Puzzle for April 22, 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.
Scratchpad
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Hint #1
Add B and D to both sides of eq.3: D - B + B + D = C - D + B + D which becomes 2×D = C + B Subtract C from both sides of the equation above: 2×D - C = C + B - C which becomes eq.3a) 2×D - C = B
Hint #2
In eq.4, replace B with 2×D - C (from eq.3a): 2×D - C + F = C + D In the above equation, add C to both sides, and subtract D from both sides: 2×D - C + F + C - D = C + D + C - D which becomes eq.4a) D + F = 2×C
Hint #3
In eq.5, replace D + F with 2×C (from eq.4a): C + E = 2×C Subtract C from each side of the above equation: C + E - C = 2×C - C which makes E = C
Hint #4
In eq.2, substitute E for C: eq.2a) A = E + F
Hint #5
Add (C + D) and (E + F) to both sides of eq.6: E + F - (C + D) + (C + D) + (E + F) = A + B - (E + F) + (C + D) + (E + F) which becomes E + F + (E + F) = A + B + (C + D) which may be written as eq.6a) 2×(E + F) = A + B + C + D
Hint #6
In eq.6a, replace (E + F) with A (from eq.2a): 2×A = A + B + C + D Subtract A from each side of the equation above: 2×A - A = A + B + C + D - A which becomes eq.6b) A = B + C + D
Hint #7
In eq.1, substitute A for B + C + D (from eq.6b), and A for E + F (from eq.2a): A + A + A = 27 which makes 3×A = 27 Divide both sides of the above equation by 3: 3×A ÷ 3 = 27 ÷ 3 which makes A = 9
Hint #8
Substitute 9 for A in eq.6b: eq.6c) 9 = B + C + D
Hint #9
Add C and D to both sides of eq.3a: 2×D - C + C + D = B + C + D which becomes 3×D = B + C + D Substitute 9 for B + C + D (from eq.6c) into the above equation: 3×D = 9 Divide both sides by 3: 3×D ÷ 3 = 9 ÷ 3 which makes D = 3
Hint #10
Substitute 9 for A in eq.2: 9 = C + F Subtract C from each side of the equation above: 9 - C = C + F - C which becomes eq.2b) 9 - C = F
Hint #11
Substitute 3 for D, and 9 - C for F (from eq.2b) in eq.4a: 3 + 9 - C = 2×C which becomes 12 - C = 2×C Add C to both sides of the above equation: 12 - C + C = 2×C + C which becomes 12 = 3×C Divide both sides by 3: 12 ÷ 3 = 3×C ÷ 3 which makes 4 = C and also makes 4 = C = E
Solution
Since D = 3 and C = 4, then: B = 2×D - C = 2×3 - 4 = 6 - 4 = 2 (from eq.3a) F = 9 - C = 9 - 4 = 5 (from eq.2b) and ABCDEF = 924345