Puzzle for November 20, 2022 ( )
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.
Scratchpad
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Hint #1
Add A and E to both sides of eq.6: D + E - A + A + E = C - E + F + A + E which becomes D + 2×E = C + F + A which may be written as eq.6a) D + 2×E = A + C + F
Hint #2
In eq.6a, replace A + C with B + D (from eq.4): D + 2×E = B + D + F Subtract D from each side of the equation above: D + 2×E - D = B + D + F - D which becomes eq.6b) 2×E = B + F
Hint #3
eq.5 may be written as: C + E = D + B + F In the above equation, replace B + F with 2×E (from eq.6b): C + E = D + 2×E Subtract E from each side: C + E - E = D + 2×E - E which becomes eq.5a) C = D + E
Hint #4
In eq.2, substitute D + E for C (from eq.5a): D + E = B + E Subtract E from each side of the equation above: D + E - E = B + E - E which makes D = B
Hint #5
eq.6a may be written as: D + E + E = A + C + F Substitute C for D + E (from eq.5a) in the equation above: C + E = A + C + F Subtract C from both sides: C + E - C = A + C + F - C which becomes eq.6c) E = A + F
Hint #6
Substitute A + F for E (from eq.6c) into eq.5a: eq.5b) C = D + A + F
Hint #7
Add A and F to both sides of eq.3: F - A + A + F = D - F + A + F which becomes 2×F = D + A Substitute 2×F for D + A in eq.5b: C = 2×F + F which makes C = 3×F
Hint #8
Multiply both sides of eq.5b by 3: 3 × C = 3 × (D + A + F) which becomes 3×C = 3×D + 3×A + 3×F Substitute C for 3×F in the above equation: 3×C = 3×D + 3×A + C Subtract C from both sides: 3×C - C = 3×D + 3×A + C - C which becomes eq.5c) 2×C = 3×D + 3×A
Hint #9
Substitute D for B in eq.4: D + D = A + C which becomes 2×D = A + C Subtract A from each side of the above equation: 2×D - A = A + C - A which becomes eq.4a) 2×D - A = C
Hint #10
In eq.5c, substitute (2×D - A) for C (from eq.4a): 2×(2×D - A) = 3×D + 3×A which becomes 4×D - 2×A = 3×D + 3×A In the equation above, add 2×A to both sides, and subtract 3×D from both sides: 4×D - 2×A + 2×A - 3×D = 3×D + 3×A + 2×A - 3×D which simplifies to D = 5×A and also makes B = D = 5×A
Hint #11
Substitute (5×A) for D in eq.4a: 2×(5×A) - A = C which becomes 10×A - A = C which makes 9×A = C
Hint #12
Substitute 9×A for C, and 5×A for D in eq.5a: 9×A = 5×A + E Subtract 5×A from both sides of the above equation: 9×A - 5×A = 5×A + E - 5×A which makes 4×A = E
Hint #13
Substitute 4×A for E in eq.6c: 4×A = A + F Subtract A from both sides of the equation above: 4×A - A = A + F - A which becomes 3×A = F
Solution
Substitute 5×A for B and D, 9×A for C, 4×A for E, and 3×A for F in eq.1: A + 5×A + 9×A + 5×A + 4×A + 3×A = 27 which simplifies to 27×A = 27 Divide both sides of the above equation by 27: 27×A ÷ 27 = 27 ÷ 27 which means A = 1 making B = D = 5×A = 5 × 1 = 5 C = 9×A = 9 × 1 = 9 E = 4×A = 4 × 1 = 4 F = 3×A = 3 × 1 = 3 and ABCDEF = 159543