Puzzle for October 4, 2020 ( )
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.1 may be written as: eq.1a) A + D + F + B + C + E = 34 In eq.1a, replace B + C + E with A + D + F (from eq.5): A + D + F + A + D + F = 34 which is the same as 2×(A + D + F) = 34 Divide both sides of the above equation by 2: 2×(A + D + F) ÷ 2 = 34 ÷ 2 which becomes eq.1b) A + D + F = 17
Hint #2
In eq.1a, replace A + D + F with 17 (from eq.1b): 17 + B + C + E = 34 Subtract 17 from each side of the equation above: 17 + B + C + E – 17 = 34 – 17 which becomes eq.1c) B + C + E = 17
Hint #3
eq.1b may be written as: A + F + D = 17 In the above equation, substitute B + D for A + F (from eq.2): B + D + D = 17 which becomes B + 2×D = 17 Subtract 2×D from both sides: B + 2×D – 2×D = 17 – 2×D which becomes eq.1d) B = 17 – 2×D
Hint #4
Substitute B + F for C + E (from eq.3) in eq.1c: B + B + F = 17 which becomes 2×B + F = 17 Subtract 2×B from each side of the equation above: 2×B + F – 2×B = 17 – 2×B which becomes eq.1e) F = 17 – 2×B
Hint #5
Substitute (17 – 2×D) for B (from eq.1d) in eq.1e: F = 17 – 2×(17 – 2×D) which becomes F = 17 – 34 + 4×D which becomes F = –17 + 4×D which may be written as eq.1f) F = 4×D – 17
Hint #6
Substitute 4×D – 17 for F (from eq.1f) in eq.4: D = E + 4×D – 17 In the equation above, subtract 4×D from both sides, and add 17 to both sides: D – 4×D + 17 = E + 4×D – 17 – 4×D + 17 which becomes –3×D + 17 = E which may be written as eq.1g) 17 – 3×D = E
Hint #7
Substitute 4×D – 17 for F (from eq.1f) in eq.1b: A + D + 4×D – 17 = 17 which becomes A + 5×D – 17 = 17 In the equation above, subtract 5×D from each side, and add 17 to each side: A + 5×D – 17 – 5×D + 17 = 17 – 5×D + 17 which becomes eq.1h) A = 34 – 5×D
Hint #8
Substitute 17 – 2×D for B (from eq.1d), and 17 – 3×D for E (from eq.1g) in eq.1c: 17 – 2×D + C + 17 – 3×D = 17 which becomes 34 – 5×D + C = 17 In the equation above, subtract 34 from each side, and add 5×D to each side: 34 – 5×D + C – 34 + 5×D = 17 – 34 + 5×D which becomes C = –17 + 5×D which may be written as eq.1i) C = 5×D – 17
Solution
Substitute 5×D – 17 for C (from eq.1i), 4×D – 17 for F (from eq.1f), 34 – 5×D for A (from eq.1h), and 17 – 3×D for E (from eq.1g) in eq.6: 5×D – 17 + 4×D – 17 = 34 – 5×D + 17 – 3×D which becomes 9×D – 34 = 51 – 8×D Add 34 and 8×D to both sides of the above equation: 9×D – 34 + 34 + 8×D = 51 – 8×D + 34 + 8×D which makes 17×D = 85 Divide both sides by 17: 17×D ÷ 17 = 85 ÷ 17 which means D = 5 making A = 34 – 5×D = 34 – 5×5 = 34 – 25 = 9 (from eq.1h) B = 17 – 2×D = 17 – 2×5 = 17 – 10 = 7 (from eq.1d) C = 5×D – 17 = 5×5 – 17 = 25 – 17 = 8 (from eq.1i) E = 17 – 3×D = 17 – 3×5 = 17 – 15 = 2 (from eq.1g) F = 4×D – 17 = 4×5 – 17 = 20 – 17 = 3 (from eq.1f) and ABCDEF = 978523