Puzzle for December 5, 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
Help Area
Hint #1
In eq.4, replace F with D – B (from eq.2): E + D – B = C – B Add B to both sides of the above equation: E + D – B + B = C – B + B which becomes eq.2a) E + D = C
Hint #2
In eq.3, replace C with E + D (from eq.2a): E + D + E = D + F which becomes 2×E + D = D + F Subtract D from each side of the above equation: 2×E + D – D = D + F – D which makes eq.3a) 2×E = F
Hint #3
Substitute E + D for C (from eq.2a) in eq.6: D × E = E + D + E which becomes D × E = D + 2×E Subtract D from each side of the equation above: D × E – D = D + 2×E – D which becomes D × (E – 1) = 2×E Divide both sides by (E – 1), which requires that E ≠ 1: D × (E – 1) ÷ (E – 1) = 2×E ÷ (E – 1) which becomes eq.6a) D = 2×E ÷ (E – 1)
Hint #4
Begin confirming: E ≠ 1 ... Substituting 1 for E in eq.6 would yield: D × 1 = C + 1 which would make eq.6b) D = C + 1 Substituting 1 for E in eq.3a would yield: 2×1 = F which would make 2 = F
Hint #5
Finish confirming: E ≠ 1 ... Substituting 1 for E, C + 1 for D (from eq.6b), and 2 for F in eq.3 would yield: C + 1 = C + 1 + 2 which would make C + 1 = C + 3 Subtracting C from both sides would yield: C + 1 – C = C + 3 – C which would make 1 = 3 Since 1 ≠ 3, then E ≠ 1
Hint #6
Substitute 2×E for F (from eq.3a), and (2×E ÷ (E – 1)) for D (from eq.6a) in eq.2: 2×E = (2×E ÷ (E – 1)) – B In the equation above, subtract 2×E from both sides, and add B to both sides: 2×E – 2×E + B = (2×E ÷ (E – 1)) – B – 2×E + B which becomes eq.3b) B = (2×E ÷ (E – 1)) – 2×E
Hint #7
Substitute (2×E ÷ (E – 1)) for D (from eq.6a) into eq.2a: eq.2b) (2×E ÷ (E – 1)) + E = C
Hint #8
Add A and B to both sides of eq.5: A – B + A + B = C + D – A + A + B which becomes 2×A = C + D + B In the above equation, substitute (2×E ÷ (E – 1)) + E for C (from eq.2b), (2×E ÷ (E – 1)) for D (from eq.6a), and (2×E ÷ (E – 1)) – 2×E for B (from eq.3b): 2×A = (2×E ÷ (E – 1)) + E + (2×E ÷ (E – 1)) + (2×E ÷ (E – 1)) – 2×E which becomes 2×A = (6×E ÷ (E – 1)) – E Divide both sides by 2: 2×A ÷ 2 = ((6×E ÷ (E – 1)) – E) ÷ 2 which becomes eq.5a) A = (3×E ÷ (E – 1)) – ½×E
Hint #9
In eq.1, substitute: (3×E ÷ (E – 1)) – ½×E for A (from eq.5a), (2×E ÷ (E – 1)) – 2×E for B (from eq.3b), (2×E ÷ (E – 1)) + E for C (from eq.2b), (2×E ÷ (E – 1)) for D (from eq.6a), and 2×E for F: (3×E ÷ (E – 1)) – ½×E + (2×E ÷ (E – 1)) – 2×E + (2×E ÷ (E – 1)) + E + (2×E ÷ (E – 1)) + E + 2×E = 21 which simplifies to (9×E ÷ (E – 1)) + 1½×E = 21 Subtract 1½×E from each side of the above equation: (9×E ÷ (E – 1)) + 1½×E – 1½×E = 21 – 1½×E which becomes 9×E ÷ (E – 1) = 21 – 1½×E Multiply both sides by (E – 1): (9×E ÷ (E – 1)) × (E – 1) = (21 – 1½×E) × (E – 1) which becomes 9×E = 21×E – 21 – 1½×E×E + 1½×E which becomes 9×E = 22½×E – 21 – 1½×E² Subtract 9×E from both sides: 9×E – 9×E = 22½×E – 21 – 1½×E² – 9×E which becomes 0 = 13½×E – 21 – 1½×E² which may be written as eq.1a) 0 = –1½×E² + 13½×E – 21
Hint #10
eq.1a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for E in eq.1a yields: E = { (–1)×(13½) ± sq.rt.[(13½)² – (4 × (–1½) × (–21))] } ÷ (2 × (–1½)) which becomes E = {–13½ ± sq.rt.(182¼ – 126)} ÷ (–3) which becomes E = {–13½ ± sq.rt.(56¼)} ÷ (–3) which becomes E = (–13½ ± 7½) ÷ (–3) In the above equation, either: E = (–13½ + 7½) ÷ (–3) = –6 ÷ (–3) = 2 or: E = (–13½ – 7½) ÷ (–3) = –21 ÷ (–3) = 7
Hint #11
Check: E = 7 ... Substituting 7 for E in eq.3a would yield: 2×7 = F which would make 14 = F Since F must be a one-digit integer, then F ≠ 14 which means E ≠ 7 and therefore makes E = 2
Hint #12
Substitute 2 for E in eq.5a: A = (3×2 ÷ (2 – 1)) – ½×2 which becomes A = (6 ÷ 1) – 1 which makes A = 6 – 1 which means A = 5
Hint #13
Substitute 2 for E in eq.3b: B = (2×2 ÷ (2 – 1)) – 2×2 which becomes B = (4 ÷ 1) – 4 which makes B = 4 – 4 which means B = 0
Hint #14
Substitute 2 for E in eq.2b: (2×2 ÷ (2 – 1)) + 2 = C which becomes (4 ÷ 1) + 2 = C which becomes 4 + 2 = C which makes 6 = C
Hint #15
Substitute 2 for E in eq.6a: D = 2×2 ÷ (2 – 1) which becomes D = 4 ÷ (1) which makes D = 4
Solution
Substitute 2 for E in eq.3a: 2×2 = F which makes 4 = F and makes ABCDEF = 506424