Puzzle for February 14, 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.
* "F mod E" equals the remainder of F divided by E.
Scratchpad
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Hint #1
Add D and A to both sides of eq.4: A – D + D + A = B + F – A + D + A which becomes 2×A = B + F + D which may be written as eq.4a) 2×A = B + D + F In eq.4a, replace B + D + F with C + E (from eq.1): eq.4b) 2×A = C + E
Hint #2
In eq.5, substitute 2×A for C + E (from eq.4b): B = (2×A) ÷ A which makes B = 2
Hint #3
Substitute 2 for B in eq.2: eq.2a) F = A + 2
Hint #4
Substitute 2 for B in eq.3: eq.3a) E = 2 + C
Hint #5
Substitute 2 for B, and A + 2 for F (from eq.2a) in eq.4a: 2×A = 2 + D + A + 2 which becomes 2×A = 4 + D + A Subtract 4 and A from both sides of the above equation: 2×A – 4 – A = 4 + D + A – 4 – A which makes eq.4c) A – 4 = D
Hint #6
Substitute 2 for B, A – 4 for D (from eq.4c), A + 2 for F (from eq.2a), and 2 + C for E (from eq.3a) in eq.1: 2 + A – 4 + A + 2 = C + 2 + C which becomes 2×A = 2×C + 2 Divide both sides of the above equation by 2: 2×A ÷ 2 = (2×C + 2) ÷ 2 which becomes A = C + 1 Subtract 1 from each side: A – 1 = C + 1 – 1 which makes eq.1a) A – 1 = C
Hint #7
Substitute A – 1 for C (from eq.1a) in eq.3a: E = 2 + A – 1 which makes eq.3b) E = A + 1
Hint #8
Add 4 to both sides of eq.4c: A – 4 + 4 = D + 4 which makes A = D + 4 Since D ≥ 1 (from eq.6), then the above equation makes: A ≥ 5
Hint #9
Substitute (A + 2) for F (from eq.2a), (A + 1) for E (from eq.3b), 2 for B, and (A – 4) for D (from eq.4c) in eq.6: eq.6a) (A + 2) mod (A + 1) = 2 ÷ (A – 4)
Hint #10
In eq.6a: A + 2 = A + 1 + 1 which means remainder of (A + 2) ÷ (A + 1) = 1 (whenever A ≥ 1) which means (A + 2) mod (A + 1) = 1 (whenever A ≥ 1) Since A ≥ 5, substitute 1 for (A + 2) mod (A + 1) in eq.6a: eq.6b) 1 = 2 ÷ (A – 4)
Solution
Multiply both sides of eq.6b by (A – 4): 1 × (A – 4) = 2 ÷ (A – 4) × (A – 4) which becomes A – 4 = 2 Add 4 to both sides of the above equation: A – 4 + 4 = 2 + 4 which means A = 6 making C = A – 1 = 6 – 1 = 5 (from eq.1a) D = A – 4 = 6 – 4 = 2 (from eq.4c) E = A + 1 = 6 + 1 = 7 (from eq.3b) F = A + 2 = 6 + 2 = 8 (from eq.2a) and ABCDEF = 625278