Puzzle for October 30, 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.
* "A mod C" equals the remainder of (A ÷ C).
Scratchpad
Help Area
Hint #1
eq.2 may be written as: C = (B + D + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × C = 3 × (B + D + F) ÷ 3 which becomes 3×C = B + D + F Subtract D from both sides: 3×C - D = B + D + F - D which becomes eq.2a) 3×C - D = B + F
Hint #2
eq.3 may be written as: D = (B + E + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × D = 3 × (B + E + F) ÷ 3 which becomes 3×D = B + E + F which may be written as eq.3a) 3×D = B + F + E
Hint #3
In eq.3a, replace B + F with 3×C - D (from eq.2a): 3×D = 3×C - D + E In the above equation, subtract 3×C from both sides, and add D to both sides: 3×D - 3×C + D = 3×C - D + E - 3×C + D which becomes eq.3b) 4×D - 3×C = E
Hint #4
eq.4 may be written as: C + D = (A + B + C + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × (C + D) = 4 × (A + B + C + E) ÷ 4 which becomes 4×C + 4×D = A + B + C + E Subtract C from each side of the equation above: 4×C + 4×D - C = A + B + C + E - C which becomes eq.4a) 3×C + 4×D = A + B + E
Hint #5
In eq.4a, substitute 4×D - 3×C for E (from eq.3b): 3×C + 4×D = A + B + 4×D - 3×C In the above equation, subtract 4×D from both sides, and add 3×C to both sides: 3×C + 4×D - 4×D + 3×C = A + B + 4×D - 3×C - 4×D + 3×C which becomes eq.4b) 6×C = A + B
Hint #6
Multiply both sides of eq.2a by 2: 2×(3×C - D) = 2×(B + F) which becomes 6×C - 2×D = 2×B + 2×F In the above equation, substitute A + B for 6×C (from eq.4b): A + B - 2×D = 2×B + 2×F Subtract B from both sides, and add 2×D to both sides: A + B - 2×D - B + 2×D = 2×B + 2×F - B + 2×D which becomes eq.2b) A = B + 2×F + 2×D
Hint #7
Substitute B + 2×F + 2×D for A (from eq.2b) in eq.1: B + E + F = B + 2×F + 2×D + D - F which becomes B + E + F = B + F + 3×D Subtract B and F from each side of the equation above: B + E + F - B - F = B + F + 3×D - B - F which makes eq.1a) E = 3×D
Hint #8
Substitute 3×D for E in eq.3b: 4×D - 3×C = 3×D In the equation above, add 3×C to both sides, and subtract 3×D from both sides: 4×D - 3×C + 3×C - 3×D = 3×D + 3×C - 3×D which makes D = 3×C
Hint #9
Substitute (3×C) for D in eq.1a: E = 3×(3×C) which makes E = 9×C
Hint #10
Substitute 3×C for D in eq.2a: 3×C - 3×C = B + F which becomes 0 = B + F Since B and F must both be non-negative, the above equation makes: B = 0 and F = 0
Hint #11
Substitute 0 for B in eq.4b: 6×C = A + 0 which makes 6×C = A
Hint #12
Substitute 9×C for E, and 3×C for D in eq.5: C × 9×C = (A × 3×C) - 9×C Add 9×C to both sides of the above equation: (C × 9×C) + 9×C = (A × 3×C) - 9×C + 9×C which becomes (C × 9×C) + 9×C = A × 3×C Since C ≠ 0 (from eq.6), divide both sides by 3×C: ((C × 9×C) + 9×C) ÷ 3×C = (A × 3×C) ÷ 3×C which becomes ((C × 9×C) ÷ 3×C) + ((9×C) ÷ 3×C) = A which becomes (C × 3) + 3 = A which may be written as eq.5a) 3×C + 3 = A
Hint #13
Substitute 6×C for A in eq.5a: 3×C + 3 = 6×C Subtract 3×C from each side of the equation above: 3×C + 3 - 3×C = 6×C - 3×C which makes 3 = 3×C Divide both sides by 3: 3 ÷ 3 = 3×C ÷ 3 which makes 1 = C
Solution
Since C = 1, then: A = 6×C = 6 × 1 = 6 D = 3×C = 3 × 1 = 3 E = 9×C = 9 × 1 = 9 and ABCDEF = 601390