Puzzle for December 11, 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.
* "B mod C" equals the remainder of B divided by C.
Scratchpad
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Hint #1
In eq.4, replace C with D + F (from eq.3): A + B + D = D + F + E + F which becomes A + B + D = D + E + 2×F Subtract B and D from both sides of the equation above: A + B + D – B – D = D + E + 2×F – B – D which becomes eq.4a) A = E + 2×F – B
Hint #2
In eq.1, replace A with E + 2×F – B (from eq.4a): B + E = E + 2×F – B + F which becomes B + E = E + 3×F – B In the equation above, subtract E from both sides, and add B to both sides: B + E – E + B = E + 3×F – B – E + B which becomes 2×B = 3×F Divide both sides by 2: 2×B ÷ 2 = 3×F ÷ 2 which makes eq.1a) B = 1½×F
Hint #3
In eq.2, substitute D + F for C (from eq.3), and 1½×F for B: D + F + D = 1½×F which becomes 2×D + F = 1½×F Subtract F from each side of the above equation: 2×D + F – F = 1½×F – F which becomes 2×D = ½×F Multiply both sides by 2: 2 × 2×D = 2 × ½×F which makes 4×D = F
Hint #4
Substitute (4×D) for F in eq.1a: B = 1½×(4×D) which makes B = 6×D
Hint #5
Substitute 4×D for F in eq.3: C = D + 4×D which makes C = 5×D
Hint #6
Substitute 6×D for B, and 5×D for C in eq.6: 6×D mod 5×D = (6×D – 5×D) ÷ D which becomes 6×D mod 5×D = D ÷ D which becomes eq.6a) 6×D mod 5×D = 1
Hint #7
One property of the mod operation is -- if both numbers (or terms) have a factor in common, it may be factored out. Mathematically, the above property is expressed as: (m×D) mod (n×D) = (m mod n) × D Applying the above property to eq.6a yields: (6 mod 5) × D = 1 which becomes 1 × D = 1 which means D = 1 making B = 6×D = 6 × 1 = 6 C = 5×D = 5 × 1 = 5 F = 4×D = 4 × 1 = 4
Hint #8
Substitute 6 for B, and 4 for F in eq.1: 6 + E = A + 4 Subtract 4 from both sides of the above equation: 6 + E – 4 = A + 4 – 4 which makes eq.1b) 2 + E = A
Solution
Substitute 4 for F, 2 + E for A (from eq.1b), 6 for B, 5 for C, and 1 for D in eq.5: E × 4 = 2 + E + 6 + 5 – 1 which is equivalent to 4×E = E + 12 Subtract E from each side of the equation above: 4×E – E = E + 12 – E which makes 3×E = 12 Divide both sides by 3: 3×E ÷ 3 = 12 ÷ 3 which means E = 4 making A = 2 + E = 2 + 4 = 6 (from eq.1b) and ABCDEF = 665144