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question:What is the least prime factor of 11^5 11^4?
answer:First, factor out the common term 11^4 in the given expression: [ 11^5 - 11^4 = 11^4(11-1) = 11^4 cdot 10 ] [ = 11^4 cdot (2 cdot 5) ] So, the factorization yields 2 cdot 5 cdot 11^4. The smallest prime factor in the expression is boxed{2}.
question:a plane, six points ( A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2} ) are given. Prove that if the circumcircles of the triangles ( A_{1}B_{1}C_{1}, A_{1}B_{2}C_{2}, A_{2}B_{1}C_{2}, A_{2}B_{2}C_{1} ) pass through one point, then the circumcircles of the triangles ( A_{2}B_{2}C_{2}, A_{2}B_{1}C_{1}, A_{1}B_{2}C_{1}, A_{1}B_{1}C_{2} ) also pass through one point.
answer:1. **Initial Setup and Given Conditions:** The problem states that we are given six points on a plane: ( A_1, A_2, B_1, B_2, C_1, C_2 ). 2. **Noting the Circles and Intersections:** We need to prove that if the circles circumscripted around triangles ( A_1B_1C_1 ), ( A_1B_2C_2 ), ( A_2B_1C_2 ), and ( A_2B_2C_1 ) pass through a common point, then the circles around triangles ( A_2B_2C_2 ), ( A_2B_1C_1 ), ( A_1B_2C_1 ), and ( A_1B_1C_2 ) also intersect at a single point. 3. **Using Inversion to Simplify the Geometry:** Consider an inversion centered at the common intersection point of the circumcircles of triangles ( A_1B_1C_1 ), ( A_1B_2C_2 ), ( A_2B_1C_2 ), and ( A_2B_2C_1 ). 4. **Effect of the Inversion:** An inversion in geometry transforms circles passing through the center of inversion into lines. Therefore, the circumcircles mentioned will transform into straight lines upon inversion. 5. **Result of the Inversion:** Post inversion, we obtain four lines instead of the four circles. We should now consider the arrangement of these lines: - Let these lines be denoted as the set ( L ) formed by the inversion of the circles around ( A_1B_1C_1 ), ( A_1B_2C_2 ), ( A_2B_1C_2 ), and ( A_2B_2C_1 ). 6. **Applying Known Theorem (referenced as problem 2.83):** According to a geometric theorem stated as problem 2.83(a) (likely from a reference geometry text), if four lines (which were originally four circles passing through one point) form a specific configuration after inversion, the circumcircles of the triangles formed by the intersecting lines in the configuration will pass through a single point. 7. **Inverting Back to the Original Configuration:** Apply inversion again to transform the lines back into circles. Therefore, the circumcircles from the newly formed triangles will also pass through a common point: - These triangles are ( A_2B_2C_2 ), ( A_2B_1C_1 ), ( A_1B_2C_1 ), and ( A_1B_1C_2 ). 8. **Conclusion:** Given the above steps, through inversion and the referenced theorem, we confirm that if the circumcircles of ( A_1B_1C_1 ), ( A_1B_2C_2 ), ( A_2B_1C_2 ), and ( A_2B_2C_1 ) pass through a common point, then indeed the circumcircles of ( A_2B_2C_2 ), ( A_2B_1C_1 ), ( A_1B_2C_1 ), and ( A_1B_1C_2 ) will also pass through a common point. [ boxed{} ]
question:Let a and b be nonzero real numbers such that [(3 4i)(a + bi)] is pure imaginary. Find frac{a}{b}.
answer:Expanding (3 - 4i)(a + bi), we get [3a - 4ai + 3bi - 4bi^2.] Since i^2 = -1, we simplify this to: [3a - 4ai + 3bi + 4b = (3a + 4b) + (-4a + 3b)i.] For the number to be pure imaginary, the real part must be zero: [3a + 4b = 0.] Solving for frac{a}{b}, we isolate a: [a = -frac{4}{3}b.] Thus, the ratio is: [frac{a}{b} = boxed{-frac{4}{3}}.]
question:Let ( S = left{ p_1 p_2 cdots p_n mid p_1, p_2, ldots, p_n text{ are distinct primes and } p_1, ldots, p_n < 30 right} ). Assume 1 is in ( S ). Let ( a_1 ) be an element of ( S ). We define, for all positive integers ( n ): [ a_{n+1} = frac{a_n}{n+1} quad text{if } a_n text{ is divisible by } n+1; ] [ a_{n+1} = (n+2) a_n quad text{if } a_n text{ is not divisible by } n+1. ] How many distinct possible values of ( a_1 ) are there such that ( a_j = a_1 ) for infinitely many ( j )'s?
answer:1. **Understanding the Set S:** The set S consists of products of distinct primes that are less than 30. Let's list these primes: [ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ] If 1 is included in S, then any subset of {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} where the product of elements in the subset can be an element of S. 2. **Behavior Analysis for Odd a_1:** If (a_1) is odd, then all numbers (a_j) will retain an odd nature in changing sequence states. - When ( j ) is even: [ a_j = (j+1) a_1 ] because ( a_1 ) is never divisible by ( j+1 ) if ( j+1 ) is even. - When ( j ) is odd: [ a_j = a_1 ] because ( a_1 ) remains odd and is constant through odd indices. Therefore, for infinitely many ( j )-values, ( a_j = a_1 ). 3. **Behavior Analysis for Even a_1):** If (a_1) is an even number greater than 2, the position change leads to a non-repetition if it contains more factors: - Let ( p = min({p_1, p_2, ldots, p_n } backslash { 2 }) ). - For even (a_1 > 2), (a_2 = frac{a_1}{2}), retaining oddness. - Now, generalizing: [ a_j = frac{(j+1) a_1}{2} ] for odd ( j ). This ensures ( a_i neq a_1 ) for all values exceeding 1. Specifically: - (a_p = frac{a_1}{2p}) - For indexes (k), there will be differences: [ a_{p+k} = a_p text{ when } k text{ is even} ] Adding: [ a_{p+k} = (p+k+1) a_p text{ when } k text{ is odd} ] No infinite (j) values hold (a_j = a_1 ). 4. **Edge Case a_1 = 2:** - We demonstrate through induction: [ a_j = j+1 text{ for } j text{ odd} ] [ a_j = 1 text{ for } j text{ even} ] But no repetition leads want ( a_j = a_1 ) infinitely. 5. **Counting the Valid Values:** The set S should be represented per required terms: [ text{We only count odd elements of } {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} ] Giving us 9 odd primes. These primes form subsets each providing a product valid for distinct values: [ text{Number of odd subsets in } S = 2^9 = 512 ] # Conclusion: [ boxed{512} ]