in eq.2: 2×B + 3×B - A = B + C which becomes 5×B - A = B + C Subtract B from both sides of the equation above: 5×B - A - B = B + C - B which becomes eq.2a) 4×B - A = C
Hint #6
Substitute 4×B - A for C (from eq.2a), 2×B for E, and 3×B - A for F (from eq.5b) in eq.3: B + 4×B - A + 3×B - A = D + 2×B which becomes 8×B - 2×A = D + 2×B Subtract 2×B from both sides of the above equation: 8×B - 2×A - 2×B = D + 2×B - 2×B which becomes eq.3a) 6×B - 2×A = D
Hint #7
eq.6 may be written as: 10×A + B + 10×B + C = 10×C + D + 10×D + E - F which becomes 10×A + 11×B = 9×C + 11×D + E - F Substitute (4×B - A) for C (from eq.2a), (6×B - 2×A) for D (from eq.3a), 2×B for E, and (3×B - A) for F (from eq.5b) into the above equation: 10×A + 11×B = 9×(4×B - A) + 11×(6×B - 2×A) + 2×B - (3×B - A) which becomes 10×A + 11×B = 36×B - 9×A + 66×B - 22×A + 2×B - 3×B + A which becomes 10×A + 11×B = 101×B - 30×A Add (30×A - 11×B) to both sides: 10×A + 11×B + (30×A - 11×B) = 101×B - 30×A + (30×A - 11×B) which simplifies to 40×A = 90×B Divide each side by 40: 40×A ÷ 40 = 90×B ÷ 40 which means A = 2¼×B
Hint #8
In eq.5b, replace A with 2¼×B: F = 3×B - 2¼×B = ¾×B
Hint #9
In eq.2a, replace A with 2¼×B: 4×B - 2¼×B = C which makes 1¾×B = C
Hint #10
In eq.3a, replace A with 2¼×B: 6×B - 2×(2¼×B) = D which becomes 6×B - 4½×B = D which makes 1½×B = D
Solution
Substitute 2¼×B for A, 1¾×B for C, 1½×B for D, 2×B for E, and ¾×B for F in eq.1: A + B + C + D + E + F = 37 2¼×B + B + 1¾×B + 1½×B + 2×B + ¾×B = 37 which simplifies to 9¼×B = 37 Divide both sides by 9¼: 9¼×B ÷ 9¼ = 37 ÷ 9¼ which makes B = 4 making A = 2¼×B = 2¼ × 4 = 9 C = 1¾×B = 1¾ × 4 = 7 D = 1½×B = 1½ × 4 = 6 E = 2×B = 2 × 4 = 8 F = ¾×B = ¾ × 4 = 3 and ABCDEF = 947683