Puzzle for June 4, 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 re-written as: A – F = B + E – C In the equation above, replace B + E with C + F (from eq.3): A – F = C + F – C which becomes A – F = F Add F to both sides: A – F + F = F + F which makes eq.4a) A = 2×F
Hint #2
In eq.2, replace A with 2×F (from eq.4a): F – E = 2×F – F which becomes F – E = F Subtract F from each side of the above equation: F – E – F = F – F which makes –E = 0 which means E = 0
Hint #3
In eq.5, substitute 0 for E, and 2×F for A (from eq.4a): C + 0 – F = 2×F + D – 0 which becomes C – F = 2×F + D Add F to both sides of the equation above: C – F + F = 2×F + D + F which becomes eq.5a) C = 3×F + D
Hint #4
Substitute 3×F + D for C (from eq.5a), and 0 for E in eq.3: 3×F + D + F = B + 0 which becomes eq.3a) 4×F + D = B
Hint #5
Substitute 4×F + D for B (from eq.3a), 2×F for A (from eq.4a), and (3×F + D) for C (from eq.5a) in eq.6: 4×F + D + D + F = (2×F × (3×F + D)) – F which becomes 5×F + 2×D = (2×F × 3×F) + (2×F × D) – F Add F to both sides of the above equation: 5×F + 2×D + F = (2×F × 3×F) + (2×F × D) – F + F which becomes eq.6a) 6×F + 2×D = (2×F × 3×F) + (2×F × D)
Hint #6
eq.6a is equivalent to: 6×F + 2×D = 6×F² + 2×F×D which may be re-written as 6×F + 2×D = F × (6×F + 2×D) Divide both sides by (6×F + 2×D) (assumes D and F cannot both = 0): (6×F + 2×D) ÷ (6×F + 2×D) = F × (6×F + 2×D) ÷ (6×F + 2×D) which makes 1 = F
Hint #7
Begin checking: D = 0, and F = 0 ... Substituting 0 for D and F in eq.5a would yield: C = 3×0 + 0 which would make C = 0 Substituting 0 for D and F in eq.3a would yield: 4×0 + 0 = B which would make 0 = B Substituting 0 for F in eq.4a would yield: A = 2×0 A = 0
Hint #8
Finish checking: D = 0, and F = 0 ... Substituting 0 for A, B, C, D, E, and F in eq.1 would yield: 0 + 0 + 0 + 0 + 0 + 0 = 25 which would make 0 = 25 However, 0 ≠ 25 which means D and F cannot both = 0
Hint #9
Substitute 1 for F in eq.4a: A = 2×1 which makes A = 2
Hint #10
Substitute 1 for F in eq.5a: C = 3×1 + D which becomes eq.5b) C = 3 + D
Hint #11
Substitute 1 for F in eq.3a: 4×1 + D = B which becomes eq.3b) 4 + D = B
Solution
Substitute 2 for A, 4 + D for B (from eq.3b), 3 + D for C (from eq.5b), 0 for E, and 1 for F in eq.1: 2 + 4 + D + 3 + D + D + 0 + 1 = 25 which becomes 10 + 3×D = 25 Subtract 10 from both sides of the equation above: 10 + 3×D – 10 = 25 – 10 which makes 3×D = 15 Divide both sides by 3: 3×D ÷ 3 = 15 ÷ 3 which means D = 5 making B = 4 + D = 4 + 5 = 9 (from eq.3b) C = 3 + D = 3 + 5 = 8 (from eq.5b) and ABCDEF = 298501