answer：First, identify the coordinates of point ( A ), which are ( (-2, 1) ).The ( y )-axis is the line where the ( x )-coordinate is 0.To find the distance from point ( A ) to the ( y )-axis, take the absolute value of the ( x )-coordinate of point ( A ).Calculate the absolute value: ( | -2 | = 2 ).Therefore, the distance from point ( A(-2, 1) ) to the ( y )-axis is 2 units.

answer：First, I need to determine the breadth of the rectangular plot. The problem states that the length is 50 meters more than the breadth, and the length is given as 75 meters.So, I can set up the equation:75 = breadth + 50Solving for breadth:breadth = 75 - 50breadth = 25 metersNext, I'll calculate the perimeter of the plot using the formula for the perimeter of a rectangle:Perimeter = 2 × (length + breadth)Perimeter = 2 × (75 + 25)Perimeter = 2 × 100Perimeter = 200 metersFinally, to find the total cost of fencing, I'll multiply the perimeter by the cost per meter:Total Cost = Perimeter × Cost per meterTotal Cost = 200 meters × 26.50Total Cost = 5300Therefore, the total cost of fencing the plot is 5300 currency units.

answer：First, I need to determine the number of innings before the one where the batsman scored 85. Let's denote this number as ( n ).The batsman's average increased by 3 after scoring 85, so his average before this innings was ( 37 - 3 = 34 ).The total runs scored in ( n ) innings would be ( 34n ).After scoring 85 in the ( (n+1)^{th} ) innings, his average becomes 37. Therefore, the total runs scored after ( (n+1) ) innings would be ( 37(n+1) ).The difference between the total runs after ( (n+1) ) innings and the total runs before this innings is the score of the ( (n+1)^{th} ) innings, which is 85.Setting up the equation:[37(n+1) = 34n + 85]Solving for ( n ):[37n + 37 = 34n + 85][37n - 34n = 85 - 37][3n = 48][n = frac{48}{3} = 16]Therefore, the batsman scored 85 in his ( 16 + 1 = 17^{th} ) innings.

answer：Alright, so I have this problem where I need to find the smallest natural number that can be expressed as the sum of 2002 natural addends, each with the same sum of digits, and also as the sum of 2003 natural addends, each with the same sum of digits. Hmm, okay, let me try to unpack this step by step.First, let me make sure I understand the problem correctly. I need a number ( n ) such that:1. ( n ) can be written as the sum of 2002 numbers, say ( a_1, a_2, ldots, a_{2002} ), where each ( a_i ) has the same sum of digits. Let's denote this common sum of digits as ( r ).2. Similarly, ( n ) can also be written as the sum of 2003 numbers, say ( b_1, b_2, ldots, b_{2003} ), where each ( b_i ) has the same sum of digits. Let's denote this common sum of digits as ( s ).So, ( n = a_1 + a_2 + ldots + a_{2002} ) and ( n = b_1 + b_2 + ldots + b_{2003} ).Now, I recall that the sum of the digits of a number is congruent to the number modulo 9. This is because 10 is congruent to 1 modulo 9, so each digit contributes its face value times 1 to the modulo 9 result. Therefore, the sum of the digits of a number is congruent to the number itself modulo 9.Given this, each ( a_i ) is congruent to ( r ) modulo 9, and each ( b_i ) is congruent to ( s ) modulo 9. Therefore, the sum ( n ) must satisfy:1. ( n equiv 2002r mod 9 )2. ( n equiv 2003s mod 9 )So, both expressions must be equal modulo 9:( 2002r equiv 2003s mod 9 )Now, let me compute 2002 and 2003 modulo 9 to simplify this equation.Calculating 2002 modulo 9:2002 divided by 9: 9*222=1998, so 2002-1998=4. Therefore, 2002 ≡ 4 mod 9.Similarly, 2003 divided by 9: 2003-1998=5. So, 2003 ≡ 5 mod 9.Substituting these into the equation:4r ≡ 5s mod 9So, 4r - 5s ≡ 0 mod 9.I need to find integers r and s (each between 0 and 8, since they are sums of digits modulo 9) such that 4r ≡ 5s mod 9.Let me list possible values for r and s:r can be 0,1,2,3,4,5,6,7,8Similarly, s can be 0,1,2,3,4,5,6,7,8I need to find pairs (r,s) such that 4r ≡ 5s mod 9.Let me compute 4r for each r:r=0: 0r=1:4r=2:8r=3:12≡3r=4:16≡7r=5:20≡2r=6:24≡6r=7:28≡1r=8:32≡5Similarly, compute 5s for each s:s=0:0s=1:5s=2:10≡1s=3:15≡6s=4:20≡2s=5:25≡7s=6:30≡3s=7:35≡8s=8:40≡4Now, I need to find pairs where 4r ≡ 5s mod9.Looking for matching residues:Looking at 4r:0,4,8,3,7,2,6,1,5Looking at 5s:0,5,1,6,2,7,3,8,4So, let's see where they match:For r=0: 4r=0. 5s=0 when s=0. So (0,0) is a solution.r=1:4. 5s=4 when s=8. So (1,8)r=2:8. 5s=8 when s=7. So (2,7)r=3:3. 5s=3 when s=6. So (3,6)r=4:7. 5s=7 when s=5. So (4,5)r=5:2. 5s=2 when s=4. So (5,4)r=6:6. 5s=6 when s=3. So (6,3)r=7:1. 5s=1 when s=2. So (7,2)r=8:5. 5s=5 when s=1. So (8,1)So, all possible pairs (r,s) are:(0,0), (1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1)Now, for each of these pairs, we can find n such that:n ≡ 2002r ≡ 2003s mod9, which is satisfied by the pairs above.But we need to find the smallest n that can be expressed both as 2002 numbers each with digit sum r and as 2003 numbers each with digit sum s.So, n must be at least the maximum of (2002 * r, 2003 * s). But since r and s are related via 4r ≡5s mod9, and r and s are between 0 and 8, we need to find the minimal n that is a multiple of both 2002 and 2003 in some way, but considering the digit sum constraints.Wait, actually, n must be equal to 2002 * k, where each a_i has digit sum r, and also equal to 2003 * m, where each b_i has digit sum s.But k and m are numbers such that the digit sums are r and s respectively.But n must be equal to both 2002 * k and 2003 * m, so n must be a common multiple of 2002 and 2003. However, 2002 and 2003 are consecutive integers, so they are coprime. Therefore, the least common multiple of 2002 and 2003 is 2002*2003.But that would make n=2002*2003, which is a very large number. However, the problem is asking for the smallest n that can be expressed in both ways, not necessarily that n is the least common multiple.Wait, perhaps I'm overcomplicating. Let me think differently.Each a_i has digit sum r, so each a_i is congruent to r mod9, so n=2002*a_i ≡2002*r mod9, which we already have.Similarly, n=2003*b_i ≡2003*s mod9.But n must satisfy both congruences, which we have already considered.Now, to find the minimal n, we need to find the minimal n such that n is a multiple of 2002 and 2003 in terms of their digit sum constraints.But perhaps a better approach is to consider the minimal n that is a multiple of both 2002 and 2003, but considering the digit sum constraints.Wait, but n must be expressible as 2002 numbers each with digit sum r, and as 2003 numbers each with digit sum s.So, n must be at least 2002*r and 2003*s.But since r and s are related by 4r ≡5s mod9, and r and s are between 0 and 8, we can find the minimal n by choosing the smallest possible r and s that satisfy the congruence and then compute n as the least common multiple or something similar.Wait, perhaps it's better to consider that n must be a multiple of both 2002 and 2003, but adjusted for the digit sum constraints.Alternatively, perhaps n must be such that n is divisible by both 2002 and 2003 in terms of their digit sum properties.Wait, maybe I should look for the minimal n such that n is divisible by 2002 and 2003, but considering the digit sum conditions.But 2002 and 2003 are coprime, so their least common multiple is 2002*2003, which is 2002*2003=4,006,006. That's a very large number, but perhaps the minimal n is smaller because of the digit sum constraints.Wait, perhaps the minimal n is the least common multiple of 2002 and 2003 divided by some factor related to the digit sums.Alternatively, perhaps n must be a multiple of both 2002 and 2003, but considering the digit sum constraints, which might allow for a smaller n.Wait, let me think differently. Since each a_i has digit sum r, and each b_i has digit sum s, and n=2002*a_i=2003*b_i, then n must be a multiple of both 2002 and 2003, but the digit sums r and s must satisfy 4r ≡5s mod9.So, perhaps the minimal n is the least common multiple of 2002 and 2003, but adjusted for the digit sum constraints.But 2002 and 2003 are coprime, so their LCM is 2002*2003=4,006,006.But maybe we can find a smaller n by choosing appropriate r and s.Wait, let's consider the pairs (r,s) we found earlier:(0,0), (1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1)For each pair, we can compute the minimal n that is a multiple of both 2002 and 2003, but with the digit sums r and s.But since 2002 and 2003 are coprime, the minimal n that is a multiple of both is 2002*2003=4,006,006.But perhaps for some pairs (r,s), n can be smaller because the digit sums allow for smaller numbers.Wait, for example, if r=0 and s=0, then n=0, but we need a natural number, so n=0 is not allowed. So the next possible is r=1, s=8.Wait, but n must be at least 2002*1=2002 and 2003*8=16,024. So n must be at least 16,024.But 16,024 is less than 4,006,006, so maybe that's a candidate.Wait, but n must be expressible as 2002 numbers each with digit sum 1, and as 2003 numbers each with digit sum 8.Is 16,024 expressible as 2002 numbers each with digit sum 1? Well, each a_i would be 1, since the digit sum is 1, so n=2002*1=2002, but 2002≠16,024. So that's a problem.Wait, so perhaps I'm misunderstanding. If each a_i has digit sum r, then a_i can be any number with digit sum r, not necessarily r itself.So, for example, if r=1, then a_i could be 1, 10, 100, etc., as long as their digit sum is 1.Similarly, for s=8, b_i could be 8, 17, 26, etc., as long as their digit sum is 8.Therefore, n=2002*a_i, where each a_i has digit sum r, and n=2003*b_i, where each b_i has digit sum s.So, n must be a multiple of 2002 and 2003, but the digit sums r and s must satisfy 4r≡5s mod9.Wait, but n must be the same in both cases, so n must be a multiple of both 2002 and 2003, but also satisfy the digit sum conditions.So, perhaps the minimal n is the least common multiple of 2002 and 2003, which is 4,006,006, but adjusted for the digit sums.But maybe there's a smaller n that satisfies the digit sum conditions.Wait, let's consider the pair (r=5, s=4). Then, n must be ≡2002*5 mod9 and ≡2003*4 mod9.But 2002≡4 mod9, so 4*5=20≡2 mod9.Similarly, 2003≡5 mod9, so 5*4=20≡2 mod9.So, n≡2 mod9.Now, n must be a multiple of both 2002 and 2003, so n=2002*2003*k, where k is a positive integer.But 2002*2003=4,006,006, which is already quite large.Wait, but maybe n can be smaller if we choose k=1, but n must be expressible as 2002 numbers each with digit sum 5 and as 2003 numbers each with digit sum 4.Is 4,006,006 expressible in both ways?Well, 4,006,006 divided by 2002 is 2003, so each a_i=2003, which has digit sum 2+0+0+3=5. So that works.Similarly, 4,006,006 divided by 2003 is 2002, which has digit sum 2+0+0+2=4. So that also works.Therefore, n=4,006,006 is expressible as both 2002 numbers each with digit sum 5 and as 2003 numbers each with digit sum 4.But is this the minimal n?Wait, perhaps there's a smaller n that satisfies the conditions.Let me check for smaller k. If k=1, n=4,006,006.If k=0, n=0, which is not a natural number.So, perhaps 4,006,006 is the minimal n.But wait, earlier I considered the pair (r=5, s=4) leading to n≡2 mod9, and n=4,006,006 is indeed 4,006,006 mod9.Let me check 4,006,006 mod9:Sum the digits: 4+0+0+6+0+0+6=16, 1+6=7. So 4,006,006≡7 mod9.But earlier, we had n≡2 mod9 for r=5, s=4. Wait, that's a contradiction.Wait, no, let me recalculate.Wait, n=4,006,006.Sum of digits: 4+0+0+6+0+0+6=16, 1+6=7. So n≡7 mod9.But earlier, for r=5, s=4, we had n≡2 mod9.This suggests a mistake in my reasoning.Wait, let's go back.We had 4r≡5s mod9.For r=5, s=4:4*5=20≡2 mod95*4=20≡2 mod9So, n≡2 mod9.But n=4,006,006≡7 mod9, which contradicts.Therefore, n=4,006,006 cannot be expressed as 2002 numbers each with digit sum 5 and 2003 numbers each with digit sum 4, because it doesn't satisfy the congruence.Wait, that's a problem. So my earlier conclusion was wrong.So, perhaps n must be a multiple of 2002 and 2003, but also satisfy n≡2 mod9.So, n=2002*2003*k, and n≡2 mod9.We need to find the smallest k such that 2002*2003*k≡2 mod9.First, compute 2002*2003 mod9.2002≡4 mod9, 2003≡5 mod9.So, 4*5=20≡2 mod9.Therefore, 2002*2003≡2 mod9.So, n=2002*2003*k≡2k mod9.We need 2k≡2 mod9.So, 2k≡2 mod9 ⇒ k≡1 mod9.Therefore, the smallest k is 1.Thus, n=2002*2003*1=4,006,006.But earlier, we saw that 4,006,006≡7 mod9, which contradicts n≡2 mod9.Wait, that can't be. There must be a mistake in my calculation.Wait, let me recalculate 4,006,006 mod9.4+0+0+6+0+0+6=16, 1+6=7. So 4,006,006≡7 mod9.But according to the earlier calculation, n=2002*2003*k≡2k mod9.If k=1, n≡2 mod9, but 4,006,006≡7 mod9.This inconsistency suggests an error in my reasoning.Wait, perhaps I made a mistake in calculating 2002*2003 mod9.Let me recalculate:2002 divided by 9: 9*222=1998, 2002-1998=4, so 2002≡4 mod9.2003=2002+1≡4+1=5 mod9.So, 2002*2003≡4*5=20≡2 mod9.Therefore, 2002*2003≡2 mod9.Thus, n=2002*2003*k≡2k mod9.We need n≡2 mod9, so 2k≡2 mod9 ⇒ k≡1 mod9.Therefore, the smallest k is 1, so n=2002*2003=4,006,006.But 4,006,006≡7 mod9, which contradicts n≡2 mod9.This suggests that my earlier assumption that n=2002*2003*k is incorrect.Wait, perhaps n is not necessarily a multiple of both 2002 and 2003, but rather, n must be expressible as 2002 numbers each with digit sum r and as 2003 numbers each with digit sum s, where r and s satisfy 4r≡5s mod9.Therefore, n must be equal to 2002*a_i and 2003*b_i, where a_i has digit sum r and b_i has digit sum s.Thus, n must be a multiple of both 2002 and 2003, but considering the digit sum constraints.Wait, but 2002 and 2003 are coprime, so n must be a multiple of their least common multiple, which is 2002*2003=4,006,006.But as we saw, 4,006,006≡7 mod9, which contradicts n≡2 mod9 for r=5, s=4.Therefore, perhaps there's no solution with r=5, s=4.Wait, but earlier we had n=4,006,006≡7 mod9, but we needed n≡2 mod9.Therefore, perhaps we need to choose a different pair (r,s) where n≡2 mod9.Wait, let's look back at the pairs:(0,0): n≡0 mod9(1,8): n≡4*1=4≡4 mod9 and n≡5*8=40≡4 mod9. So n≡4 mod9.(2,7): n≡4*2=8≡8 mod9 and n≡5*7=35≡8 mod9. So n≡8 mod9.(3,6): n≡4*3=12≡3 mod9 and n≡5*6=30≡3 mod9. So n≡3 mod9.(4,5): n≡4*4=16≡7 mod9 and n≡5*5=25≡7 mod9. So n≡7 mod9.(5,4): n≡4*5=20≡2 mod9 and n≡5*4=20≡2 mod9. So n≡2 mod9.(6,3): n≡4*6=24≡6 mod9 and n≡5*3=15≡6 mod9. So n≡6 mod9.(7,2): n≡4*7=28≡1 mod9 and n≡5*2=10≡1 mod9. So n≡1 mod9.(8,1): n≡4*8=32≡5 mod9 and n≡5*1=5≡5 mod9. So n≡5 mod9.So, for each pair (r,s), n must be ≡ a certain value mod9.Now, for the pair (5,4), n≡2 mod9.But 2002*2003=4,006,006≡7 mod9, which doesn't match.Therefore, perhaps n must be 4,006,006 plus some multiple of 9 to make it ≡2 mod9.So, 4,006,006≡7 mod9. We need n≡2 mod9. So, 7 + k*9 ≡2 mod9 ⇒ k≡-5≡4 mod9.Therefore, the smallest n is 4,006,006 + 4*9=4,006,006 +36=4,006,042.But now, n=4,006,042.Is this expressible as 2002 numbers each with digit sum 5 and as 2003 numbers each with digit sum 4?Well, 4,006,042 divided by 2002 is 2003.000... something? Wait, 2002*2003=4,006,006, so 4,006,042-4,006,006=36.So, 4,006,042=2002*2003 +36.Wait, but 36 is 2002* something? No, 36 is less than 2002.Wait, perhaps I'm approaching this incorrectly.Alternatively, perhaps n must be a multiple of 2002 and 2003, but adjusted to satisfy the congruence.Wait, but 2002 and 2003 are coprime, so n must be a multiple of 2002*2003.But if n=2002*2003*k, then n must be ≡2k mod9.We need n≡2 mod9, so 2k≡2 mod9 ⇒k≡1 mod9.Thus, the smallest k is 1, so n=2002*2003=4,006,006.But as we saw, 4,006,006≡7 mod9, which contradicts n≡2 mod9.This suggests that there is no solution with k=1, but perhaps with k=10, which is 1 mod9.Wait, k=10: n=2002*2003*10=40,060,060.40,060,060 mod9: sum the digits:4+0+0+6+0+0+6+0=16≡7 mod9.Still 7 mod9, which is not 2 mod9.Wait, this suggests that n=2002*2003*k≡2k mod9, and we need 2k≡2 mod9 ⇒k≡1 mod9.But 2002*2003≡2 mod9, so n=2002*2003*k≡2k mod9.Thus, for k=1, n≡2 mod9.But when we compute 2002*2003=4,006,006, which is 7 mod9, not 2 mod9.This inconsistency suggests that my initial assumption that n=2002*2003*k is incorrect.Wait, perhaps n is not necessarily a multiple of both 2002 and 2003, but rather, n must be expressible as 2002 numbers each with digit sum r and as 2003 numbers each with digit sum s, where r and s satisfy 4r≡5s mod9.Therefore, n must be equal to 2002*a_i and 2003*b_i, where a_i has digit sum r and b_i has digit sum s.Thus, n must be a multiple of both 2002 and 2003, but considering the digit sum constraints.Wait, but 2002 and 2003 are coprime, so n must be a multiple of their least common multiple, which is 2002*2003=4,006,006.But as we saw, 4,006,006≡7 mod9, which contradicts n≡2 mod9 for r=5, s=4.Therefore, perhaps there's no solution with r=5, s=4.Wait, but earlier we had n=4,006,006≡7 mod9, which doesn't match n≡2 mod9.Therefore, perhaps we need to choose a different pair (r,s) where n≡7 mod9.Looking back at the pairs:(4,5): n≡7 mod9.So, for r=4, s=5, n≡7 mod9.Thus, n=4,006,006≡7 mod9, which matches.Therefore, n=4,006,006 can be expressed as 2002 numbers each with digit sum 4 and as 2003 numbers each with digit sum 5.Wait, let's check:n=4,006,006.Divide by 2002: 4,006,006 /2002=2003.So, each a_i=2003, which has digit sum 2+0+0+3=5. Wait, but we needed digit sum 4.Hmm, that's a problem.Wait, no, if r=4, then each a_i must have digit sum 4, but 2003 has digit sum 5.Therefore, this doesn't work.Wait, perhaps I made a mistake in choosing r and s.Wait, for the pair (r=4, s=5), n≡7 mod9, which matches n=4,006,006≡7 mod9.But n=4,006,006=2002*2003.So, each a_i=2003, which has digit sum 5, but we needed r=4.This suggests that it's not possible to express n=4,006,006 as 2002 numbers each with digit sum 4.Therefore, perhaps n must be larger.Wait, perhaps n must be a multiple of 2002 and 2003, but also satisfy the digit sum conditions.Wait, but 2002 and 2003 are coprime, so n must be a multiple of 2002*2003.But as we saw, n=4,006,006 cannot be expressed as 2002 numbers each with digit sum 4, because each a_i would have to be 2003, which has digit sum 5.Therefore, perhaps we need to choose a different pair (r,s) where n can be expressed as 2002 numbers each with digit sum r and 2003 numbers each with digit sum s.Wait, let's consider the pair (r=5, s=4), which requires n≡2 mod9.But n=4,006,006≡7 mod9, which doesn't match.Therefore, perhaps we need to find the smallest multiple of 2002*2003 that is ≡2 mod9.So, n=2002*2003*k≡2k mod9.We need 2k≡2 mod9 ⇒k≡1 mod9.Thus, the smallest k is 1, but n=4,006,006≡7 mod9, which doesn't work.Therefore, the next k is 10, which is 1 mod9.n=2002*2003*10=40,060,060.40,060,060 mod9: sum the digits:4+0+0+6+0+0+6+0=16≡7 mod9.Still 7 mod9, which is not 2 mod9.Wait, this suggests that no multiple of 2002*2003 will satisfy n≡2 mod9, because 2002*2003≡2 mod9, so n=2002*2003*k≡2k mod9.We need 2k≡2 mod9 ⇒k≡1 mod9.But 2002*2003≡2 mod9, so n=2002*2003*k≡2k mod9.Thus, for k=1, n≡2 mod9.But when we compute 2002*2003=4,006,006, which is 7 mod9, not 2 mod9.This suggests that there's a mistake in my earlier calculation.Wait, let me recalculate 2002*2003 mod9.2002≡4 mod9, 2003≡5 mod9.4*5=20≡2 mod9.Therefore, 2002*2003≡2 mod9.Thus, n=2002*2003*k≡2k mod9.Therefore, for k=1, n≡2 mod9.But when I compute 4,006,006 mod9, I get 7.Wait, this is a contradiction.Wait, perhaps I made a mistake in calculating 4,006,006 mod9.Let me recalculate:4,006,006.Digits: 4,0,0,6,0,0,6.Sum:4+0+0+6+0+0+6=16.16 mod9: 1+6=7.So, 4,006,006≡7 mod9.But according to the earlier calculation, 2002*2003≡2 mod9.This suggests that 2002*2003≡2 mod9, but 4,006,006≡7 mod9.This is a contradiction.Therefore, there must be a mistake in my reasoning.Wait, perhaps I made a mistake in calculating 2002*2003 mod9.Wait, 2002≡4 mod9, 2003≡5 mod9.4*5=20≡2 mod9.So, 2002*2003≡2 mod9.But 4,006,006≡7 mod9.This suggests that 2002*2003≡2 mod9, but 4,006,006≡7 mod9.This is impossible because 2002*2003=4,006,006.Therefore, my earlier calculation must be wrong.Wait, let me check 2002*2003.2002*2003=2002*(2000+3)=2002*2000 +2002*3=4,004,000 +6,006=4,010,006.Wait, no, that's incorrect.Wait, 2002*2003=2002*(2000+3)=2002*2000 +2002*3=4,004,000 +6,006=4,010,006.But 4,010,006 mod9:Sum the digits:4+0+1+0+0+0+6=11, 1+1=2.So, 4,010,006≡2 mod9.Ah, so I must have made a mistake earlier in calculating 2002*2003.Yes, 2002*2003=4,010,006, not 4,006,006.Therefore, n=4,010,006≡2 mod9, which matches the requirement for r=5, s=4.Therefore, n=4,010,006 is the minimal n that satisfies the conditions.But let me verify:n=4,010,006.Divide by 2002: 4,010,006 /2002=2003.Each a_i=2003, which has digit sum 2+0+0+3=5. So, r=5.Similarly, divide by 2003: 4,010,006 /2003=2002.Each b_i=2002, which has digit sum 2+0+0+2=4. So, s=4.Therefore, n=4,010,006 can be expressed as 2002 numbers each with digit sum 5 and as 2003 numbers each with digit sum 4.Thus, n=4,010,006 is the minimal such number.But wait, is there a smaller n?Let me check if there's a smaller n that satisfies the conditions.Since 2002 and 2003 are coprime, the minimal n must be their least common multiple, which is 2002*2003=4,010,006.Therefore, n=4,010,006 is indeed the minimal number.But wait, earlier I thought 2002*2003=4,006,006, but that was incorrect. The correct product is 4,010,006.Therefore, the minimal n is 4,010,006.But let me check if there's a smaller n by choosing a different pair (r,s).For example, let's consider the pair (r=1, s=8), which requires n≡4 mod9.Is there a smaller n than 4,010,006 that satisfies n≡4 mod9 and is expressible as 2002 numbers each with digit sum 1 and as 2003 numbers each with digit sum 8.But n must be at least 2002*1=2002 and 2003*8=16,024.So, n must be at least 16,024.But 16,024 is much smaller than 4,010,006.But can n=16,024 be expressed as 2002 numbers each with digit sum 1 and as 2003 numbers each with digit sum 8?Wait, n=16,024.Divide by 2002: 16,024 /2002≈8.004.So, each a_i≈8.004, but a_i must be integers with digit sum 1.But 8.004 is not an integer, so n=16,024 cannot be expressed as 2002 integers each with digit sum 1.Therefore, n must be a multiple of 2002 and 2003, so the minimal n is 4,010,006.Thus, the smallest natural number n is 4,010,006.

answer：Okay, so I have this function f(x) = (1/3)x³ - (1/2)a x² + (a - 1)x + 1, and I need to find the range of a such that the function is monotonically decreasing on the interval (1, 4) and monotonically increasing on the interval (6, +∞). Hmm, okay. Let me try to figure this out step by step.First, I remember that to determine where a function is increasing or decreasing, I need to look at its first derivative. So, let me find f'(x). The derivative of (1/3)x³ is x², right? Then, the derivative of -(1/2)a x² is -a x. Next, the derivative of (a - 1)x is (a - 1), and the derivative of the constant 1 is 0. So putting it all together, f'(x) = x² - a x + (a - 1). Let me write that down:f'(x) = x² - a x + (a - 1)Okay, so f'(x) is a quadratic function. Quadratics can be a bit tricky, but since the coefficient of x² is positive (1), the parabola opens upwards. That means it has a minimum point. So, the derivative will be positive outside the roots and negative between the roots.But wait, I need f(x) to be decreasing on (1, 4) and increasing on (6, +∞). So, f'(x) should be ≤ 0 on (1, 4) and ≥ 0 on (6, +∞). Let me think about what that means for the quadratic f'(x).Since f'(x) is a quadratic opening upwards, if it's negative on (1, 4), that interval must lie between the two roots of the quadratic. Similarly, for f'(x) to be positive on (6, +∞), the interval (6, +∞) must lie to the right of the larger root.So, let me denote the roots of f'(x) as r1 and r2, where r1 < r2. Then, for f'(x) ≤ 0 on (1, 4), we need 1 ≥ r1 and 4 ≤ r2. Wait, no, actually, since f'(x) is negative between r1 and r2, to have f'(x) ≤ 0 on (1, 4), the interval (1, 4) must be entirely within (r1, r2). So, that would mean r1 ≤ 1 and r2 ≥ 4.Similarly, for f'(x) ≥ 0 on (6, +∞), since the quadratic opens upwards, f'(x) is positive when x is greater than r2. So, to have f'(x) ≥ 0 on (6, +∞), we need r2 ≤ 6. Wait, no, that can't be right. If r2 is the larger root, then f'(x) is positive when x > r2. So, if we want f'(x) ≥ 0 on (6, +∞), we need r2 ≤ 6. Because if r2 is less than or equal to 6, then for all x > 6, f'(x) is positive.But wait, hold on. If r2 is the larger root, then for x > r2, f'(x) is positive. So, if we want f'(x) ≥ 0 on (6, +∞), we need r2 ≤ 6. Because if r2 is less than or equal to 6, then for all x > 6, which is greater than r2, f'(x) is positive. Conversely, if r2 were greater than 6, then f'(x) would be negative between r2 and infinity, which is not what we want. So, yes, r2 must be ≤ 6.But earlier, we had that r2 must be ≥ 4 because (1, 4) is between r1 and r2. So, combining these two, r2 must be between 4 and 6. So, 4 ≤ r2 ≤ 6.Similarly, for r1, since (1, 4) is between r1 and r2, and we want f'(x) ≤ 0 on (1, 4), which is within (r1, r2). So, r1 must be ≤ 1, because otherwise, if r1 were greater than 1, then f'(x) would be negative starting at r1, which is greater than 1, but we need it to be negative starting at 1. So, r1 ≤ 1.So, in summary, the roots of f'(x) must satisfy r1 ≤ 1 and 4 ≤ r2 ≤ 6.Now, let's find the roots of f'(x). The quadratic equation is x² - a x + (a - 1) = 0. Let's solve for x.Using the quadratic formula, x = [a ± sqrt(a² - 4*(1)*(a - 1))]/2Simplify the discriminant:sqrt(a² - 4(a - 1)) = sqrt(a² - 4a + 4) = sqrt((a - 2)²) = |a - 2|So, the roots are:x = [a ± |a - 2|]/2Hmm, okay. So, let's consider two cases based on the absolute value.Case 1: a - 2 ≥ 0, which implies a ≥ 2.Then, |a - 2| = a - 2, so the roots are:x = [a + (a - 2)]/2 = (2a - 2)/2 = a - 1andx = [a - (a - 2)]/2 = (2)/2 = 1So, in this case, the roots are x = 1 and x = a - 1.Case 2: a - 2 < 0, which implies a < 2.Then, |a - 2| = 2 - a, so the roots are:x = [a + (2 - a)]/2 = (2)/2 = 1andx = [a - (2 - a)]/2 = (2a - 2)/2 = a - 1Wait, that's the same as Case 1. So, regardless of whether a is greater than or equal to 2 or less than 2, the roots are x = 1 and x = a - 1.Interesting. So, the roots are always 1 and a - 1.Therefore, the quadratic f'(x) factors as (x - 1)(x - (a - 1)).So, f'(x) = (x - 1)(x - (a - 1)).So, the roots are x = 1 and x = a - 1.Given that, let's note that the quadratic opens upwards, so it's positive outside the interval (1, a - 1) and negative inside (1, a - 1).Wait, but earlier, I thought that the quadratic is negative between the roots. Since it opens upwards, yes, it's negative between the smaller root and the larger root.So, if a - 1 is greater than 1, which it is if a > 2, then the quadratic is negative between 1 and a - 1.If a - 1 is less than 1, which is when a < 2, then the quadratic is negative between a - 1 and 1.But in our case, we need f'(x) ≤ 0 on (1, 4). So, the interval (1, 4) must lie within the interval where f'(x) is negative.So, if a - 1 > 1, which is a > 2, then f'(x) is negative between 1 and a - 1. So, to have (1, 4) within (1, a - 1), we need a - 1 ≥ 4, which implies a ≥ 5.Alternatively, if a - 1 < 1, which is a < 2, then f'(x) is negative between a - 1 and 1. But since a - 1 < 1, the interval where f'(x) is negative is (a - 1, 1). However, we need f'(x) ≤ 0 on (1, 4). But in this case, f'(x) is positive on (1, +∞), which contradicts our requirement. So, a cannot be less than 2.Therefore, a must be greater than or equal to 5.Wait, but hold on. Let me think again. If a - 1 is greater than 1, so a > 2, then f'(x) is negative between 1 and a - 1. So, to have f'(x) ≤ 0 on (1, 4), we need a - 1 ≥ 4, so a ≥ 5.But also, we have the condition that f'(x) ≥ 0 on (6, +∞). Since f'(x) is positive outside (1, a - 1), so for x > a - 1, f'(x) is positive. So, to have f'(x) ≥ 0 on (6, +∞), we need a - 1 ≤ 6, so a ≤ 7.Therefore, combining both conditions, a must be between 5 and 7, inclusive.Wait, let me verify this.If a = 5, then f'(x) = (x - 1)(x - 4). So, the roots are at 1 and 4. So, f'(x) is negative between 1 and 4, which satisfies the decreasing condition on (1, 4). And for x > 4, f'(x) is positive, so on (6, +∞), it's positive, which satisfies the increasing condition.Similarly, if a = 7, then f'(x) = (x - 1)(x - 6). So, the roots are at 1 and 6. So, f'(x) is negative between 1 and 6, which includes (1, 4), so it's decreasing there. And for x > 6, f'(x) is positive, so it's increasing on (6, +∞).What if a is between 5 and 7, say a = 6. Then, f'(x) = (x - 1)(x - 5). So, roots at 1 and 5. So, f'(x) is negative between 1 and 5, which includes (1, 4), so decreasing there. And for x > 5, f'(x) is positive, so on (6, +∞), it's positive, hence increasing.Similarly, if a is 5.5, then f'(x) = (x - 1)(x - 4.5). So, negative between 1 and 4.5, which includes (1, 4), and positive beyond 4.5, so on (6, +∞), positive.Therefore, it seems that a must be between 5 and 7, inclusive.Wait, but let me check the endpoints. If a = 5, then f'(x) is zero at x = 4, but since the interval is open (1,4), it's fine. Similarly, at a = 7, f'(x) is zero at x = 6, but the interval is open (6, +∞), so it's okay.What if a is exactly 5? Then, f'(4) = 0, but since the interval is (1,4), which doesn't include 4, so f'(x) is ≤ 0 on [1,4], but since it's open, it's still decreasing on (1,4). Similarly, for a =7, f'(6)=0, but the interval is (6, +∞), so it's increasing there.Therefore, the range of a is [5,7].I think that makes sense. So, the answer should be a is between 5 and 7, inclusive.

answer：Okay, so I need to figure out how to write 3.56 as a simplified fraction. Hmm, I remember that decimals can be converted into fractions by looking at the place value. Let me think about this step by step.First, 3.56 has a whole number part, which is 3, and a decimal part, which is 0.56. Maybe I can separate them and work with each part individually. So, 3 is straightforward; it's just 3. Now, 0.56 is the part that needs to be converted into a fraction.I recall that the first place after the decimal is the tenths place, and the second is the hundredths place. So, 0.56 is 56 hundredths. That means I can write 0.56 as 56/100. Okay, so now I have 3 and 56/100.But I need to simplify this fraction. Simplifying a fraction means dividing both the numerator and the denominator by their greatest common divisor (GCD). So, what's the GCD of 56 and 100? Let me think. The factors of 56 are 1, 2, 4, 7, 8, 14, 28, 56, and the factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, 100. The common factors are 1, 2, and 4. The greatest one is 4.So, I'll divide both 56 and 100 by 4. 56 divided by 4 is 14, and 100 divided by 4 is 25. That means 56/100 simplifies to 14/25. Okay, so now I have 3 and 14/25.To write this as an improper fraction, I need to combine the whole number and the fraction. 3 is the same as 3/1. To add 3/1 and 14/25, I need a common denominator. The denominator for 3/1 is 1, and for 14/25 it's 25. So, I'll convert 3/1 to 75/25 by multiplying both the numerator and denominator by 25. Now, I can add 75/25 and 14/25 to get 89/25.Let me double-check to make sure I didn't make any mistakes. Starting with 3.56, I separated it into 3 and 0.56. Converted 0.56 to 56/100, simplified that to 14/25, and then combined it with 3 to get 89/25. That seems right.Wait, just to be sure, I'll convert 89/25 back to a decimal to see if it equals 3.56. 25 goes into 89 three times, which is 75, leaving a remainder of 14. Bringing down a zero, 25 goes into 140 five times, which is 125, leaving a remainder of 15. Bringing down another zero, 25 goes into 150 six times exactly. So, 89/25 is indeed 3.56. Okay, that checks out.I think I've got it right. So, 3.56 as a simplified fraction is 89/25.