Puzzle for June 13, 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.
* BC is a 2-digit number (not B×C).
** DE is a 2-digit number (not D×E). DE is an angle expressed in degrees. "sq.rt." means positive square root.
Our thanks go out to Tom H for sending us a very challenging puzzle. Thank you, Tom!
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Hint #1
There are three possible 2-digit integer values for DE that could make eq.4 true: If DE = 00, then tangent (00) = 0, making F – A = 0, which means F = A If DE = 45, then tangent (45) = 1, making F – A = 1, which means F = A + 1 If DE = 60, then tangent (60) = sq.rt.(3), making F – A = 3, which means F = A + 3
Hint #2
Since D ≠ 0 (from eq.2), then DE ≠ 00 which means F ≠ A Therefore: D = 4 and E = 5 and eq.4a) F = A + 1 or: D = 6 and E = 0 and eq.4b) F = A + 3
Hint #3
Multiply both sides of eq.2 by D: B × D = C ÷ D × D which makes eq.2a) B × D = C eq.3 may be written as: F = 10×B + C – F Add F to both sides of the equation above: F + F = 10×B + C – F + F which makes eq.3a) 2×F = 10×B + C
Hint #4
Since F is a one-digit non-negative integer, then: 0 ≤ F ≤ 9 which means 0 ≤ 2×F ≤ 18 Substitute 10×B + C for 2×F (from eq.3a) in the above inequality: 0 ≤ 10×B + C ≤ 18 which makes 0 ≤ B ≤ 1 Since B is a one-digit non-negative integer, then: B = 0 or 1
Hint #5
Begin checking: B = 0 ... Substituting (B × D) for C (from eq.2a) in eq.3a would yield: 2×F = 10×B + (B × D) which could be written as eq.3b) 2×F = B × (10 + D) Substituting 0 for B in the above equation would yield: 2×F = 0 × (10 + D) which would become 2×F = 0 which would make F = 0
Hint #6
Finish checking: B = 0 ... Substituting 0 for F in eq.4a would yield: 0 = A + 1 which would make A = –1 Substituting 0 for F in eq.4b would yield: 0 = A + 3 which would make A = –3 Since A is a non-negative integer, then A ≠ –1 or –3 which means B ≠ 0 and therefore means B = 1
Hint #7
Substitute 1 for B into eq.2a: 1 × D = C which makes D = C
Hint #8
Substitute 1 for B in eq.3b: 2×F = 1 × (10 + D) which becomes 2×F = 10 + D Subtract 10 from each side of the above equation: 2×F – 10 = 10 + D – 10 which becomes eq.3c) 2×F – 10 = D
Hint #9
eq.1 may be written as: eq.1a) A = (C + D + E + F) ÷ 4
Hint #10
Begin checking: D = 4, and E = 5 ... Substituting 4 for D in eq.3c would yield: 2×F – 10 = 4 Adding 10 to both sides of the above equation would yield: 2×F – 10 + 10 = 4 + 10 which would become 2×F = 14 Dividing both sides by 2 would yield: 2×F ÷ 2 = 14 ÷ 2 which would make F = 7 Substituting 7 for F in eq.4a would yield: 7 = A + 1 which would make eq.3d) A = 6
Hint #11
Finish checking: D = 4, and E = 5 ... Substituting 4 for C and D, 5 for E, and 7 for F in eq.1a: A = (4 + 4 + 5 + 7) ÷ 4 which would make A = (20) ÷ 4 which would mean A = 5 Since A ≠ both 5 (from the above equation) and 6 (from eq.3d), then: D ≠ 4 and E ≠ 5 which therefore means D = 6 and E = 0 and also means C = D = 6
Hint #12
Substitute 6 for D in eq.3c: 2×F – 10 = 6 Add 10 to both sides of the equation above: 2×F – 10 + 10 = 6 + 10 which makes 2×F = 16 Divide both sides by 2: 2×F ÷ 2 = 16 ÷ 2 which makes F = 8
Solution
Substitute 6 for C and D, 0 for E, and 8 for F in eq.1a: A = (6 + 6 + 0 + 8) ÷ 4 which becomes A = (20) ÷ 4 which makes A = 5 and makes ABCDEF = 516608