Puzzle for March 14, 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 positive integer.
* BC is a 2-digit number (not B×C).
** BCD is a 3-digit angle in degrees (not B×C×D). "sq.rt." means positive square root. "abs" means absolute value.
Many thanks to Tom H. for submitting this intriguing and challenging puzzle. Thank you, Tom!
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Hint #1
eq.1 may be written as: eq.1a) 10×B + C = A + F Since B, C, A, and F are all one-digit positive integers, then: 1 ≤ B ≤ 9 and 1 ≤ C ≤ 9 and 1 ≤ A ≤ 9 and 1 ≤ F ≤ 9 Combining the last two of the above inequalities yields: 1 + 1 ≤ A + F ≤ 9 + 9 which makes ie.1) 2 ≤ A + F ≤ 18
Hint #2
In ie.1, replace A + F with BC (from eq.1): 2 ≤ BC ≤ 18 Since B is a one-digit positive integer, the only number for B that makes the above inequality true is: B = 1
Hint #3
Since B = 1, and since C and D are one-digit positive integers, then for the 3-digit angle BCD: ie.3) 111 ≤ BCD ≤ 199 In eq.3, all mathematical terms on the right side of the equal sign are positive. This makes the sum and division of those terms positive. This means: sine (BCD) > 0 Since the sine (BCD) is positive, and since the sine of an angle between 180 and 360 is negative, then: BCD < 180 Combining the above inequality with ie.3 means: ie.3a) 111 ≤ BCD < 180
Hint #4
For angles between 0 and 180 degrees, the sine function is symmetric about 90 degrees. For angle BCD, this relationship may be mathematically written as: sine (BCD) = sine (180 – BCD) Combining the equation above with inequality ie.3a yields: 180 – 180 < 180 – BCD ≤ 180 – 111 which means ie.3b) 0 < 180 – BCD ≤ 79
Hint #5
Many angles have sines that are fractional ratios of integers and/or square roots of integers. However, there is only one angle between 0 and 79 degrees (from ie.3b) having a sine with a fractional ratio that has 1 + sq.rt. in its numerator (as required by eq.3). This angle is 54 degrees, and its sine is: eq.3a) sine (54) = (1 + sq.rt.(5)) ÷ 4 which makes eq.3b) 54 = 180 – BCD
Hint #6
In eq.3b, add BCD to both sides, and subtract 54 from both sides: 54 + BCD – 54 = 180 – BCD + BCD – 54 which makes BCD = 126 In eq.3a, replace 54 with 126: sine (126) = (1 + sq.rt.(5)) ÷ 4 In eq.3, this makes: B = 1 C = 2 D = 6 E = 5
Hint #7
In eq.2, substitute 2 for C, 6 for D, 5 for E, and 1 for B: 2 = log [base A] (6 + 5 + F – 1 – 2) which becomes 2 = log [base A] (8 + F) According to the laws of logarithms and exponents, the equation above may be re-written as: A ^ 2 = 8 + F which is the same as eq.2a) A² = 8 + F
Hint #8
In eq.1a, substitute 1 for B, and 2 for C: 10×1 + 2 = A + F which becomes 12 = A + F Subtract A from both sides of the above equation: 12 – A = A + F – A which makes eq.1b) 12 – A = F
Hint #9
Substitute 12 – A for F (from eq.1b) into eq.2a: A² = 8 + 12 – A which becomes A² = 20 – A In the above equation, subtract 20 from both sides, and add A to both sides: A² – 20 + A = 20 – A – 20 + A which becomes A² – 20 + A = 0 which may be written as eq.2b) A² + A – 20 = 0
Hint #10
eq.2b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for A in eq.2b yields: A = { (–1)×1 ± sq.rt.[1² – (4 × 1 × (–20))] } ÷ (2 × 1) which becomes A = {–1 ± sq.rt.(1 – (–80))} ÷ 2 which becomes A = (–1 ± sq.rt.(81)) ÷ 2 which becomes A = (–1 ± 9) ÷ 2 In the above equation, either A = (–1 + 9) ÷ 2 = 8 ÷ 2 = 4 or A = (–1 – 9) ÷ 2 = –10 ÷ 2 = –5 Since A must be a positive integer, then A ≠ –5 and therefore makes A = 4
Solution
Substitute 4 for A in eq.1b: 12 – 4 = F which makes 8 = F and makes ABCDEF = 412658