Puzzle for June 26, 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.
Many thanks to Abby S (age 10) for sharing this puzzle with us! Thank you, Abby!
Scratchpad
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Hint #1
eq.6 may be re-written as the following equivalent equation: E + F = A × (C ÷ D) In the equation above, replace (C ÷ D) with E (from eq.2): E + F = A × E Subtract E from each side: E + F – E = (A × E) – E which becomes F = (A × E) – E which may be written as eq.6a) F = E × (A – 1)
Hint #2
In eq.4, substitute (E × (A – 1)) for F (from eq.6a): A – B = (E × (A – 1)) ÷ E which becomes A – B = A – 1 Subtract A from each side of the above equation: A – B – A = A – 1 – A which makes –B = –1 which means B = 1
Hint #3
In eq.3, replace B with 1: D – 1 = E ÷ 1 which means eq.3a) D – 1 = E
Hint #4
Substitute D – 1 for E (from eq.3a) in eq.2: D – 1 = C ÷ D Multiply both sides of the above equation by D: D × (D – 1) = D × (C ÷ D) which becomes eq.2a) D² – D = C
Hint #5
Substitute D – 1 for E (from eq.3a), and 1 for B in eq.5: D + D – 1 = A ÷ 1 which becomes eq.5a) 2×D – 1 = A
Hint #6
Substitute (D – 1) for E (from eq.3a), and 2×D – 1 for A (from eq.5a) in eq.6a: F = (D – 1) × (2×D – 1 – 1) which is equivalent to F = (D – 1) × (2×D – 2) which becomes F = D×2×D – D×2 – 2×D + 2 which becomes eq.6a) F = 2×D² – 4×D + 2
Hint #7
Substitute 1 for B, D² – D for C (from eq.2a), D – 1 for E (from eq.3a), and 2×D² – 4×D + 2 for F (from eq.6a) in eq.1: 1 + D² – D + D = D – 1 + 2×D² – 4×D + 2 which becomes 1 + D² = 2×D² – 3×D + 1 Subtract 1 and D² from each side of the equation above: 1 + D² – 1 – D² = 2×D² – 3×D + 1 – 1 – D² which becomes 0 = D² – 3×D which may be written as eq.1a) 0 = D×(D – 3)
Solution
In eq.1a, either: D = 0 or D = 3 Since D ≠ 0 (from eq.2 or eq.6), then: D = 3 making A = 2×D – 1 = 2×3 – 1 = 6 – 1 = 5 (from eq.5a) C = D² – D = 3² – 3 = 9 – 3 = 6 (from eq.2a) E = 3 – 1 = 2 (from eq.3a) F = 2×D² – 4×D + 2 = 2×3² – 4×3 + 2 = 2×9 – 12 + 2 = 18 – 10 = 8 (from eq.6a) and ABCDEF = 516328