Puzzle for December 29, 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.
* CD, DE, and EF are 2-digit numbers (not C×D, D×E, or E×F).
Scratchpad
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Hint #1
Subtract the left and right sides of eq.2 from the left and right sides of eq.3, respectively: A + B + F – (B + C) = C + D + E – (A + D) which is the same as A + B + F – B – C = C + D + E – A – D which becomes A + F – C = C + E – A Add A and C to both sides of the above equation: A + F – C + A + C = C + E – A + A + C which becomes eq.3a) 2×A + F = 2×C + E
Hint #2
Subtract E from both sides of eq.4: E + F – E = A + C – E which becomes F = A + C – E In eq.3a, replace F with A + C – E: 2×A + A + C – E = 2×C + E which becomes 3×A + C – E = 2×C + E Subtract C and E from both sides of the equation above: 3×A + C – E – C – E = 2×C + E – C – E which becomes eq.3b) 3×A – 2×E = C
Hint #3
In eq.4, replace C with 3×A – 2×E (from eq.3b): E + F = A + 3×A – 2×E which becomes E + F = 4×A – 2×E Subtract E from both sides of the equation above: E + F – E = 4×A – 2×E – E which becomes eq.4a) F = 4×A – 3×E
Hint #4
Subtract C from each side of eq.2: B + C – C = A + D – C which becomes B = A + D – C In eq.5, substitute A + D – C for B: A + D – C + D = E which becomes eq.2a) A – C + 2×D = E
Hint #5
Substitute (3×A – 2×E) for C (from eq.3b) in eq.2a: A – (3×A – 2×E) + 2×D = E which is the same as A – 3×A + 2×E + 2×D = E which becomes –2×A + 2×E + 2×D = E In the above equation, add 2×A to both sides, and subtract 2×E from both sides: –2×A + 2×E + 2×D + 2×A – 2×E = E + 2×A – 2×E which becomes eq.5a) 2×D = 2×A – E
Hint #6
eq.6 may be written as: 10×C + D – (10×E + F) = D + 10×D + E + E which becomes 10×C + D – 10×E – F = 11×D + 2×E In the equation above, subtract D from each side, and add 10×E and F to each side: 10×C + D – 10×E – F – D + 10×E + F = 11×D + 2×E – D + 10×E + F which simplifies to 10×C = 10×D + 12×E + F which may be written as eq.6a) 10×C = 5×(2×D) + 12×E + F
Hint #7
Substitute 3×A – 2×E for C (from eq.3b), 2×A – E for 2×D (from eq.5a), and 4×A – 3×E for F (from eq.4a) in eq.6a: 10×(3×A – 2×E) = 5×(2×A – E) + 12×E + 4×A – 3×E which becomes 30×A – 20×E = 10×A – 5×E + 12×E + 4×A – 3×E which becomes 30×A – 20×E = 14×A + 4×E In the above equation, add 20×E to both sides, and subtract 14×A from both sides: 30×A – 20×E + 20×E – 14×A = 14×A + 4×E + 20×E – 14×A which simplifies to 16×A = 24×E Divide both sides by 16: 16×A ÷ 16 = 24×E ÷ 16 which makes A = 1½×E
Hint #8
In eq.5a, replace A with (1½×E): 2×D = 2×(1½×E) – E which becomes 2×D = 3×E – E = 2×E which makes 2×D = 2×E Divide both sides of the equation above by 2: 2×D ÷ 2 = 2×E ÷ 2 which means D = E
Hint #9
In eq.5, replace D with E: B + E = E Subtract E from each side of the equation above: B + E – E = E – E which makes B = 0
Hint #10
In eq.4a, substitute (1½×E) for A: 4×(1½×E) – 3×E = F which becomes 6×E – 3×E = F which makes 3×E = F
Hint #11
In eq.3b, replace A with (1½×E): 3×(1½×E) – 2×E = C which becomes 4½×E – 2×E = C which makes 2½×E = C
Solution
Substitute 1½×E for A, 0 for B, 2½×E for C, E for D, and 3×E for F in eq.1: 1½×E + 0 + 2½×E + E + E + 3×E = 18 which simplifies to 9×E = 18 Divide both sides of the equation above by 9: 9×E ÷ 9 = 18 ÷ 9 which means E = 2 making A = 1½×E = 1½ × 2 = 3 C = 2½×E = 2½ × 2 = 5 D = E = 2 F = 3×E = 3 × 2 = 6 and ABCDEF = 305226