Puzzle for May 30, 2021 ( )
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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.
We thank Lily S (age 12) for sending us this challenging puzzle! Thank you, Lily!
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Hint #1
In eq.5, substitute (B × E) for C (from eq.4): (B × E) ÷ B = A – F which becomes E = A – F Add F to both sides of the equation above: E + F = A – F + F which becomes eq.5a) E + F = A
Hint #2
In eq.3, replace A with E + F (from eq.5a): C + F = E + F + B + E which becomes C + F = 2×E + F + B Subtract F from each side of the above equation: C + F – F = 2×E + F + B – F which becomes eq.3a) C = 2×E + B
Hint #3
Subtract F and D from each side of eq.2: E + F – F – D = B + D – F – D which becomes eq.2a) E – D = B – F Subtract F, A, and E from each side of eq.3: C + F – F – A – E = A + B + E – F – A – E which becomes eq.3b) C – A – E = B – F
Hint #4
In eq.2a, substitute C – A – E for B – F (from eq.3b): E – D = C – A – E Add A and E to both sides of the equation above: E – D + A + E = C – A – E + A + E which becomes eq.2b) 2×E – D + A = C
Hint #5
eq.6 may be written as: D + F = (A + C) ÷ 2 Multiply both sides of the above equation by 2: 2 × (D + F) = 2 × ((A + C) ÷ 2) which becomes eq.6a) 2×D + 2×F = A + C
Hint #6
Substitute 2×E – D + A for C (from eq.2b) into eq.6a: 2×D + 2×F = A + 2×E – D + A which becomes 2×D + 2×F = 2×A + 2×E – D Add D to both sides of the equation above: 2×D + 2×F + D = 2×A + 2×E – D + D which becomes 3×D + 2×F = 2×A + 2×E Divide both sides by 2: (3×D + 2×F) ÷ 2 = (2×A + 2×E) ÷ 2 which becomes eq.6b) 1½×D + F = A + E
Hint #7
eq.3 may be written as: C + F = A + E + B Substitute 1½×D + F for A + E (from eq.6b) in the equation above: C + F = 1½×D + F + B Subtract F from each side: C + F – F = 1½×D + F + B – F which becomes eq.3c) C = 1½×D + B
Hint #8
Substitute 1½×D + B for C (from eq.3c) into eq.3a: 1½×D + B = 2×E + B Subtract B from both sides of the above equation: 1½×D + B – B = 2×E + B – B which makes 1½×D = 2×E Divide both sides by 2: 1½×D ÷ 2 = 2×E ÷ 2 which makes ¾×D = E
Hint #9
Substitute (¾×D) for E in eq.4: eq.4a) B × (¾×D) = C
Hint #10
Substitute ¾×D for E in eq.2: ¾×D + F = B + D Subtract ¾×D from each side of the equation above: ¾×D + F – ¾×D = B + D – ¾×D which becomes eq.2c) F = B + ¼×D
Hint #11
Substitute B + D for E + F (from eq.2) in eq.5a: eq.5b) B + D = A
Hint #12
In eq.1, substitute B + D for A (from eq.5b), 1½×D + B for C (from eq.3c), ¾×D for E, and B + ¼×D for F (from eq.2c): B + D + B + 1½×D + B + D + ¾×D + B + ¼×D = 30 which becomes 4×B + 4½×D = 30 Subtract 4½×D from each side of the above equation: 4×B + 4½×D – 4½×D = 30 – 4½×D which becomes eq.1a) 4×B = 30 – 4½×D
Hint #13
Substitute B × (¾×D) for C (from eq.4a) in eq.3c: B × (¾×D) = 1½×D + B Subtract B from both sides of the above equation: B × (¾×D) – B = 1½×D + B – B which becomes B × (¾×D) – B = 1½×D which may be written as B × (¾×D – 1) = 1½×D Divide both sides by (¾×D – 1): (B × (¾×D – 1)) ÷ (¾×D – 1) = 1½×D ÷ (¾×D – 1) which makes eq.3d) B = 1½×D ÷ (¾×D – 1) (assumes that (¾×D – 1) ≠ 0)
Hint #14
Confirm: (¾×D – 1) ≠ 0 ... Assume (for a moment) that: ¾×D – 1 = 0 Adding 1 to both sides of the above equation would then yield: ¾×D – 1 + 1 = 0 + 1 which would make ¾×D = 1 Multiplying both sides of the above equation by 1⅓ would then yield: 1⅓ × ¾×D = 1⅓ × 1 which would make D = 1⅓ Since D must be an integer, then: D ≠ 1⅓ which means (¾×D – 1) ≠ 0
Hint #15
In eq.1a, substitute (1½×D ÷ (¾×D – 1)) for B (from eq.3d): 4×(1½×D ÷ (¾×D – 1)) = 30 – 4½×D which becomes 6×D ÷ (¾×D – 1) = 30 – 4½×D Multiply both sides of the above equation by (¾×D – 1): (6×D ÷ (¾×D – 1)) × (¾×D – 1) = (30 – 4½×D) × (¾×D – 1) which becomes 6×D = 22½×D – 3⅜×D² – 30 + 4½×D which becomes 6×D = 27×D – 3⅜×D² – 30 Subtract 6×D from both sides: 6×D – 6×D = 27×D – 3⅜×D² – 30 – 6×D which becomes 0 = 21×D – 3⅜×D² – 30 which may be written as eq.1b) 0 = –3⅜×D² + 21×D – 30
Hint #16
eq.1b is a quadratic equation in standard form: Using the quadratic equation solution formula to solve for D in eq.1b yields: D = { (–1)×(21) ± sq.rt.[(21)² – (4 × (–3⅜) × (–30))] } ÷ (2 × (–3⅜)) which becomes D = {–21 ± sq.rt.(441 – 405)} ÷ (–6¾) which becomes D = {–21 ± sq.rt.(36)} ÷ (–6¾) which becomes D = (–21 ± 6) ÷ (–6¾) In the above equation, either D = (–21 + 6) ÷ (–6¾) = –15 ÷ (–6¾) = 2.22222222222 or D = (–21 – 6) ÷ (–6¾) = –27 ÷ (–6¾) = 4 Since D must be an integer, then D ≠ 2.22222222222 and therefore makes D = 4 and also makes E = ¾×D = ¾ × 4 = 3
Solution
Substitute 4 for D in eq.3d: B = 1½×4 ÷ (¾×4 – 1) which becomes B = 6 ÷ (3 – 1) which becomes B = 6 ÷ 2 which means B = 3 making A = B + D = 3 + 4 = 7 (from eq.5b) C = 1½×D + B = 1½×4 + 3 = 6 + 3 = 9 (from eq.3c) F = B + ¼×D = 3 + ¼×4 = 3 + 1 = 4 (from eq.2c) and ABCDEF = 739434