Puzzle for April 10, 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
eq.4 may be written as: B = (C + D + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × B = 3 × (C + D + E) ÷ 3 which becomes eq.4a) 3×B = C + D + E In eq.1, replace C + D + E with 3×B (from eq.4a): A + B + 3×B + F = 29 which becomes eq.1a) A + 4×B + F = 29
Hint #2
eq.5 may be written as: A = (B + D + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × A = 3 × (B + D + F) ÷ 3 which becomes 3×A = B + D + F Subtract D from both sides: 3×A – D = B + D + F – D which becomes eq.5a) 3×A – D = B + F
Hint #3
eq.1a may be written as: A + 3×B + B + F = 29 In the above equation, replace B + F with 3×A – D (from eq.5a): A + 3×B + 3×A – D = 29 which becomes eq.1b) 4×A + 3×B – D = 29
Hint #4
Add A and B to both sides of eq.2: D – B + A + B = B – A + A + B which becomes eq.2a) D + A = 2×B Multiply both sides of eq.1b by 2: 2(4×A + 3×B – D) = 2(29) which becomes 8×A + 6×B – 2×D = 58 which may be written as eq.1c) 8×A + 3×(2×B) – 2×D = 58
Hint #5
In eq.1c, substitute D + A for 2×B (from eq.2a): 8×A + 3×(D + A) – 2×D = 58 which is equivalent to 8×A + 3×D + 3×A – 2×D = 58 which becomes 11×A + D = 58 Subtract 11×A from each side of the equation above: 11×A + D – 11×A = 58 – 11×A which becomes eq.1d) D = 58 – 11×A
Hint #6
To make eq.1d true, check several possible values for A and D: If A = 5, then D = 58 – 11×5 = 58 – 55 = 3 If A = 4, then D = 58 – 11×4 = 58 – 44 = 12 If A < 4, then D > 12 If A = 6, then D = 58 – 11×6 = 58 – 66 = –8 If A > 6, then D < –8 Since D must be a non-negative one-digit integer, then the above equations make: D = 3 which makes A = 5
Hint #7
Substitute 3 for D, and 5 for A in eq.2a: 3 + 5 = 2×B which becomes 8 = 2×B Divide both sides of the above equation by 2: 8 ÷ 2 = 2×B ÷ 2 which makes 4 = B
Hint #8
Substitute 5 for A, and 4 for B in eq.1a: 5 + 4×4 + F = 29 which becomes 5 + 16 + F = 29 which becomes 21 + F = 29 Subtract 21 from each side of the equation above: 21 + F – 21 = 29 – 21 which makes F = 8
Hint #9
Substitute 3 for D, 4 for B, and 8 for F in eq.3: C + 3 – E = 4 + E + 8 which becomes C + 3 – E = 12 + E In the above equation, subtract 3 from both sides, and add E to both sides: C + 3 – E – 3 + E = 12 + E – 3 + E which becomes eq.3a) C = 9 + 2×E
Hint #10
Substitute 4 for B, 9 + 2×E for C (from eq.3a), and 3 for D in eq.4a: 3×4 = 9 + 2×E + 3 + E which becomes 12 = 12 + 3×E Subtract 12 from each side of the above equation: 12 – 12 = 12 + 3×E – 12 which becomes 0 = 3×E which means 0 = E
Solution
Substitute 0 for E into eq.3a: C = 9 – 2×0 which becomes C = 9 – 0 which makes C = 9 and makes ABCDEF = 549308