Puzzle for March 7, 2019 ( )
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.
* BC, CD, and EF are 2-digit numbers (not B×C, C×D, or E×F).
Scratchpad
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Hint #1
eq.6 may be written as: 10×C + D - F = C + D + F Add (F - C - D) to both sides of the above equation: 10×C + D - F + (F - C - D) = C + D + F + (F - C - D) which simplifies to 9×C = 2×F Divide both sides by 2: 9×C ÷ 2 = 2×F ÷ 2 which makes 4½×C = F
Hint #2
Add B + D to both sides of eq.4: F - D + B + D = A - B + C + B + D which becomes F + B = A + C + D In eq.2, replace A + C + D with F + B: F + B = E + F Subtract F from both sides of the above equation: F + B - F = E + F - F which makes B = E
Hint #3
eq.5 may be written as: A + 10×B + C + D = 10*E + F In the equation above, replace E with B: A + 10×B + C + D = 10*B + F Subtract 10×B from both sides: A + 10×B + C + D - 10×B = 10*B + F - 10×B which becomes eq.5a) A + C + D = F
Hint #4
In eq.5a, substitute E + F for A + C + D (from eq.2): E + F = F Subtract F from both sides: E + F - F = F - F which makes E = 0 and which also makes B = E = 0
Hint #5
Add A + D to both sides of eq.3: C - A + A + D = A - D + A + D which becomes eq.3a) C + D = 2×A In eq.2, substitute 2×A for C + D, 0 for E, and 4½×C for F: A + 2×A = 0 + 4½×C which becomes 3×A = 4½×C Divide both sides of the above equation by 3: 3×A ÷ 3 = 4½×C ÷ 3 which makes A = 1½×C
Hint #6
In eq.3a, replace A with 1½×C: C + D = 2×1½×C which becomes C + D = 3×C Subtract C from each side: C + D - C = 3×C - C which makes D = 2×C
Solution
Substitute 1½×C for A, 0 for B and E, 2×C for D, and 4½×C for F in eq.1: 1½×C + 0 + C + 2×C + 0 + 4½×C = 18 which simplifies to 9×C = 18 Divide both sides by 9: 9×C ÷ 9 = 18 ÷ 9 which means C = 2 making A = 1½×C = 1½ × 2 = 3 D = 2×C = 2 × 2 = 4 F = 4½×C = 4½ × 2 = 9 and ABCDEF = 302409