Puzzle for April 24, 2020 ( )
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
Subtract the left and right sides of eq.5 from the left and right sides of eq.2, respectively: A + C – (A – C) = D – (D – E – F) which is equivalent to A + C – A + C = D – D + E + F which becomes eq.2a) 2×C = E + F
Hint #2
Add the left and right sides of eq.3 to the left and right sides of eq.4, respectively: E + F = C – B + A + B which becomes E + F = C + A In the above equation, replace E + F with 2×C (from eq.2a): 2×C = C + A Subtract C from each side: 2×C – C = C + A – C which makes C = A
Hint #3
In eq.2, replace A with C: C + C = D which means eq.2b) 2×C = D
Hint #4
In eq.6, substitute 2×C for D, and C for A: 2×C ÷ C = B – E which becomes 2 = B – E Add E to each side of the equation above: 2 + E = B – E + E which becomes eq.6a) 2 + E = B
Hint #5
Substitute (2 + E) for B (from eq.6a) in eq.4: E = C – (2 + E) which may be written as: E = C – 2 – E Add 2 and E to both sides of the above equation: E + 2 + E = C – 2 – E + 2 + E which makes 2 + 2×E = C and also makes eq.4a) A = C = 2 + 2×E
Hint #6
Substitute (2 + 2×E) for C (from eq.4a) in eq.2a: 2×(2 + 2×E) = E + F which becomes 4 + 4×E = E + F Subtract E from each side of the above equation: 4 + 4×E – E = E + F – E which means eq.2c) 4 + 3×E = F
Hint #7
Substitute (2 + 2×E) for C (from eq.4a) in eq.2b: 2×(2 + 2×E) = D which makes eq.2d) 4 + 4×E = D
Solution
Substitute 2 + 2×E for A and C (from eq.4a), 2 + E for B (from eq.6a), 4 + 4×E for D (from eq.2d), and 4 + 3×E for F (from eq.2c) in eq.1: 2 + 2×E + 2 + E + 2 + 2×E + 4 + 4×E + E + 4 + 3×E = 27 which simplifies to 13×E + 14 = 27 Subtract 14 from both sides of the equation above: 13×E + 14 – 14 = 27 – 14 which makes 13×E = 13 Divide both sides by 13: 13×E ÷ 13 = 13 ÷ 13 which means E = 1 making A = C = 2 + 2×E = 2 + 2×1 = 4 (from eq.4a) B = 2 + E = 2 + 1 = 3 (from eq.6a) D = 4 + 4×E = 4 + 4×1 = 8 (from eq.2d) F = 4 + 3×E = 4 + 3×1 = 7 (from eq.2c) and ABCDEF = 434817