Puzzle for May 6, 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.
* "B mod C" equals the remainder of B divided by C.
Scratchpad
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Hint #1
eq.5 may be written as: (A + D + F) ÷ 3 = B – C Multiply both sides of the above equation by 3: 3 × (A + D + F) ÷ 3 = 3 × (B – C) which becomes eq.5a) A + D + F = 3×B – 3×C
Hint #2
In eq.1, replace A + D + F with 3×B – 3×C (from eq.5a): B + C – D + E – F = 3×B – 3×C Subtract B and C from each side of the equation above: B + C – D + E – F – B – C = 3×B – 3×C – B – C which becomes eq.1a) –D + E – F = 2×B – 4×C
Hint #3
Substitute B + D for E – F (from eq.2) in eq.1a: –D + B + D = 2×B – 4×C which becomes B = 2×B – 4×C In the equation above, subtract B from both sides, and add 4×C to both sides: B – B + 4×C = 2×B – 4×C – B + 4×C which makes 4×C = B
Hint #4
Substitute (4×C) for B in eq.4: D + F = (4×C) mod C which means D + F = remainder of ((4×C) ÷ C) which makes D + F = 0 Since D and F are non-negative integers, the above equation makes D = 0 and F = 0
Hint #5
Substitute 0 for F and D, and 4×C for B in eq.2: E – 0 = 4×C + 0 which makes E = 4×C
Hint #6
Substitute 0 for D and F, and (4×C) for B in eq.5a: A + 0 + 0 = 3×(4×C) – 3×C which becomes A = 12×C – 3×C which makes A = 9×C
Solution
Substitute 9×C for A, and 4×C for B and E in eq.3: 9×C + 4×C – C + 4×C = 4×C × 4×C which becomes 16×C = 16×C² Since C ≠ 0 (from eq.4), divide both sides of the equation above by 16×C: 16×C ÷ 16×C = 16×C² ÷ 16×C which makes 1 = C making A = 9×C = 9 × 1 = 9 B = E = 4×C = 4 × 1 = 4 and ABCDEF = 941040