Puzzle for March 19, 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.
Scratchpad
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Hint #1
eq.4 may be written as: A + E = (A + B + D + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × (A + E) = 4 × (A + B + D + E) ÷ 4 which becomes 4×A + 4×E = A + B + D + E Subtract A and E from each side: 4×A + 4×E - A - E = A + B + D + E - A - E which becomes eq.4a) 3×A + 3×E = B + D
Hint #2
Add E to both sides of eq.2: B - E + E = C + E - A + E which becomes eq.2a) B = C + 2×E - A
Hint #3
In eq.4a, replace B with C + 2×E - A (from eq.2a): 3×A + 3×E = C + 2×E - A + D In the above equation, subtract 3×E from both sides, and add A to both sides: 3×A + 3×E - 3×E + A = C + 2×E - A + D - 3×E + A which becomes 4×A = C - E + D which may be written as eq.4b) 4×A = C + D - E
Hint #4
In eq.3, add C and D to both sides, and subtract E from both sides: C + D - E + C + D - E = B - C + E - A - D + C + D - E which becomes 2×C + 2×D - 2×E = B - A which may be written as eq.3a) 2×(C + D - E) = B - A
Hint #5
In eq.3a, replace C + D - E with 4×A (from eq.4b): 2×(4×A) = B - A which becomes 8×A = B - A Add A to both sides of the above equation: 8×A + A = B - A + A which makes 9×A = B
Hint #6
In eq.2a, substitute 9×A for B: 9×A = C + 2×E - A Add A to both sides of the above equation: 9×A + A = C + 2×E - A + A which becomes eq.2b) 10×A = C + 2×E
Hint #7
eq.5 may be written as: E - A = (A + B + C + D + E + F) ÷ 6 Multiply both sides of the above equation by 6: 6 × (E - A) = 6 × (A + B + C + D + E + F) ÷ 6 which becomes 6×E - 6×A = A + B + C + D + E + F Subtract A and E from each side: 6×E - 6×A - A - E = A + B + C + D + E + F - A - E which becomes 5×E - 7×A = B + C + D + F which may be written as eq.5a) 5×E - 7×A = B + D + C + F
Hint #8
Substitute 3×A + 3×E for B + D (from eq.4a) in eq.5a: 5×E - 7×A = 3×A + 3×E + C + F Subtract 3×A and 3×E from both sides of the equation above: 5×E - 7×A - 3×A - 3×E = 3×A + 3×E + C + F - 3×A - 3×E which becomes eq.5b) 2×E - 10×A = C + F
Hint #9
Substitute (C + 2×E) for 10×A (from eq.2b) into eq.5b: 2×E - (C + 2×E) = C + F which becomes 2×E - C - 2×E = C + F which becomes -C = C + F Add C to both sides of the above equation: -C + C = C + F + C which becomes 0 = 2×C + F Since C and F are non-negative integers, the above equation makes: C = 0 and F = 0
Hint #10
Substitute 0 for C in eq.2b: 10×A = 0 + 2×E which becomes 10×A = 2×E Divide both sides of the above equation by 2: 10×A ÷ 2 = 2×E ÷ 2 which makes 5×A = E
Hint #11
Substitute 0 for C, and 5×A for E in eq.4b: 4×A = 0 + D - 5×A which becomes 4×A = D - 5×A Add 5×A to both sides of the equation above: 4×A + 5×A = D - 5×A + 5×A which makes 9×A = D
Solution
Substitute 9×A for B and D, 0 for C and F, and 5×A for E in eq.1: A + 9×A + 0 + 9×A + 5×A + 0 = 24 which simplifies to 24×A = 24 Divide both sides of the above equation by 24: 24×A ÷ 24 = 24 ÷ 24 which means A = 1 making B = D = 9×A = 9 × 1 = 9 E = 5×A = 5 × 1 = 5 and ABCDEF = 190950