Puzzle for November 10, 2019 ( )
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, BC, CD, DE, and EF are 2-digit numbers (not A×B, B×C, C×D, D×E, or E×F).
** CD is an angle expressed in degrees.
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Hint #1
E and F are integers, which makes E – F (from eq.6) an integer. Since the sine of an angle is a real number between 0 and 1 inclusive, then E – F = 0 or 1. To make eq.6 true, check the possible values for E – F, C, and D: If E – F = 0, then sine (CD) = 0, making CD = 00 If E – F = 1, then sine (CD) = 1, making CD = 90 Therefore, C = 0 or 9 and D = 0
Hint #2
In eq.3, replace D with 0: B + 0 + F = A which is the same as eq.3a) B + F = A
Hint #3
In eq.2, replace A with B + F (from eq.3a): B + F + B = C which becomes eq.2a) 2×B + F = C
Hint #4
eq.4 may be written as: 10×E + F = 10×B + C + 10×D + E In the above equation, substitute 0 for D, and subtract E from both sides: 10×E + F – E = 10×B + C + 10×0 + E – E which becomes eq.4a) 9×E + F = 10×B + C
Hint #5
Substitute 2×B + F for C (from eq.2a) in eq.4a: 9×E + F = 10×B + 2×B + F Subtract F from both sides of the equation above: 9×E + F – F = 10×B + 2×B + F – F which becomes 9×E = 12×B Divide both sides by 12: 9×E ÷ 12 = 12×B ÷ 12 which makes ¾×E = B
Hint #6
eq.5 may be written as: 10×C + D + E + F = 10×A + B – C – D + 10×E + F Subtract E and F from both sides of the above equation: 10×C + D + E + F – E – F = 10×A + B – C – D + 10×E + F – E – F which becomes 10×C + D = 10×A + B – C – D + 9×E Add C and D to both sides: 10×C + D + C + D = 10×A + B – C – D + 9×E + C + D which becomes eq.5a) 11×C + 2×D = 10×A + B + 9×E
Hint #7
Substitute (A + B) for C (from eq.2), and 0 for D in eq.5a: 11×(A + B) + 2×0 = 10×A + B + 9×E which becomes 11×A + 11×B = 10×A + B + 9×E Subtract both 10×A and 11×B from each side of the above equation: 11×A + 11×B – 10×A – 11×B = 10×A + B + 9×E – 10×A – 11×B which simplifies to eq.5b) A = 9×E – 10×B
Hint #8
Substitute (¾×E) for B in eq.5b: A = 9×E – 10×(¾×E) which is equivalent to A = 9×E – 7½×E which makes A = 1½×E
Hint #9
Substitute 1½×E for A, and ¾×E for B in eq.2: 1½×E + ¾×E = C which makes 2¼×E = C
Hint #10
Substitute ¾×E for B, and 1½×E for A in eq.3a: ¾×E + F = 1½×E Subtract ¾×E from both sides of the above equation: ¾×E + F – ¾×E = 1½×E – ¾×E which makes F = ¾×E
Solution
Substitute 1½×E for A, ¾×E for B and F, 2¼×E for C, and 0 for D in eq.1: 1½×E + ¾×E + 2¼×E + 0 + E + ¾×E = 25 which simplifies to 6¼×E = 25 Divide both sides of the equation above by 6¼: 6¼×E ÷ 6¼ = 25 ÷ 6¼ which means E = 4 making A = 1½×E = 1½ × 4 = 6 B = F = ¾×E = ¾ × 4 = 3 C = 2¼×E = 2¼ × 4 = 9 and ABCDEF = 639043