Puzzle for March 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.
* "D mod E" equals the remainder of (D ÷ E).
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Hint #1
Since E ≠ 0 (from eq.6), divide both sides of eq.5 by E: D ÷ E = (C × E) ÷ E which becomes D ÷ E = C Since C is an integer, the above equation implies that: eq.5a) D ÷ E = an integer
Hint #2
eq.6 means that: B = remainder of (D ÷ E) Since D ÷ E is an integer (from eq.5a), then: remainder of (D ÷ E) = 0 which means B = 0
Hint #3
In eq.2, replace B with 0: C = 0 + E which makes C = E
Hint #4
In eq.5, substitute C for E: D = C × C which may be written as eq.5b) D = C²
Hint #5
Substitute C for E, and 0 for B in eq.3: C + F – 0 = A + 0 which becomes C + F = A Subtract F from each side of the above equation: C + F – F = A – F which becomes eq.3a) C = A – F
Hint #6
Substitute C for E, 0 for B, and C² for D (from eq.5b) in eq.4: A + C + C – 0 = 0 + C² + F which becomes A + 2×C = C² + F Subtract 2×C and F from each side of the equation above: A + 2×C – 2×C – F = C² + F – 2×C – F which becomes eq.4a) A – F = C² – 2×C
Hint #7
In eq.4a, substitute C for A – F (from eq.3a): C = C² – 2×C Add 2×C to both sides of the equation above: C + 2×C = C² – 2×C + 2×C which becomes 3×C = C² Since C ≠ 0 (C = E, and E ≠ 0 (from eq.6)), divide both sides of the above equation by C: 3×C ÷ C = C² ÷ C which makes 3 = C and also makes E = C = 3
Hint #8
Substitute 3 for C in eq.5b: D = 3² which makes D = 9
Hint #9
Substitute 3 for C and E, 0 for B, and 9 for D in eq.4: A + 3 + 3 – 0 = 0 + 9 + F which becomes A + 6 = 9 + F Subtract 9 from both sides of the equation above: A + 6 – 9 = 9 + F – 9 which makes eq.4b) A – 3 = F
Solution
Substitute 0 for B, 3 for C and E, 9 for D, and A – 3 for F (from eq.4b) in eq.1: A + 0 + 3 + 9 + 3 + A – 3 = 20 which becomes 2×A + 12 = 20 Subtract 12 from each side of the above equation: 2×A + 12 – 12 = 20 – 12 which becomes 2×A = 8 Divide both sides by 2: 2×A ÷ 2 = 8 ÷ 2 which means A = 4 making F = A – 3 = 4 – 3 = 1 (from eq.4b) and ABCDEF = 403931