Puzzle for December 26, 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.
* "A mod E" is the remainder of A ÷ E.
** CD is a 2-digit number (not C×D). CD is an angle expressed in degrees.
Many thanks to Tom H for sending us this interesting puzzle. Thanks, Tom!
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Hint #1
The cosine of any angle is a real number from –1 thru 1. Since A and F are integers, then in eq.5: A – F = an integer which means cosine (CD) = an integer and makes cosine (CD) = –1 or 0 or 1
Hint #2
In eq.5, the only non-negative two-digit integers for CD that yield an integer value for cosine are: If CD = 00, then cosine (00) = 1 If CD = 90, then cosine (90) = 0 making D = 0 and C = 0 or 9
Hint #3
Begin checking: C = 9 ... Substituting 9 for C in eq.4 would yield: A mod E = 9 which would mean ie.4a) remainder of (A ÷ E) = 9 The remainder of any division must be less than the divisor. This means: remainder of (A ÷ E) < E Since E ≤ 9 (E must be a one-digit non-negative integer), then: ie.4b) remainder of (A ÷ E) < 9
Hint #4
Finish checking: C = 9 ... In ie.4b, replacing remainder of (A ÷ E) with 9 (from ie.4a) would make: 9 < 9 Since 9 cannot be < 9, then: C ≠ 9 and therefore makes C = 0
Hint #5
In eq.5, replace C and D with 0: cosine (00) = A – F which makes 1 = A – F Add F to both sides of the above equation: 1 + F = A – F + F which makes eq.5a) 1 + F = A
Hint #6
eq.2 may be written as: B + D = (C + E + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × (B + D) = 3 × (C + E + F) ÷ 3 which becomes 3 × (B + D) = C + E + F Substitute 0 for C and D: 3 × (B + 0) = 0 + E + F which becomes eq.2a) 3×B = E + F
Hint #7
eq.3 may be written as: C + F = (A + B + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × (C + F) = 3 × (A + B + E) ÷ 3 which becomes 3 × (C + F) = A + B + E Substitute 0 for C, and 1 + F for A (from eq.5a): 3 × (0 + F) = 1 + F + B + E which becomes 3×F = 1 + F + B + E which may be written as eq.3a) 3×F = 1 + B + E + F
Hint #8
Substitute 3×B for E + F (from eq.2a) in eq.3a: 3×F = 1 + B + 3×B which becomes 3×F = 1 + 4×B Subtract 1 from both sides of the equation above: 3×F – 1 = 1 + 4×B – 1 which becomes 3×F – 1 = 4×B Divide both sides by 4: (3×F – 1) ÷ 4 = 4×B ÷ 4 which becomes eq.3b) ¾×F – ¼ = B
Hint #9
Substitute (¾×F – ¼) for B (from eq.3b) in eq.2a: 3×(¾×F – ¼) = E + F which becomes 2¼×F – ¾ = E + F Subtract F from each side of the equation above: 2¼×F – ¾ – F = E + F – F which becomes eq.2b) 1¼×F – ¾ = E
Solution
Substitute 1 + F for A (from eq.5a), ¾×F – ¼ for B (from eq.3b), 0 for C and D, and 1¼×F – ¾ for E (from eq.2b) in eq.1: 1 + F + ¾×F – ¼ + 0 + 0 + 1¼×F – ¾ + F = 28 which simplifies to 4×F = 28 Divide both sides of the above equation by 4: 4×F ÷ 4 = 28 ÷ 4 which means F = 7 making A = 1 + F = 1 + 7 = 8 (from eq.5a) B = ¾×F – ¼ = (¾ × 7) – ¼ = 5¼ – ¼ = 5 (from eq.3b) E = 1¼×F – ¾ = (1¼ × 7) – ¾ = 8¾ – ¾ = 8 (from eq.2b) and ABCDEF = 850087