answer:# Problem: Solve in natural numbers the equation [ 2^{x} + 5^{y} + 63 = z! ] where ( z! ) denotes the factorial of ( z ), that is, the product of all natural numbers from 1 to ( z ). 1. First, we observe that ( z! > 63 ). Since the factorial of any number ( z ) greater than or equal to 5 is greater than 63, we have: [ z geq 5 ] 2. Notice that ( z! ) is divisible by 3 and 5 for ( z geq 5 ): - Since ( z! ) must include the terms 3 and 5 in its product, ( 2^x + 5^y + 63 ) is also divisible by 5. - Therefore, we look at the equation modulo 5: [ 2^x + 5^y + 63 equiv 0 pmod{5} ] - Since ( 63 equiv 3 pmod{5} ), we can reduce our equation modulo 5 to: [ 2^x equiv 2 pmod{5} ] - This implies: [ 2^x equiv 2 pmod{5} ] - We know that the powers of 2 modulo 5 cycle every 4 terms: ( 2, 4, 3, 1 ). Thus, ( 2^x equiv 2 pmod{5} ) if and only if ( x equiv 1 pmod{4} ). Therefore: [ x = 4k + 1 text{ for some integer } k ] 3. Next, consider that the left-hand side of our original equation must be congruent to ( -1 pmod{3} ): - Checking the equation modulo 3, where ( 63 equiv 0 pmod{3} ), we get: [ 2^x + 5^y + 63 equiv 0 pmod{3} ] [ 2^x + (2)^y equiv 0 pmod{3} ] Since ( 5 equiv 2 pmod{3} ). - Therefore: [ 2^x + 2^y equiv 0 pmod{3} ] - Examining the powers of 2 modulo 3, we note that: ( 2^1 equiv 2 ), ( 2^2 equiv 1 ), ( 2^3 equiv 2 ), ( 2^4 equiv 1 ), etc. - Hence, ( 2^x + 2^y equiv 0 pmod{3} ) implies: [ (-1)^x + (-1)^y = (-1)^y - 1 equiv 0 pmod{3} ] - This requires ( y ) to be even: [ y = 2n text{ for some integer } n ] 4. Now, let’s consider the possible values of ( z ): - If ( z = 5 ), then: [ 2^x + 5^y + 63 = 120 implies 2^x + 25^n = 57 ] The only possible value for ( n ) is 1 (because for ( n geq 2 ), ( 25^n ) would be too large). Solving for ( x ): [ 2^x = 32 implies x = 5 ] Therefore, one solution is: [ (x, y, z) = (5, 2, 5) ] - If ( z = 6 ), then: [ 2^x + 5^y + 63 = 720 implies 2^x + 25^n = 657 ] Here, admissible values for ( n ) are only 1 and 2. When ( n = 2 ), we have: [ 2^x = 625 implies x = 4 ] Therefore, the second solution is: [ (x, y, z) = (4, 4, 6) ] # Conclusion: The solutions to the equation ( 2^{x} + 5^{y} + 63 = z! ) in natural numbers are: [ boxed{(5, 2, 5), (4, 4, 6)} ]

answer:1. **Define the labels and cumulative sums:** Let ( r_1, r_2, ldots, r_N ) be the labels of the red cards and ( b_1, b_2, ldots, b_N ) be the labels of the blue cards. Define the cumulative sums: [ x_i = sum_{j=1}^i r_j quad text{for} quad 1 le i le N ] [ y_i = sum_{j=1}^i b_j quad text{for} quad 1 le i le N ] 2. **Construct the set of points ( S ):** Consider the set of points ( S = {0, 1, 2, ldots, N} times {0, 1, 2, ldots, N} ). For each point ((a, b) in S), label it with ( x_a + y_b ). 3. **Apply the Pigeonhole Principle:** We claim that some two points in ( S ) have the same label. To show this, we will use the Pigeonhole Principle. 4. **Define the sets ( S_i ):** For each ( 0 le i le N ), let ( S_i ) be the set of points ((a, b) in S) such that ( max(|a-N|, |b|) = i ). This means ( S_i ) contains points on the boundary of the square grid at distance ( i ) from the point ((N, 0)). 5. **Label the points and apply discrete continuity:** Let ( ell ) be the label of the point ((N, 0)), which is ( x_N + y_0 = x_N ). For every ( 0 le i le N ), note that since ((N-i, 0)) and ((N, i)) are both in ( S_i ), by "discrete continuity" there must exist some ((a, b) in S_i) such that the label of ((a, b)) is in the interval ([ell, ell + N)). 6. **Count the distinct labels:** Since the ( S_i )'s are mutually disjoint and there are ( N+1 ) of them, we now know the existence of ( N+1 ) distinct points in ( S ) which all have labels in ([ell, ell+N)). By the Pigeonhole Principle, there must be at least two points in ( S ) with the same label. 7. **Identify the points with the same label:** Hence, we can pick ((a, b), (c, d) in S) such that ( a < c ) and ( d < b ) and ((a, b)) and ((c, d)) have the same label. 8. **Derive the subset sums:** By the definition of our table, we have: [ x_c + y_d = x_a + y_b ] Rearranging, we get: [ x_c - x_a = y_b - y_d ] This implies that the sum of the numbers on the chosen red cards (from ( a+1 ) to ( c )) equals the sum of the numbers on the chosen blue cards (from ( d+1 ) to ( b )). Therefore, we have shown that it is possible to choose a (non-empty) subset of the red cards and a (non-empty) subset of the blue cards such that the sum of the numbers on the chosen red cards equals the sum of the numbers on the chosen blue cards. (blacksquare)

answer:To find the equation of the line passing through the points (-1, 1) and (0, 3), we use the formula frac {y-3}{1-3}= frac {x-0}{-1-0}, which simplifies to y=2x+3. Setting y=0, we solve for x to get x=-frac {3}{2}. Therefore, the x-intercept of the line is -frac {3}{2}. So, the answer is -frac {3}{2}. This problem primarily tests the ability to derive the equation of a line using two points and to calculate intercepts, which is quite basic. Thus, the x-intercept of the line is boxed{-frac{3}{2}}.

answer:Solution: (Ⅰ) When a = -2, f(x) = ln x - 2x^2 + 3x, Therefore, f'(x) = frac{1}{x} - 4x + 3 = -frac{(x-1)(4x+1)}{x}, Therefore, the function is increasing in the interval (0, 1) and decreasing in the interval (1, +infty), Therefore, when x=1, the function reaches its maximum value of boxed{1}. (Ⅱ) Since f'(x) = frac{(2ax-1)(x-1)}{x}, For a leq 0, the function is decreasing in the interval (0, 1) and increasing in the interval (1, +infty), Since f(x) has only one zero in the interval (0, e), Therefore, f(e) leq 0, thus 1 + ae^2 - (2a+1)e leq 0, Therefore, a leq frac{e-1}{e(e-2)}, For frac{1}{2} > a > 0, let f'(x) = 0, x_1 = 1, x_2 = frac{1}{2a} > 1 Since f(1) < 0 and f(x) has only one zero in the interval (0, e), Therefore, f(e) geq 0, thus 1 + ae^2 - (2a+1)e geq 0, Therefore, a geq frac{e-1}{e(e-2)}, thus frac{1}{2} > a > 0, In summary, a < boxed{frac{1}{2}}.