Puzzle for April 24, 2022 ( )
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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.
* "D mod A" equals the remainder of D ÷ A.
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Hint #1
Since A ≠ 0 (from eq.5 and eq.6), divide both sides of eq.3 by A: D ÷ A = A × E ÷ A which becomes D ÷ A = E Since E must be an integer, then: D ÷ A = an integer which makes remainder D ÷ A = 0 which means eq.3a) D mod A = 0
Hint #2
In eq.3a, replace D mod A with C – F (from eq.6): C – F = 0 Add F to both sides of the above equation: C – F + F = 0 + F which makes C = F
Hint #3
eq.2 may be written as: F – A = (B + D + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × (F – A) = 3 × (B + D + E) ÷ 3 which becomes eq.2a) 3×(F – A) = B + D + E
Hint #4
In eq.1, replace C with F: B + F – E = A + D + E In the above equation, subtract B and A from both sides, and add E to both sides: B + F – E – B – A + E = A + D + E – B – A + E which simplifies to eq.1a) F – A = D + 2×E – B
Hint #5
In eq.2a, substitute D + 2×E – B for F – A (from eq.1a): 3×(D + 2×E – B) = B + D + E which becomes 3×D + 6×E – 3×B = B + D + E In the above equation, add 3×B to both sides, and subtract D and E from both sides: 3×D + 6×E – 3×B + 3×B – D – E = B + D + E + 3×B – D – E which simplifies to eq.2b) 2×D + 5×E = 4×B
Hint #6
Substitute (A × E) for D (from eq.3) in eq.5: A + E = ((A × E) + E) ÷ A which becomes A + E = E + (E ÷ A) Subtract E from each side of the above equation: A + E – E = E + (E ÷ A) – E which becomes A = E ÷ A Multiply both sides by A: A × A = (E ÷ A) × A which makes A² = E
Hint #7
Substitute A² for E in eq.3: D = A × A² which becomes eq.3b) D = A³
Hint #8
Substitute A³ for D (from eq.3b), and A² for E in eq.2b: 2×A³ + 5×A² = 4×B Divide both sides of the above equation by 4: (2×A³ + 5×A²) ÷ 4 = 4×B ÷ 4 which becomes eq.2c) ½×A³ + 1¼×A² = B
Hint #9
To make eq.3b true, check several possible values for A and D: If A = 1, then D = A³ = 1³ = 1 If A = 2, then D = A³ = 2³ = 8 If A = 3, then D = A³ = 3³ = 27 If A > 3, then D > 27 Since A and D must be one-digit integers, the above equations make: A = 1 and D = 1 or A = 2 and D = 8
Hint #10
Check: A = 1 ... Substituting 1 for A in eq.2c would yield: ½×1³ + 1¼×1² = B which would become ½ + 1¼ = B which would make 1¾ = B Since B must be an integer, then: B ≠ 1¾ which means A ≠ 1 making A = 2 and D = 8 and also making E = A² = 2² = 4
Hint #11
Substitute 2 for A in eq.2c: ½×2³ + 1¼×2² = B which becomes ½×8 + 1¼×4 = B which becomes 4 + 5 = B which makes 9 = B
Solution
Substitute 2 for A, 8 for D, 4 for E, and 9 for B in eq.1a: F – 2 = 8 + 2×4 – 9 which becomes F – 2 = 8 + 8 – 9 which becomes F – 2 = 7 Add 2 to both sides of the above equation: F – 2 + 2 = 7 + 2 which makes F = 9 and also makes C = F = 9 and makes ABCDEF = 299849