Puzzle for June 4, 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 non-negative integer.
Scratchpad
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Hint #1
eq.6 may be written as: A = (B + C + D + E + F) ÷ 5 Multiply both sides of the above equation by 5: 5 × A = 5 × (B + C + D + E + F) ÷ 5 which becomes eq.6a) 5×A = B + C + D + E + F
Hint #2
In eq.1, replace B + C + D + E + F with 5×A (from eq.6a): A + 5×A = 30 which becomes 6×A = 30 Divide both sides of the above equation by 6: 6×A ÷ 6 = 30 ÷ 6 which makes A = 5
Hint #3
eq.5 may be written as: F = (A + B + D + E) ÷ 4 Multiply both sides of the equation above by 4: 4 × F = 4 × (A + B + D + E) ÷ 4 which becomes eq.5a) 4×F = A + B + D + E
Hint #4
eq.1 may be written as: A + B + D + E + C + F = 30 In the above equation, replace A + B + D + E with 4×F (from eq.5a): 4×F + C + F = 30 which becomes 5×F + C = 30 Subtract 5×F from each side: 5×F + C – 5×F = 30 – 5×F which becomes eq.1a) C = 30 – 5×F
Hint #5
Add C and F to both sides of eq.3: C + F + C + F = B – C + D – F + C + F which becomes 2×C + 2×F = B + D In the equation above, substitute (30 – 5×F) for C (from eq.1a): 2×(30 – 5×F) + 2×F = B + D which becomes 60 – 10×F + 2×F = B + D which becomes eq.3a) 60 – 8×F = B + D
Hint #6
eq.6a may be written as: 5×A = B + D + C + E + F In the above equation, substitute 5 for A, 60 – 8×F for B + D (from eq.3a), and 30 – 5×F for C (from eq.1a): 5×5 = 60 – 8×F + 30 – 5×F + E + F which becomes 25 = 90 – 12×F + E Subtract 90 from both sides, and add 12×F to both sides: 25 – 90 + 12×F = 90 – 12×F + E – 90 + 12×F which becomes eq.6b) 12×F – 65 = E
Hint #7
Substitute 60 – 8×F for B + D (from eq.3a), 5 for A, (30 – 5×F) for C (from eq.1a), and 12×F – 65 for E (from eq.6b) in eq.2: 60 – 8×F = 5 – (30 – 5×F) + 12×F – 65 which becomes 60 – 8×F = 5 – 30 + 5×F + 12×F – 65 which becomes 60 – 8×F = 17×F – 90 Add 8×F and 90 to both sides of the above equation: 60 – 8×F + 8×F + 90 = 17×F – 90 + 8×F + 90 which becomes 150 = 25×F Divide both sides by 25: 150 ÷ 25 = 25×F ÷ 25 which makes 6 = F making E = 12×F – 65 = 12×6 – 65 = 72 – 65 = 7 (from eq.6b) C = 30 – 5×F = 30 – 5×6 = 30 – 30 = 0 (from eq.1a)
Hint #8
Substitute 0 for C, 5 for A, and 6 for F in eq.4: B + 0 + 6 – 5 = 5 – 0 + D which becomes B + 1 = 5 + D Subtract 1 from both sides of the above equation: B + 1 – 1 = 5 + D – 1 which becomes eq.4a) B = 4 + D
Solution
Substitute 6 for F, and 4 + D for B (from eq.4a) in eq.3a: 60 – 8×6 = 4 + D + D which becomes 60 – 48 = 4 + 2×D which becomes 12 = 4 + 2×D Subtract 4 from each side of the equation above: 12 – 4 = 4 + 2×D – 4 which becomes 8 = 2×D Divide both sides by 2: 8 ÷ 2 = 2×D ÷ 2 which makes 4 = D making B = 4 + D = 4 + 4 = 8 (from eq.4a) and ABCDEF = 580476