Puzzle for May 31, 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.
* AB, CD, DE, and EF are 2-digit numbers (not A×B, C×D, D×E, or E×F).
Scratchpad
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Hint #1
eq.6 may be written as: 10×A + B + C = 10×C + D + 10×D + E + B Subtract both B and C from each side of the above equation: 10×A + B + C - B - C = 10×C + D + 10×D + E + B - B - C which becomes eq.6a) 10×A = 9×C + 11×D + E
Hint #2
In eq.6a, substitute (C + D) for A (from eq.3): 10×(C + D) = 9×C + 11×D + E which is equivalent to 10×C + 10×D = 9×C + 11×D + E Subtract both 9×C and 10×D from each side of the above equation: 10×C + 10×D - 9×C - 10×D = 9×C + 11×D + E - 9×C - 10×D which simplifies to eq.6b) C = D + E
Hint #3
In eq.6b, replace D + E with C + F (from eq.2): C = C + F Subtract C from each side: C - C = C + F - C which makes 0 = F
Hint #4
eq.5 may be written as: 10×E + F = 10×C + D - (10×D + E) which is the same as 10×E + F = 10×C + D - 10×D - E which becomes 10×E + F = 10×C - 9×D - E In the above equation, substitute 0 for F, and add E to each side: 10×E + 0 + E = 10×C - 9×D - E + E which becomes eq.5a) 11×E = 10×C - 9×D
Hint #5
In eq.5a, substitute (D + E) for C (from eq.6a): 11×E = 10×(D + E) - 9×D which is equivalent to 11×E = 10×D + 10×E - 9×D which becomes 11×E = D + 10×E Subtract 10×E from each side: 11×E - 10×E = D + 10×E - 10×E which makes E = D
Hint #6
Substitute D for E in eq.6b: C = D + D which means C = 2×D
Hint #7
Substitute 2×D for C in eq.3: 2×D + D = A which makes 3×D = A
Hint #8
Substitute 3×D for A, and 2×D for C in eq.4: B - 3×D = 3×D + 2×D Add 3×D to both sides of the equation above: B - 3×D + 3×D = 3×D + 2×D + 3×D which makes B = 8×D
Solution
Substitute 3×D for A, 8×D for B, 2×D for C, D for E, and 0 for F in eq.1: 3×D + 8×D + 2×D + D + D + 0 = 15 which becomes 15×D = 15 Divide both sides by 15: 15×D ÷ 15 = 15 ÷ 15 which means D = 1 making A = 3×D = 3 × 1 = 3 B = 8×D = 8 × 1 = 8 C = 2×D = 2 × 1 = 2 E = D = 1 and ABCDEF = 382110