Puzzle for September 11, 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.
Scratchpad
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Hint #1
Add E to both sides of eq.5: C + E + E = B – E + E which becomes eq.5a) C + 2×E = B In eq.6, replace B with C + 2×E (from eq.5a): A + C – E = C + 2×E + E In the above equation, subtract C from both sides, and add E to both sides: A + C – E – C + E = C + 2×E + E – C + E which simplifies to A = 4×E
Hint #2
Add E to both sides of eq.3: B – E + E = A – C + E which becomes B = A – C + E In the above equation, replace A with 4×E: B = 4×E – C + E which becomes eq.3a) B = 5×E – C
Hint #3
In eq.5a, substitute 5×E – C for B (from eq.3a): C + 2×E = 5×E – C In the equation above, add C to both sides, and subtract 2×E from both sides: C + 2×E + C – 2×E = 5×E – C + C – 2×E which makes 2×C = 3×E Divide both sides by 2: 2×C ÷ 2 = 3×E ÷ 2 which makes C = 1½×E
Hint #4
Substitute 1½×E for C in eq.3a: B = 5×E – 1½×E which makes B = 3½×E
Hint #5
Add C to both sides of eq.4: F – C + C = C + E + C which becomes F = 2×C + E Substitute (1½×E) for C in the above equation: F = 2×(1½×E) + E which becomes F = 3×E + E which makes F = 4×E
Hint #6
Substitute 4×E for F, 3½×E for B, and 1½×E for C in eq.2: D + E + 4×E = 3½×E + 1½×E which becomes D + 5×E = 5×E Subtract 5×E from both sides of the above equation: D + 5×E – 5×E = 5×E – 5×E which makes D = 0
Solution
Substitute 4×E for A and F, 3½×E for B, 1½×E for C, and 0 for D in eq.1: 4×E + 3½×E + 1½×E + 0 + E + 4×E = 28 which simplifies to 14×E = 28 Divide both sides of the equation above by 14: 14×E ÷ 14 = 28 ÷ 14 which makes E = 2 making A = F = 4×E = 4 × 2 = 8 B = 3½×E = 3½ × 2 = 7 C = 1½×E = 1½ × 2 = 3 and ABCDEF = 873028