answer:To solve this problem, we need to find the smallest positive integer ( K ) such that every ( K )-element subset of ( {1, 2, ldots, 50} ) contains two distinct elements ( a ) and ( b ) such that ( a + b ) divides ( ab ). 1. **Rewriting the Condition**: Given ( a + b mid ab ), we can rewrite ( b ) as ( ka ) for some integer ( k ). This implies: [ a + b mid ab implies a + ka mid a cdot ka implies (k+1)a mid a^2k implies k+1 mid ak ] Since ( k+1 ) must divide ( ak ) and ( a ), we need to find pairs ( (a, b) ) such that ( a + b mid ab ). 2. **Generating Pairs**: We can generate pairs ( (a, b) ) by taking ( q ), a factor of ( a ), and letting ( b = a(q-1) ). This means ( a + b = a + a(q-1) = aq ) and ( ab = a cdot a(q-1) = a^2(q-1) ). Thus, ( aq mid a^2(q-1) ). 3. **Observations**: - Odds beyond 25 do not generate any usable pairs. - Similarly, 17, 19, and 23 do not generate anything. - Evens above 24 only generate usable pairs of the form ( (a, a) ). - Numbers like 2, 3, 14, 20, and 22 also generate pairs of the form ( (a, a) ). - 4 generates only ( (4, 4) ) and ( (4, 12) ). 4. **Counting Elements**: - We need to count the elements that do not generate usable pairs. - Odds beyond 25: 13 elements (27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49) - 17, 19, 23: 3 elements - Evens above 24: 13 elements (26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50) - 2, 3, 14, 20, 22: 5 elements - 4: 1 element - 11, 13: 2 elements Total: ( 13 + 3 + 13 + 5 + 1 + 2 = 37 ) 5. **Adding One for 1**: Adding one for the number 1, we get ( 37 + 1 = 38 ). 6. **Conclusion**: Therefore, the smallest ( K ) such that every ( K )-element subset of ( {1, 2, ldots, 50} ) contains two distinct elements ( a ) and ( b ) such that ( a + b mid ab ) is ( 39 ). The final answer is ( boxed{39} ).

answer:1. **Identify Collinear Sets**: - **Horizontal Lines**: Each of the 4 rows can have 1 set of 4 collinear dots. - **Vertical Lines**: Each of the 5 columns can have 1 set of 4 collinear dots. - **Diagonal Lines**: In a 4x5 grid, there are no sets of 4 collinear dots that cover the diagonals because the minimum diagonal that can have 4 dots must be in a square or nearly square (like 5x5 or 4x4) arrangement. Total collinear sets = 4 (horizontal) + 5 (vertical) = 9. 2. **Calculate Total Combinations**: [ {20 choose 4} = frac{20 cdot 19 cdot 18 cdot 17}{4 cdot 3 cdot 2 cdot 1} = 4845 ] 3. **Compute Probability**: [ text{Probability} = frac{9}{4845} ] Simplifying the fraction: [ text{GCD of 9 and 4845 is 1, so, Probability} = boxed{frac{9}{4845}} ]

answer:To simplify the given expression, notice the pattern of sums of powers. Using the identity for the difference of squares: [ (a^n - b^n)(a^n + b^n) = a^{2n} - b^{2n}, ] we multiply the expression by 1, expressed as (3-4) = -1, converting the sums into a difference: [ (3+4)(3^2+4^2)(3^4+4^4)(3^8+4^8)(3^{16}+4^{16})(3^{32}+4^{32})(3^{64}+4^{64}) times (3-4) ] Applying the difference of squares: - Start with (3-4)(3+4): [ (3-4)(3+4) = 3^2 - 4^2 ] - Multiply (3^2 - 4^2)(3^2 + 4^2): [ (3^2 - 4^2)(3^2 + 4^2) = 3^4 - 4^4 ] - Continue this process, increasing the exponents: [ (3^4 - 4^4)(3^4 + 4^4) = 3^8 - 4^8 ] [ (3^8 - 4^8)(3^8 + 4^8) = 3^{16} - 4^{16} ] [ (3^{16} - 4^{16})(3^{16} + 4^{16}) = 3^{32} - 4^{32} ] [ (3^{32} - 4^{32})(3^{32} + 4^{32}) = 3^{64} - 4^{64} ] [ (3^{64} - 4^{64})(3^{64} + 4^{64}) = 3^{128} - 4^{128} ] Thus, the simplified expression is: [ 3^{128 - 4^{128}} ] Conclusion: The process of applying the difference of squares simplifies the expression to 3^{128} - 4^{128}. Each step correctly follows the mathematical rules ensuring the solution's validity. The final answer is boxed{C) ~ 3^{128} - 4^{128}}

answer:1. We start with the given sequence (a_1 = k+1) and the recurrence relation (a_{n+1} = a_n^2 - ka_n + k). 2. We need to show that (a_m) and (a_n) are coprime for (m neq n). Assume (d) is a common divisor of (a_m) and (a_n) for some (m neq n). We will show that (d = 1). 3. Consider the recurrence relation: [ a_{n+1} = a_n^2 - ka_n + k ] Rearrange it to: [ a_{n+1} - k = a_n^2 - ka_n + k - k = a_n(a_n - k) ] 4. Suppose (d) divides both (a_m) and (a_n). Then (d) must also divide (a_{n+1} - k): [ d mid a_{n+1} - k = a_n(a_n - k) ] 5. By induction, we assume (d) divides (a_{n-1}). Then: [ d mid a_n(a_n - k) ] Since (d) divides (a_n), it must also divide (a_n - k). Therefore, (d) divides (a_n(a_n - k)). 6. By continuing this process, we see that (d) must divide (a_1 = k + 1). 7. Since (a_1 = k + 1) and (d) divides (a_1), we have: [ d mid k + 1 ] 8. Since (d) is a common divisor of (a_m) and (a_n) for any (m neq n), and (d) divides (k + 1), we conclude that (d) must be 1. This is because the only common divisor of all terms in the sequence is 1. 9. Therefore, (a_m) and (a_n) are coprime for (m neq n). (blacksquare)