Puzzle for December 23, 2018 ( )
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, and EF are 2-digit numbers (not A×B, C×D, or E×F). ABC, BCD, and CDE are 3-digit numbers (not A×B×C, B×C×D, or C×D×E).
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Hint #1
Substitute C + E for B (from eq.2) in eq.4: C + E - A = C - D + F Add (A - C) to both sides of the equation above: C + E - A + (A - C) = C - D + F + (A - C) which becomes eq.4a) E = -D + F + A
Hint #2
Substitute -D + F + A for E (from eq.4a) in eq.3: -D + F + A = A + D + F Subtract A, D, and F from each side: -D + F + A - A - D - F = A + D + F - A - D - F which simplifies to -2×D = 0 which means D = 0
Hint #3
Substitute 0 for D in eq.4a: E = -0 + F + A which makes eq.4b) E = F + A
Hint #4
eq.6 may be written as: 100×B + 10×C + D = 100×A + 10×B + C + 100×C + 10×D + E + 10×C + D + F which becomes 100×B + 10×C + D = 100×A + 10×B + 111×C + 11×D + E + F Subtract (10×B + 10×C + D) from both sides of the above equation: 100×B + 10×C + D - (10×B + 10×C + D) = 100×A + 10×B + 111×C + 11×D + E + F - (10×B + 10×C + D) which is the same as 100×B + 10×C + D - 10×B - 10×C - D = 100×A + 10×B + 111×C + 11×D + E + F - 10×B - 10×C - D which becomes eq.6a) 90×B = 100×A + 101×C + 10×D + E + F
Hint #5
Substitute F + A for E (from eq.4b) in eq.6a: 90×B = 100×A + 101×C + 10×D + F + A + F which becomes 90×B = 101×A + 101×C + 10×D + 2×F In the above equation, subtract 101×A from each side, and substitute 0 for D: 90×B - 101×A = 101×A + 101×C + 10×0 + 2×F - 101×A which becomes 90×B - 101×A = 101×C + 2×F which may be written as 90×B - 90×A - 11×A = 101×C + 2×F which may also be written as eq.6b) 90×(B - A) - 11×A = 101×C + 2×F
Hint #6
In eq.6b, replace B - A with C - D + F (from eq.4): 90×(C - D + F) - 11×A = 101×C + 2×F which is equivalent to 90×C - 90×D + 90×F - 11×A = 101×C + 2×F In the above equation, subtract (90×C + 2×F) from both sides, and replace D with 0: 90×C - 90×0 + 90×F - 11×A - (90×C + 2×F) = 101×C + 2×F - (90×C + 2×F) which is equivalent to 90×C + 90×F - 11×A - 90×C - 2×F = 101×C + 2×F - 90×C - 2×F which becomes 88×F - 11×A = 11×C Divide both sides by 11: (88×F - 11×A) ÷ 11 = 11×C ÷ 11 which simplifies to eq.6c) 8×F - A = C
Hint #7
In eq.4, substitute 8×F - A for C (from eq.6c), and replace D with 0: B - A = 8×F - A - 0 + F which becomes B - A = 9×F - A Add A to both sides: B - A + A = 9×F - A + A which makes B = 9×F
Hint #8
eq.5 may be written as: 10×E + F = 10×A + B + C In the above equation, substitute 9×F for B, and subtract F from both sides: 10×E + F - F = 10×A + 9×F + C - F which becomes eq.5a) 10×E = 10×A + 8×F + C
Hint #9
In eq.5a, substitute (F + A) for E (from eq.4b): 10×(F + A) = 10×A + 8×F + C which is the same as 10×F + 10×A = 10×A + 8×F + C Subtract both 10×A and 8×F from each side: 10×F + 10×A - 10×A - 8×F = 10×A + 8×F + C - 10×A - 8×F which simplifies to 2×F = C
Hint #10
Substitute 9×F for B, and 2×F for C in eq.2: 9×F = 2×F + E Subtract 2×F from each side: 9×F - 2×F = 2×F + E - 2×F which means 7×F = E
Hint #11
Substitute 7×F for E in eq.4b: 7×F = F + A Subtract F from each side: 7×F - F = F + A - F which means 6×F = A
Solution
Substitute 6×F for A, 9×F for B, 2×F for C, 0 for D, and 7×F for E in eq.1: 6×F + 9×F + 2×F + 0 + 7×F + F = 25 which simplifies to 25×F = 25 Divide each side by 25: 25×F ÷ 25 = 25 ÷ 25 which makes F = 1 making A = 6×F = 6 × 1 = 6 B = 9×F = 9 × 1 = 9 C = 2×F = 2 × 1 = 2 E = 7×F = 7 × 1 = 7 and ABCDEF = 692071