Puzzle for April 16, 2023 ( )
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.
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Hint #1
Add E and C to both sides of eq.3: B + D - E + E + C = A - C + E + E + C which becomes B + D + C = A + 2×E which may be written as eq.3a) B + C + D = A + 2×E
Hint #2
eq.6 may be written as: E = (A + B + C + D) ÷ 4 Multiply both sides of the above equation by 4: 4 × E = 4 × (A + B + C + D) ÷ 4 which becomes eq.6a) 4×E = A + B + C + D
Hint #3
In eq.6a, replace B + C + D with A + 2×E (from eq.3a): 4×E = A + A + 2×E which becomes 4×E = 2×A + 2×E Subtract 2×E from each side of the equation above: 4×E - 2×E = 2×A + 2×E - 2×E which makes 2×E = 2×A Divide both sides by 2: 2×E ÷ 2 = 2×A ÷ 2 which makes E = A
Hint #4
In eq.3a, replace E with A: B + C + D = A + 2×A which becomes eq.3b) B + C + D = 3×A
Hint #5
Add D to both sides of eq.4: A + D + E + D = B + C + F + D which is the same as eq.4a) A + 2×D + E = B + C + D + F
Hint #6
In eq.4a, substitute A for E, and 3×A for B + C + D (from eq.3b): A + 2×D + A = 3×A + F which becomes 2×A + 2×D = 3×A + F Subtract 2×A from each side of the above equation: 2×A + 2×D - 2×A = 3×A + F - 2×A which becomes eq.4b) 2×D = A + F
Hint #7
Substitute A for E in eq.5: B + D - F = F - (B - A) which becomes B + D - F = F - B + A Add F and B to both sides of the equation above: B + D - F + F + B = F - B + A + F + B which becomes eq.5a) 2×B + D = F + A + F
Hint #8
Substitute 2×D for A + F (from eq.4b) into eq.5a: 2×B + D = F + 2×D Subtract D from each side of the equation above: 2×B + D - D = F + 2×D - D which becomes eq.5b) 2×B = F + D
Hint #9
Add B and D to both sides of eq.2: F - B + B + D = B + C - D + B + D which becomes eq.2a) F + D = 2×B + C
Hint #10
Substitute 2×B for F + D (from eq.5b) into eq.2a: 2×B = 2×B + C Subtract 2×B from each side of the equation above: 2×B - 2×B = 2×B + C - 2×B which makes 0 = C
Hint #11
Substitute 0 for C in eq.3b: B + 0 + D = 3×A which becomes eq.3c) B + D = 3×A
Hint #12
eq.1 may be written as: eq.1a) A + D + E + B + C + F = 34 Substitute A + D + E for B + C + F (from eq.4) in eq.1a: A + D + E + A + D + E = 34 which becomes 2×(A + D + E) = 34 Divide both sides of the above equation by 2: 2×(A + D + E) ÷ 2 = 34 ÷ 2 which becomes eq.1b) A + D + E = 17
Hint #13
Substitute 17 for A + D + E (from eq.1b), and 0 for C in eq.1a: 17 + B + 0 + F = 34 which becomes 17 + B + F = 34 Subtract 17 from both sides of the above equation: 17 + B + F - 17 = 34 - 17 which becomes eq.1c) B + F = 17
Hint #14
eq.1 may be written as: A + B + D + C + E + F = 34 In the above equation, substitute 3×A for B + D (from eq.3c), 0 for C, and A for E: A + 3×A + 0 + A + F = 34 which becomes 5×A + F = 34 Subtract 5×A from both sides of the equation above: 5×A + F - 5×A = 34 - 5×A which becomes eq.1d) F = 34 - 5×A
Hint #15
Substitute 34 - 5×A for F (from eq.1d) in eq.1c: B + 34 - 5×A = 17 In the above equation, add 5×A to both sides, and subtract 34 from both sides: B + 34 - 5×A + 5×A - 34 = 17 + 5×A - 34 which becomes eq.1e) B = 5×A - 17
Hint #16
Substitute 5×A - 17 for B (from eq.1e) in eq.3c: 5×A - 17 + D = 3×A In the above equation, add 17 to both sides, and subtract 5×A from both sides: 5×A - 17 + D + 17 - 5×A = 3×A + 17 - 5×A which becomes eq.3d) D = 17 - 2×A
Hint #17
In eq.2a, substitute 34 - 5×A for F (from eq.1d), 17 - 2×A for D (from eq.3d), (5×A - 17) for B (from eq.1e), and 0 for C: 34 - 5×A + 17 - 2×A = 2×(5×A - 17) + 0 which becomes 51 - 7×A = 10×A - 34 Add 7×A and 34 to both sides of the above equation: 51 - 7×A + 7×A + 34 = 10×A - 34 + 7×A + 34 which simplifies to 85 = 17×A Divide both sides by 17: 85 ÷ 17 = 17×A ÷ 17 which makes 5 = A
Solution
Since A = 5, then: B = 5×A - 17 = 5×5 - 17 = 25 - 17 = 8 (from eq.1e) D = 17 - 2×A = 17 - 2×5 = 17 - 10 = 7 (from eq.3d) E = A = 5 F = 34 - 5×A = 34 - 5×5 = 34 - 25 = 9 (from eq.1d) and ABCDEF = 580759