Puzzle for February 7, 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.
* AB is a 2-digit number (not A×B).
Once again, we extend our thanks to Judah S (age 14) for sending us a challenging puzzle. Thank you, Judah!
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Hint #1
In eq.6, substitute (D + F) for AB (from eq.4): (D + F) ÷ F = C Multiply both sides of the above equation by F: ((D + F) ÷ F) × F = C × F which becomes eq.6a) D + F = C × F
Hint #2
Subtract F from both sides of eq.6a: D + F – F = C × F – F which becomes D = F × (C – 1) In the above equation, replace C with A + F (from eq.3): eq.6b) D = F × (A + F – 1)
Hint #3
In eq.2, substitute F × (A + F – 1) for D (from eq.6b): F × (A + F – 1) = E + F Subtract F from each side of the equation above: F × (A + F – 1) – F = E + F – F which becomes F × ((A + F – 1) – 1) = E which becomes eq.2a) F × (A + F – 2) = E
Hint #4
In eq.5, substitute (F × (A + F – 2)) for E (from eq.2a): (F × (A + F – 2)) ÷ F = B which becomes eq.5a) A + F – 2 = B
Hint #5
Substitute A + F – 2 for B (from eq.5a), A + F for C (from eq.3), F × (A + F – 1) for D (from eq.6b), and F × (A + F – 2) for E (from eq.2a) in eq.1: A + A + F – 2 + A + F + F × (A + F – 1) + F × (A + F – 2) + F = 25 which becomes 3×A + 3×F – 2 + F×A + F×F – F + F×A + F×F – 2×F = 25 which becomes 3×A – 2 + 2×F×A + 2×F×F = 25 which may be written as eq.1a) 3×A – 2 + 2×(F×A + F²) = 25
Hint #6
eq.6 may be written as: (10×A + B) ÷ F = C In the equation above, replace B with A + F – 2 (from eq.5a), and C with A + F (from eq.3): (10×A + A + F – 2) ÷ F = A + F which becomes (11×A + F – 2) ÷ F = A + F Multiply both sides of the equation above by F: ((11×A + F – 2) ÷ F) × F = (A + F) × F which becomes eq.6c) 11×A + F – 2 = F×A + F²
Hint #7
Substitute 11×A + F – 2 for F×A + F² (from eq.6c) into eq.1a: 3×A – 2 + 2×(11×A + F – 2) = 25 which becomes 3×A – 2 + 22×A + 2×F – 4 = 25 which becomes 25×A + 2×F – 6 = 25 In the above equation, subtract 25×A from both sides, and add 6 to both sides: 25×A + 2×F – 6 – 25×A + 6 = 25 – 25×A + 6 which becomes 2×F = 31 – 25×A Divide both sides by 2: 2×F ÷ 2 = (31 – 25×A) ÷ 2 which makes eq.1b) F = 15½ – 12½×A
Hint #8
Substitute (15½ – 12½×A) for F (from eq.1b) in eq.1a: 3×A – 2 + 2×((15½ – 12½×A)×A + (15½ – 12½×A)²) = 25 which becomes 3×A – 2 + 2×(15½×A – 12½×A² + 240¼ – 387½×A + 156¼×A²) = 25 which becomes 3×A – 2 + 2×(240¼ – 372×A + 143¾×A²) = 25 which becomes 3×A – 2 + 480½ – 744×A + 287½×A² = 25 which becomes 478½ – 741×A + 287½×A² = 25 Subtract 25 from both sides of the above equation: 478½ – 741×A + 287½×A² – 25 = 25 – 25 which becomes 453½ – 741×A + 287½×A² = 0 which may be written as eq.1c) 287½×A² – 741×A + 453½ = 0
Hint #9
eq.1c is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for A in eq.1c yields: A = { (–1)×(–741) ± sq.rt.[(–741)² – (4 × (287½) × (453½))] } ÷ (2 × (287½)) which becomes A = {741 ± sq.rt.[549081 – 521525]} ÷ 575 which becomes A = (741 ± sq.rt.[27556]) ÷ 575 which becomes A = (741 ± 166) ÷ 575 In the above equation, either: A = (741 + 166) ÷ 575 = 907 ÷ 575 = 1.57739130434783 or: A = (741 – 166) ÷ 575 = 575 ÷ 575 = 1 Since A must be an integer, then A ≠ 1.57739130434783 and therefore makes A = 1
Solution
Substitute 1 for A in eq.1b: F = 15½ – 12½×1 = 15½ – 12½ = 3 making B = A + F – 2 = 1 + 3 – 2 = 2 (from eq.5a) C = A + F = 1 + 3 = 4 (from eq.3) D = F × (A + F – 1) = 3 × (1 + 3 – 1) = 3 × (3) = 9 (from eq.6b) E = F × (A + F – 2) = 3 × (1 + 3 – 2) = 3 × (2) = 6 (from eq.2a) and ABCDEF = 124963