Puzzle for February 20, 2021 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit positive integer.
Scratchpad
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Hint #1
In eq.4, replace B + E with F – B (from eq.2): C + D = F – B + F which becomes C + D = 2×F – B Add B to both sides of the above equation: C + D + B = 2×F – B + B which becomes C + D + B = 2×F which is the same as eq.4a) B + C + D = 2×F
Hint #2
In eq.5, replace B + C + D with 2×F (from eq.4a): E × F = 2×F Divide both sides of the equation above by F: E × F ÷ F = 2×F ÷ F which makes E = 2
Hint #3
In eq.2, substitute 2 for E: F – B = B + 2 Add B to both sides of the above equation: F – B + B = B + 2 + B which makes eq.2a) F = 2×B + 2
Hint #4
In eq.3, substitute 2 for E: B + C – 2 = A + 2 Subtract 2 and B from both sides of the above equation: B + C – 2 – 2 – B = A + 2 – 2 – B which becomes eq.3a) C – 4 = A – B
Hint #5
Add the left and right sides of eq.3a to the left and right sides of eq.1, respectively: D + C – 4 = A + B + A – B which becomes D + C – 4 = 2×A which may be written as eq.1a) C + D – 4 = 2×A
Hint #6
Substitute 2 for E, and 2×B + 2 for F (from eq.2a) in eq.4: C + D = B + 2 + 2×B + 2 which becomes eq.4b) C + D = 3×B + 4
Hint #7
Substitute 3×B + 4 for C + D (from eq.4b) into eq.1a: 3×B + 4 – 4 = 2×A which becomes 3×B = 2×A Divide both sides of the equation above by 2: 3×B ÷ 2 = 2×A ÷ 2 which makes 1½×B = A
Hint #8
Substitute 1½×B for A in eq.3a: C – 4 = 1½×B – B which becomes C – 4 = ½×B Add 4 to both sides of the equation above: C – 4 + 4 = ½×B + 4 which makes eq.3b) C = ½×B + 4
Hint #9
Substitute 1½×B for A in eq.1: D = 1½×B + B which makes D = 2½×B
Hint #10
Substitute 1½×B for A, 2½×B for D, 2 for E, 2×B + 2 for F (from eq.2a), and (½×B + 4) for C (from eq.3a) in eq.6: 1½×B ÷ B = (2½×B ÷ 2) – (2×B + 2 – (½×B + 4)) which is equivalent to 1½ = 1¼×B – (2×B + 2 – ½×B – 4) which becomes 1½ = 1¼×B – 2×B – 2 + ½×B + 4 which makes eq.6a) 1½ = –¼×B + 2
Solution
Subtract 2 from each side of eq.6a: 1½ – 2 = –¼×B + 2 – 2 which makes –½ = –¼×B Multiply both sides of the equation above by (–4): (–4) × –½ = (–4) × –¼×B which makes 2 = B making A = 1½×B = 1½×2 = 3 C = ½×B + 4 = ½×2 + 4 = 1 + 4 = 5 (from eq.3b) D = 2½×B = 2½×2 = 5 F = 2×B + 2 = 2×2 + 2 = 4 + 2 = 6 (from eq.2a) and ABCDEF = 325526