Puzzle for January 8, 2021 ( )
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
In eq.4, subtract C and F from both sides: F – C – C – F = A + C – F – C – F which becomes eq.4a) –2×C = A – 2×F In eq.3, add A to both sides, and subtract E and F from both sides: C – A – E + F + A – E – F = E – F + A – E – F which simplifies to eq.3a) C – 2×E = A – 2×F
Hint #2
In eq.4a, replace A – 2×F with C – 2×E (from eq.3a): –2×C = C – 2×E Add 2×C and 2×E to both sides of the equation above: –2×C + 2×C + 2×E = C – 2×E + 2×C + 2×E which becomes 2×E = 3×C Divide both sides by 2: 2×E ÷ 2 = 3×C ÷ 2 which makes eq.4b) E = 1½×C
Hint #3
In eq.2, replace E with 1½×C: eq.2a) 1½×C = A + B
Hint #4
eq.6 may be written as: C = (A + D + E) ÷ 3 Multiply both sides of the equation above by 3: 3 × C = 3 × (A + D + E) ÷ 3 which becomes eq.6a) 3×C = A + D + E
Hint #5
In eq.6a, replace E with 1½×C (from eq.4b): 3×C = A + D + 1½×C Subtract 1½×C from each side of the above equation: 3×C – 1½×C = A + D + 1½×C – 1½×C which becomes eq.6b) 1½×C = A + D
Hint #6
In eq.6b, substitute A + B for 1½×C (from eq.2a): A + B = A + D Subtract A from each side of the equation above: A + B – A = A + D – A which makes B = D
Hint #7
Substitute 1½×C for E (from eq.4b), and B for D in eq.5: B + 1½×C – A = B + F Subtract B from each side of the equation above: B + 1½×C – A – B = B + F – B which becomes eq.5a) 1½×C – A = F
Hint #8
Substitute (1½×C – A) for F (from eq.5a) into eq.4a: –2×C = A – 2×(1½×C – A) which becomes –2×C = A – 3×C + 2×A which becomes –2×C = 3×A – 3×C Add 3×C to both sides of the equation above: –2×C + 3×C = 3×A – 3×C + 3×C which makes C = 3×A
Hint #9
Substitute (3×A) for C in eq.5a: 1½×(3×A) – A = F which becomes 4½×A – A = F which makes 3½×A = F
Hint #10
Substitute (3×A) for C in eq.6b: 1½×(3×A) = A + D which becomes 4½×A = A + D Subtract A from each side of the equation above: 4½×A – A = A + D – A which becomes 3½×A = D and also makes B = D = 3½×A
Hint #11
Substitute (3×A) for C in eq.4b: E = 1½×(3×A) which makes E = 4½×A
Solution
Substitute 3½×A for B and D and F, 3×A for C, and 4½×A for E in eq.1: A + 3½×A + 3×A + 3½×A + 4½×A + 3½×A = 38 which simplifies to 19×A = 38 Divide both sides of the above equation by 19: 19×A ÷ 19 = 38 ÷ 19 which means A = 2 making B = D = F = 3½×A = 3½ × 2 = 7 C = 3×A = 3 × 2 = 6 E = 4½×A = 4½ × 2 = 9 and ABCDEF = 276797