Puzzle for December 10, 2023 ( )
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.
* CD is a 2-digit number (not C×D).
** "D mod E" equals "remainder of D ÷ E".
Scratchpad
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Hint #1
Add B to both sides of eq.4: A - B + E + B = B + C + D + B which becomes eq.4a) A + E = 2×B + C + D eq.5 may be written as: eq.5a) B + 10×C + D = A + E + F
Hint #2
In eq.5a, replace A + E with 2×B + C + D (from eq.4a): B + 10×C + D = 2×B + C + D + F Subtract B, C, and D from both sides of the above equation: B + 10×C + D - B - C - D = 2×B + C + D + F - B - C - D which simplifies to eq.5b) 9×C = B + F
Hint #3
In eq.2, replace B + F with 9×C (from eq.5b): A + C = 9×C Subtract C from each side of the equation above: A + C - C = 9×C - C which makes A = 8×C
Hint #4
In eq.4a, substitute 8×C for A: 8×C + E = 2×B + C + D Subtract C from both sides of the equation above: 8×C + E - C = 2×B + C + D - C which becomes eq.4b) 7×C + E = 2×B + D
Hint #5
Add E to both sides of eq.3: D - E + E = B + E + E which becomes eq.3a) D = B + 2×E Subtract C from each side of eq.2: A + C - C = B + F - C which becomes eq.2a) A = B + F - C
Hint #6
Substitute B + 2×E for D (from eq.3a) into eq.4b: 7×C + E = 2×B + B + 2×E which becomes 7×C + E = 3×B + 2×E Subtract E from both sides of the above equation: 7×C + E - E = 3×B + 2×E - E which becomes eq.4c) 7×C = 3×B + E
Hint #7
Substitute B + F - C for A (from eq.2a), and C + F for D (from eq.1) in eq.4a: B + F - C + E = 2×B + C + C + F which becomes B + F - C + E = 2×B + 2×C + F In the equation above, subtract B and F from both sides, and add C to both sides: B + F - C + E - B - F + C = 2×B + 2×C + F - B - F + C which simplifies to eq.4d) E = B + 3×C
Hint #8
Substitute B + 3×C for E (from eq.4d) in eq.4c: 7×C = 3×B + B + 3×C which becomes 7×C = 4×B + 3×C Subtract 3×C from each side of the above equation: 7×C - 3×C = 4×B + 3×C - 3×C which makes 4×C = 4×B Divide both sides by 4: 4×C ÷ 4 = 4×B ÷ 4 which makes C = B
Hint #9
Substitute C for B in eq.4d: E = C + 3×C which makes E = 4×C
Hint #10
Substitute C for B, and (4×C) for E in eq.3a: D = C + 2×(4×C) which becomes D = C + 8×C which makes D = 9×C
Hint #11
Substitute 9×C for D in eq.1: 9×C = C + F Subtract C from each side of the equation above: 9×C - C = C + F - C which makes 8×C = F
Hint #12
Substitute 9×C for D, and 4×C for E in eq.6: C = 9×C mod 4×C which means C = remainder of (9×C ÷ 4×C) which becomes C = remainder of (9 ÷ 4) which makes C = 1
Solution
Since C = 1, then: A = F = 8×C = 8 × 1 = 8 B = C = 1 D = 9×C = 9 × 1 = 9 E = 4×C = 4 × 1 = 4 and ABCDEF = 811948