answer：Alright, so I have this problem where I need to select 5 individuals from a population of 20 labeled 01 to 20. The method is to use a random number table, starting from the 5th and 6th columns of the 1st row, and then sequentially select two digits from left to right. The question is asking for the label of the 5th selected individual.First, let me try to visualize the random number table provided. It's given in two rows, each containing 8 four-digit numbers. So, each row has 32 digits, right? Let me write them out for clarity.First row: 7816 6572 0802 6314 0702 4369 9728 0198Second row: 3204 9234 4935 8200 3623 4869 6938 7481So, each row is 32 digits long. Now, the instruction says to start from the 5th and 6th columns of the 1st row. Hmm, okay. So, columns are vertical, right? So, each four-digit number is a group, but columns are individual digits.Wait, maybe I need to clarify how the columns are structured. Each four-digit number is a group, so each group is four columns. So, the first four digits are 7,8,1,6, which would be columns 1-4. Then the next four digits are 6,5,7,2, which are columns 5-8. Wait, so the 5th and 6th columns would be the first two digits of the second four-digit number in the first row.Looking at the first row: 7816 6572 0802 6314 0702 4369 9728 0198Breaking it down into individual digits:1:7, 2:8, 3:1, 4:6,5:6, 6:5, 7:7, 8:2,9:0, 10:8, 11:0, 12:2,13:6, 14:3, 15:1, 16:4,17:0, 18:7, 19:0, 20:2,21:4, 22:3, 23:6, 24:9,25:9, 26:7, 27:2, 28:8,29:0, 30:1, 31:9, 32:8So, the 5th column is 6, the 6th is 5. So, starting from the 5th and 6th columns, which are 6 and 5, so the first two digits are 65.Wait, but the selection method is to sequentially select two digits from left to right. So, starting at column 5 and 6, which is 65, then moving left to right, so the next two digits would be columns 7 and 8, which are 7 and 2, so 72, then columns 9 and 10, which are 0 and 8, so 08, and so on.But hold on, the problem says to start from the 5th and 6th columns of the 1st row and then sequentially select two digits from left to right. So, does that mean we take two digits starting from column 5, then move to the right? Or do we take two digits starting at column 5 and 6, then go to the next two digits, which would be columns 7 and 8, then 9 and 10, etc.?I think it's the latter. So, starting at columns 5 and 6, which are 6 and 5, forming 65, then the next two digits are columns 7 and 8, which are 7 and 2, forming 72, then columns 9 and 10, which are 0 and 8, forming 08, and so on.So, let me list out the two-digit numbers we get by taking two digits at a time, starting from column 5:First pair: columns 5-6: 65Second pair: columns 7-8: 72Third pair: columns 9-10: 08Fourth pair: columns 11-12: 02Fifth pair: columns 13-14: 63Sixth pair: columns 15-16: 14Seventh pair: columns 17-18: 07Eighth pair: columns 19-20: 02Ninth pair: columns 21-22: 43Tenth pair: columns 23-24: 69Eleventh pair: columns 25-26: 97Twelfth pair: columns 27-28: 28Thirteenth pair: columns 29-30: 01Fourteenth pair: columns 31-32: 98Wait, so that's 14 two-digit numbers from the first row. Then, moving on to the second row, starting from column 1, right? Because after column 32, we go to the next row.So, second row: 3204 9234 4935 8200 3623 4869 6938 7481Breaking it down into individual digits:33:3, 34:2, 35:0, 36:4,37:9, 38:2, 39:3, 40:4,41:4, 42:9, 43:3, 44:5,45:8, 46:2, 47:0, 48:0,49:3, 50:6, 51:2, 52:3,53:4, 54:8, 55:6, 56:9,57:6, 58:9, 59:3, 60:8,61:7, 62:4, 63:8, 64:1So, starting from column 33, which is 3, but we need to continue from where we left off. Wait, no, actually, after the first row ends at column 32, the next number would start at column 33, but since we're moving left to right, we just continue reading the next two digits from the second row.Wait, but the problem says to start from the 5th and 6th columns of the 1st row and then sequentially select two digits from left to right. So, after the first row, do we continue into the second row?I think yes, because it's a single table, so after the first row, we go to the second row, continuing from the next column. So, after column 32, we go to column 33, which is the first column of the second row.So, continuing from where we left off, after the first row, we have:Fifteenth pair: columns 33-34: 32Sixteenth pair: columns 35-36: 04Seventeenth pair: columns 37-38: 92Eighteenth pair: columns 39-40: 34Nineteenth pair: columns 41-42: 49Twentieth pair: columns 43-44: 35Twenty-first pair: columns 45-46: 82Twenty-second pair: columns 47-48: 00Twenty-third pair: columns 49-50: 36Twenty-fourth pair: columns 51-52: 23Twenty-fifth pair: columns 53-54: 48Twenty-sixth pair: columns 55-56: 69Twenty-seventh pair: columns 57-58: 69Twenty-eighth pair: columns 59-60: 38Twenty-ninth pair: columns 61-62: 74Thirtieth pair: columns 63-64: 81So, that's 30 two-digit numbers. But since we only need 5 individuals, we might not need all of them. Let's see.But let me list all the two-digit numbers we get:From first row:65, 72, 08, 02, 63, 14, 07, 02, 43, 69, 97, 28, 01, 98From second row:32, 04, 92, 34, 49, 35, 82, 00, 36, 23, 48, 69, 69, 38, 74, 81Wait, actually, when I counted before, I had 14 from the first row and 16 from the second row, totaling 30. But let me recount:First row: 14 pairsSecond row: 16 pairsTotal: 30 pairs.But since we only need 5 individuals, we can stop once we have 5 valid labels.But let's remember that the population is only 20 individuals, labeled 01 to 20. So, any two-digit number above 20 is invalid and should be skipped. Also, duplicates should be skipped as well, since we don't want to select the same individual more than once.So, let's go through the list and pick the valid numbers, skipping duplicates and numbers above 20.Starting from the first pair:1. 65: This is above 20, so skip.2. 72: Above 20, skip.3. 08: Valid, label 08. That's our first selection.4. 02: Valid, label 02. Second selection.5. 63: Above 20, skip.6. 14: Valid, label 14. Third selection.7. 07: Valid, label 07. Fourth selection.8. 02: Already selected, skip.9. 43: Above 20, skip.10. 69: Above 20, skip.11. 97: Above 20, skip.12. 28: Above 20, skip.13. 01: Valid, label 01. Fifth selection.Wait, so the fifth selection is 01. But let me double-check.Wait, so the first four valid numbers are 08, 02, 14, 07, and the fifth is 01. So, the fifth individual is 01.But hold on, let me make sure I didn't skip any numbers.Wait, after 07, the next number is 02, which is already selected, so we skip. Then 43, 69, 97, 28, all above 20. Then 01 is next, which is valid and not selected yet. So, yes, 01 is the fifth selection.But let me check the exact order to make sure.Wait, let me list all the numbers in order, marking which are valid and not duplicates.1. 65: invalid2. 72: invalid3. 08: valid (1st)4. 02: valid (2nd)5. 63: invalid6. 14: valid (3rd)7. 07: valid (4th)8. 02: duplicate9. 43: invalid10. 69: invalid11. 97: invalid12. 28: invalid13. 01: valid (5th)So, yes, 01 is the fifth selection.But wait, let me check the exact sequence again.Starting from the 5th and 6th columns of the first row, which is 65, then 72, 08, 02, 63, 14, 07, 02, 43, 69, 97, 28, 01, 98, then moving to the second row: 32, 04, 92, 34, 49, 35, 82, 00, 36, 23, 48, 69, 69, 38, 74, 81.Wait, so after 01, the next number is 98, which is invalid, then 32, invalid, 04, invalid, 92, invalid, 34, invalid, 49, invalid, 35, invalid, 82, invalid, 00, invalid, 36, invalid, 23, invalid, 48, invalid, 69, invalid, 69, invalid, 38, invalid, 74, invalid, 81, invalid.So, actually, after 01, there are no more valid numbers in the table. So, the fifth selection is indeed 01.But wait, hold on, is 00 considered? The population is labeled from 01 to 20, so 00 is not a valid label. So, 00 is skipped.Similarly, 04, 07, 08, 01 are valid, but others like 02, 14, etc., are also valid.Wait, but in the first row, after 01, the next number is 98, which is invalid, then moving to the second row, 32, 04, 92, 34, 49, 35, 82, 00, 36, 23, 48, 69, 69, 38, 74, 81. None of these are between 01 and 20 except 04, 23, but 23 is above 20, so only 04 is valid, but 04 is already in the table? Wait, no, 04 is not selected yet.Wait, hold on, let me recount the valid numbers:From the first row:08, 02, 14, 07, 01From the second row:04, 23But 04 and 23 are in the second row, but 23 is above 20, so only 04 is valid.Wait, but in the first pass, we already selected 08, 02, 14, 07, 01 as the first five. So, the fifth is 01.But let me make sure that I didn't miss any numbers in between.Wait, after 01, the next numbers are 98, 32, 04, 92, 34, 49, 35, 82, 00, 36, 23, 48, 69, 69, 38, 74, 81.So, 04 is next, but it's after 01. So, if we were to continue beyond five, 04 would be the sixth selection, but since we only need five, 01 is the fifth.Wait, but let me think again. The problem says to select 5 individuals, starting from the 5th and 6th columns of the 1st row, moving left to right.So, the order of selection is:1. 65 (invalid)2. 72 (invalid)3. 08 (valid, 1st)4. 02 (valid, 2nd)5. 63 (invalid)6. 14 (valid, 3rd)7. 07 (valid, 4th)8. 02 (duplicate, skip)9. 43 (invalid)10. 69 (invalid)11. 97 (invalid)12. 28 (invalid)13. 01 (valid, 5th)So, yes, 01 is the fifth selection.But wait, let me check the exact sequence again, because sometimes when you're selecting, you might have to go beyond the first row if you don't find enough valid numbers.But in this case, we found five valid numbers within the first row and the transition to the second row.Wait, no, actually, the fifth valid number is 01, which is in the first row. Because after 07, the next number is 02, which is a duplicate, then 43, 69, 97, 28, all invalid, then 01 is the next valid number, which is the fifth.So, the fifth selected individual is 01.But wait, let me double-check the exact order of the two-digit numbers:Starting from columns 5-6: 65Then 72, 08, 02, 63, 14, 07, 02, 43, 69, 97, 28, 01, 98, then moving to the second row: 32, 04, 92, 34, 49, 35, 82, 00, 36, 23, 48, 69, 69, 38, 74, 81.So, the order of two-digit numbers is:65, 72, 08, 02, 63, 14, 07, 02, 43, 69, 97, 28, 01, 98, 32, 04, 92, 34, 49, 35, 82, 00, 36, 23, 48, 69, 69, 38, 74, 81.Now, let's go through each number and mark the valid ones:1. 65: invalid2. 72: invalid3. 08: valid (1st)4. 02: valid (2nd)5. 63: invalid6. 14: valid (3rd)7. 07: valid (4th)8. 02: duplicate9. 43: invalid10. 69: invalid11. 97: invalid12. 28: invalid13. 01: valid (5th)14. 98: invalid15. 32: invalid16. 04: valid (6th)17. 92: invalid18. 34: invalid19. 49: invalid20. 35: invalid21. 82: invalid22. 00: invalid23. 36: invalid24. 23: invalid25. 48: invalid26. 69: invalid27. 69: invalid28. 38: invalid29. 74: invalid30. 81: invalidSo, yes, the fifth valid number is 01.But wait, let me make sure that I didn't skip any numbers between 07 and 01. After 07, the next number is 02, which is a duplicate, then 43, 69, 97, 28, all invalid, then 01 is next. So, yes, 01 is the fifth.Therefore, the label of the fifth selected individual is 01.But just to be thorough, let me write out the sequence of numbers and their validity:1. 65: invalid2. 72: invalid3. 08: valid (1)4. 02: valid (2)5. 63: invalid6. 14: valid (3)7. 07: valid (4)8. 02: duplicate9. 43: invalid10. 69: invalid11. 97: invalid12. 28: invalid13. 01: valid (5)So, yes, 01 is the fifth.Therefore, the answer is 01.

answer：Alright, so I have this problem where I need to find the maximum value of ( w = sqrt{2p - q} + sqrt{3q - 2p} + sqrt{6 - 2q} ). Hmm, okay. Let me try to figure this out step by step.First, I know that for square roots to be real numbers, the expressions inside them must be non-negative. So, I should write down the conditions for each square root:1. ( 2p - q geq 0 )2. ( 3q - 2p geq 0 )3. ( 6 - 2q geq 0 )Let me rewrite these inequalities to make them clearer.From the first inequality: ( 2p - q geq 0 ) implies ( q leq 2p ).From the second inequality: ( 3q - 2p geq 0 ) implies ( q geq frac{2p}{3} ).From the third inequality: ( 6 - 2q geq 0 ) implies ( q leq 3 ).So, combining these, ( frac{2p}{3} leq q leq 2p ) and ( q leq 3 ). Hmm, so ( q ) is bounded above by the minimum of ( 2p ) and 3, and bounded below by ( frac{2p}{3} ).I think it might help to visualize this region or maybe express ( p ) in terms of ( q ) or vice versa. Let me see.From the first two inequalities, ( frac{2p}{3} leq q leq 2p ). So, if I solve for ( p ), from ( q leq 2p ), we get ( p geq frac{q}{2} ). From ( q geq frac{2p}{3} ), we get ( p leq frac{3q}{2} ). So, ( frac{q}{2} leq p leq frac{3q}{2} ).So, for a given ( q ), ( p ) is between ( frac{q}{2} ) and ( frac{3q}{2} ). Also, ( q leq 3 ).I wonder if I can express ( w ) in terms of ( q ) only by substituting ( p ) in terms of ( q ). But since ( p ) is bounded between ( frac{q}{2} ) and ( frac{3q}{2} ), maybe I can express ( w ) as a function of ( q ) with ( p ) chosen optimally within that range. Hmm, perhaps calculus can be used here.Alternatively, maybe I can set variables to simplify the expressions. Let me try substitution.Let me denote ( a = sqrt{2p - q} ), ( b = sqrt{3q - 2p} ), and ( c = sqrt{6 - 2q} ). Then, ( w = a + b + c ).Now, let's square ( w ) to see if that helps:( w^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc ).But ( a^2 = 2p - q ), ( b^2 = 3q - 2p ), and ( c^2 = 6 - 2q ).So, adding these up: ( a^2 + b^2 + c^2 = (2p - q) + (3q - 2p) + (6 - 2q) ).Simplify:( 2p - q + 3q - 2p + 6 - 2q = (2p - 2p) + (-q + 3q - 2q) + 6 = 0 + 0 + 6 = 6 ).So, ( w^2 = 6 + 2ab + 2ac + 2bc ).Hmm, that's interesting. So, ( w^2 = 6 + 2(ab + ac + bc) ). Therefore, to maximize ( w ), we need to maximize ( ab + ac + bc ).But I'm not sure if this is helpful yet. Maybe I can find expressions for ( ab ), ( ac ), and ( bc ).Let me compute ( ab ):( ab = sqrt{(2p - q)(3q - 2p)} ).Hmm, that looks a bit complicated. Let me compute the product inside:( (2p - q)(3q - 2p) = 6pq - 4p^2 - 3q^2 + 2pq = (6pq + 2pq) - 4p^2 - 3q^2 = 8pq - 4p^2 - 3q^2 ).Hmm, not sure if that helps. Maybe another approach.Alternatively, perhaps I can use the Cauchy-Schwarz inequality or AM-QM or something like that.But before that, maybe I can consider variables substitution.Let me set ( x = 2p - q ) and ( y = 3q - 2p ). Then, ( x geq 0 ) and ( y geq 0 ).Also, let me note that ( x + y = (2p - q) + (3q - 2p) = 2q ). So, ( x + y = 2q ).Also, from the third term, ( c = sqrt{6 - 2q} ), so ( 6 - 2q geq 0 ) implies ( q leq 3 ).So, ( x + y = 2q leq 6 ). So, ( x + y leq 6 ).So, ( x ) and ( y ) are non-negative numbers adding up to at most 6.Also, from ( x = 2p - q ) and ( y = 3q - 2p ), we can solve for ( p ) and ( q ):Adding ( x + y = 2q ), so ( q = frac{x + y}{2} ).Substituting into ( x = 2p - q ):( x = 2p - frac{x + y}{2} ).Multiply both sides by 2:( 2x = 4p - x - y ).Bring terms with ( x ) and ( y ) to the left:( 2x + x + y = 4p ).So, ( 3x + y = 4p ).Similarly, from ( y = 3q - 2p ), substitute ( q = frac{x + y}{2} ):( y = 3 cdot frac{x + y}{2} - 2p ).Multiply both sides by 2:( 2y = 3(x + y) - 4p ).Simplify:( 2y = 3x + 3y - 4p ).Bring terms to the left:( 2y - 3x - 3y + 4p = 0 ).Simplify:( -3x - y + 4p = 0 ).But from earlier, ( 3x + y = 4p ). So, substituting ( 4p = 3x + y ) into this equation:( -3x - y + (3x + y) = 0 ).Simplify:( (-3x + 3x) + (-y + y) = 0 ) => 0 = 0.So, that doesn't give any new information. So, perhaps I need another approach.Wait, so I have ( x + y = 2q leq 6 ), and ( c = sqrt{6 - 2q} = sqrt{6 - (x + y)} ).So, ( c = sqrt{6 - (x + y)} ).So, ( w = sqrt{x} + sqrt{y} + sqrt{6 - x - y} ).So, now, ( w ) is expressed in terms of ( x ) and ( y ), with ( x geq 0 ), ( y geq 0 ), and ( x + y leq 6 ).So, the problem reduces to maximizing ( w = sqrt{x} + sqrt{y} + sqrt{6 - x - y} ) with ( x, y geq 0 ) and ( x + y leq 6 ).That seems more manageable. So, perhaps I can use calculus here.Let me denote ( z = 6 - x - y ), so ( z geq 0 ), and ( x + y + z = 6 ).So, ( w = sqrt{x} + sqrt{y} + sqrt{z} ), with ( x + y + z = 6 ), and ( x, y, z geq 0 ).So, now, it's a problem of maximizing the sum of square roots of variables that add up to 6.I remember that for such optimization problems, the maximum occurs when the variables are equal, due to the concavity of the square root function. So, perhaps the maximum occurs when ( x = y = z ).Let me test that.If ( x = y = z ), then each is equal to 2, since ( 3 times 2 = 6 ).So, ( w = sqrt{2} + sqrt{2} + sqrt{2} = 3sqrt{2} approx 4.2426 ).But wait, is that really the maximum? Let me see.Alternatively, maybe it's higher when one variable is larger and the others are smaller.Wait, the square root function is concave, so by Jensen's inequality, the maximum of the sum is achieved when all variables are equal.But wait, actually, Jensen's inequality for concave functions states that the maximum is achieved at the endpoints, but for sums, it's a bit different.Wait, no, for a concave function, the maximum of the sum is achieved when the variables are equal because of the concavity. Wait, maybe I should think in terms of Lagrange multipliers.Let me set up the Lagrangian.Let me denote ( f(x, y, z) = sqrt{x} + sqrt{y} + sqrt{z} ), subject to the constraint ( g(x, y, z) = x + y + z - 6 = 0 ).The Lagrangian is ( mathcal{L} = sqrt{x} + sqrt{y} + sqrt{z} - lambda(x + y + z - 6) ).Taking partial derivatives:( frac{partial mathcal{L}}{partial x} = frac{1}{2sqrt{x}} - lambda = 0 )Similarly,( frac{partial mathcal{L}}{partial y} = frac{1}{2sqrt{y}} - lambda = 0 )( frac{partial mathcal{L}}{partial z} = frac{1}{2sqrt{z}} - lambda = 0 )And the constraint:( x + y + z = 6 )From the first three equations, we have:( frac{1}{2sqrt{x}} = lambda )( frac{1}{2sqrt{y}} = lambda )( frac{1}{2sqrt{z}} = lambda )Therefore, ( frac{1}{2sqrt{x}} = frac{1}{2sqrt{y}} = frac{1}{2sqrt{z}} ).Which implies ( sqrt{x} = sqrt{y} = sqrt{z} ), so ( x = y = z ).Therefore, ( x = y = z = 2 ), as before.So, the maximum occurs when all three variables are equal, giving ( w = 3sqrt{2} ).But wait, hold on. Is this really the case? Because in the original problem, ( x = 2p - q ) and ( y = 3q - 2p ), so ( x ) and ( y ) are not independent variables. They are related through ( p ) and ( q ). So, does setting ( x = y = z ) correspond to feasible ( p ) and ( q )?Let me check.If ( x = 2 ), ( y = 2 ), ( z = 2 ), then:From ( x = 2p - q = 2 )From ( y = 3q - 2p = 2 )So, we have the system:1. ( 2p - q = 2 )2. ( 3q - 2p = 2 )Let me solve this system.From equation 1: ( 2p - q = 2 ) => ( q = 2p - 2 )Substitute into equation 2:( 3(2p - 2) - 2p = 2 )Simplify:( 6p - 6 - 2p = 2 )Combine like terms:( 4p - 6 = 2 ) => ( 4p = 8 ) => ( p = 2 )Then, ( q = 2p - 2 = 4 - 2 = 2 )So, ( p = 2 ), ( q = 2 )Check if this satisfies all the original conditions:1. ( 2p - q = 4 - 2 = 2 geq 0 ) ✔️2. ( 3q - 2p = 6 - 4 = 2 geq 0 ) ✔️3. ( 6 - 2q = 6 - 4 = 2 geq 0 ) ✔️So, yes, ( p = 2 ), ( q = 2 ) is a feasible solution, and at this point, ( w = 3sqrt{2} ).But wait, is this the maximum? Let me test another point.Suppose I set ( q = 3 ), which is the upper bound from the third inequality.Then, ( c = sqrt{6 - 2q} = sqrt{6 - 6} = 0 ).So, ( w = sqrt{2p - 3} + sqrt{9 - 2p} + 0 ).Let me denote ( a = sqrt{2p - 3} ) and ( b = sqrt{9 - 2p} ).So, ( w = a + b ), with ( a^2 = 2p - 3 ) and ( b^2 = 9 - 2p ).Adding these: ( a^2 + b^2 = 6 ).So, ( w = a + b ), with ( a^2 + b^2 = 6 ).To maximize ( a + b ) given ( a^2 + b^2 = 6 ), the maximum occurs when ( a = b ), so ( a = b = sqrt{3} ), giving ( w = 2sqrt{3} approx 3.464 ), which is less than ( 3sqrt{2} approx 4.2426 ).So, that's lower. So, ( q = 3 ) gives a lower ( w ).What if I set ( q = 0 )?Then, ( c = sqrt{6 - 0} = sqrt{6} ).Also, ( 2p - q = 2p geq 0 ), so ( p geq 0 ).( 3q - 2p = -2p geq 0 ) implies ( p leq 0 ).So, ( p = 0 ).Thus, ( w = sqrt{0} + sqrt{0} + sqrt{6} = sqrt{6} approx 2.449 ), which is even lower.So, seems like ( q = 2 ), ( p = 2 ) gives a higher ( w ).Alternatively, maybe somewhere in between.Wait, let me try ( q = 1 ).Then, ( c = sqrt{6 - 2} = sqrt{4} = 2 ).Also, ( 2p - 1 geq 0 ) => ( p geq 0.5 ).( 3(1) - 2p geq 0 ) => ( 3 - 2p geq 0 ) => ( p leq 1.5 ).So, ( p ) is between 0.5 and 1.5.Let me choose ( p = 1 ).Then, ( a = sqrt{2(1) - 1} = sqrt{1} = 1 ).( b = sqrt{3(1) - 2(1)} = sqrt{1} = 1 ).So, ( w = 1 + 1 + 2 = 4 ).Hmm, that's less than ( 3sqrt{2} approx 4.2426 ).Alternatively, let me try ( p = 1.2 ), ( q = 1 ).Then, ( a = sqrt{2.4 - 1} = sqrt{1.4} approx 1.1832 ).( b = sqrt{3 - 2.4} = sqrt{0.6} approx 0.7746 ).( c = 2 ).So, ( w approx 1.1832 + 0.7746 + 2 approx 3.9578 ), still less than ( 3sqrt{2} ).Alternatively, let me try ( q = 2.5 ).Then, ( c = sqrt{6 - 5} = sqrt{1} = 1 ).Also, ( 2p - 2.5 geq 0 ) => ( p geq 1.25 ).( 3(2.5) - 2p geq 0 ) => ( 7.5 - 2p geq 0 ) => ( p leq 3.75 ).So, ( p ) is between 1.25 and 3.75.Let me choose ( p = 2 ).Then, ( a = sqrt{4 - 2.5} = sqrt{1.5} approx 1.2247 ).( b = sqrt{7.5 - 4} = sqrt{3.5} approx 1.8708 ).( c = 1 ).So, ( w approx 1.2247 + 1.8708 + 1 approx 4.0955 ), still less than ( 3sqrt{2} ).Hmm, seems like ( 3sqrt{2} ) is indeed the maximum.Alternatively, let me consider another approach.Since ( w = sqrt{x} + sqrt{y} + sqrt{z} ) with ( x + y + z = 6 ), and ( x, y, z geq 0 ).We can use the Cauchy-Schwarz inequality.Recall that ( (sqrt{x} + sqrt{y} + sqrt{z})^2 leq (1 + 1 + 1)(x + y + z) = 3 times 6 = 18 ).Therefore, ( w^2 leq 18 ) => ( w leq sqrt{18} = 3sqrt{2} ).So, the maximum value of ( w ) is ( 3sqrt{2} ), achieved when ( x = y = z = 2 ).Therefore, the maximum value of ( w ) is ( 3sqrt{2} ).**Final Answer**The maximum value of ( w ) is boxed{3sqrt{2}}.

answer：Alright, so I have this problem where Ilya Muromets is battling a three-headed dragon named Zmey Gorynych. Every minute, Ilya chops off one head. Depending on some probabilities, new heads might grow back. The goal is to find the probability that Ilya will eventually defeat the dragon, meaning the dragon ends up with no heads.Let me break this down. The dragon starts with three heads. Each time Ilya chops off a head, there are three possible outcomes:1. With probability 1/4, two new heads grow back. So, the total number of heads increases by one (since one was chopped off, but two grew back).2. With probability 1/3, only one new head grows back. So, the total number of heads remains the same (one chopped off, one grew back).3. With probability 5/12, no new heads grow back. So, the total number of heads decreases by one.I need to model this as a Markov chain, where each state represents the number of heads the dragon has. The states can be 0, 1, 2, 3, and so on. However, since the dragon starts with three heads, we can focus on states 0, 1, 2, 3, and maybe higher, but I suspect the probabilities might stabilize or something.The key here is to find the probability that starting from state 3, the process will eventually reach state 0. This is essentially the probability of absorption in state 0, which is an absorbing state.Let me denote ( p_n ) as the probability of eventually defeating the dragon (reaching state 0) starting from state ( n ). So, we need to find ( p_3 ).First, let's note that ( p_0 = 1 ) because if there are no heads, the dragon is already defeated. For ( n geq 1 ), we can write recursive equations based on the transitions.From any state ( n geq 1 ), Ilya chops off one head, so we transition to state ( n - 1 ) with some probability, or maybe ( n ) or ( n + 1 ) depending on the outcome.Wait, actually, when Ilya chops off a head, the number of heads can either decrease by one, stay the same, or increase by one, depending on whether 0, 1, or 2 new heads grow back.So, from state ( n ), chopping off one head leads to:- ( n - 1 ) with probability 5/12,- ( n ) with probability 1/3,- ( n + 1 ) with probability 1/4.Therefore, the recursive equation for ( p_n ) is:( p_n = frac{5}{12} p_{n - 1} + frac{1}{3} p_n + frac{1}{4} p_{n + 1} )Wait, that seems a bit confusing. Let me write it properly.Starting from state ( n ), after one chop, the probability of moving to ( n - 1 ) is 5/12, to ( n ) is 1/3, and to ( n + 1 ) is 1/4. Therefore, the probability ( p_n ) can be expressed as:( p_n = frac{5}{12} p_{n - 1} + frac{1}{3} p_n + frac{1}{4} p_{n + 1} )Hmm, okay, so let's rearrange this equation to solve for ( p_{n + 1} ).First, subtract ( frac{1}{3} p_n ) from both sides:( p_n - frac{1}{3} p_n = frac{5}{12} p_{n - 1} + frac{1}{4} p_{n + 1} )Simplify the left side:( frac{2}{3} p_n = frac{5}{12} p_{n - 1} + frac{1}{4} p_{n + 1} )Now, multiply both sides by 12 to eliminate denominators:( 8 p_n = 5 p_{n - 1} + 3 p_{n + 1} )Rearranging terms:( 3 p_{n + 1} - 8 p_n + 5 p_{n - 1} = 0 )So, this is a linear recurrence relation. Let me write it as:( 3 p_{n + 1} - 8 p_n + 5 p_{n - 1} = 0 )To solve this recurrence, I can find the characteristic equation. Let's assume a solution of the form ( p_n = r^n ). Plugging into the recurrence:( 3 r^{n + 1} - 8 r^n + 5 r^{n - 1} = 0 )Divide both sides by ( r^{n - 1} ):( 3 r^2 - 8 r + 5 = 0 )So, the characteristic equation is:( 3 r^2 - 8 r + 5 = 0 )Let's solve for ( r ):Using quadratic formula:( r = frac{8 pm sqrt{64 - 60}}{6} = frac{8 pm 2}{6} )So, the roots are:( r = frac{8 + 2}{6} = frac{10}{6} = frac{5}{3} )and( r = frac{8 - 2}{6} = frac{6}{6} = 1 )Therefore, the general solution is:( p_n = A left( frac{5}{3} right)^n + B (1)^n = A left( frac{5}{3} right)^n + B )Where ( A ) and ( B ) are constants to be determined by boundary conditions.Now, let's think about boundary conditions. We know that ( p_0 = 1 ). Also, as ( n ) approaches infinity, what happens to ( p_n )?In the long run, if the number of heads keeps increasing, the probability of eventually defeating the dragon should approach 0. So, ( lim_{n to infty} p_n = 0 ).Given that, let's analyze the general solution. The term ( A left( frac{5}{3} right)^n ) will dominate as ( n ) grows. Since ( frac{5}{3} > 1 ), this term will go to infinity unless ( A = 0 ). But we need ( p_n ) to approach 0 as ( n ) approaches infinity, so we must have ( A = 0 ).Therefore, the solution simplifies to:( p_n = B )But wait, if ( p_n = B ) for all ( n ), then ( p_n ) is a constant. Let's check if this satisfies the recurrence.Plugging ( p_n = B ) into the recurrence:( 3 B - 8 B + 5 B = 0 )Simplifies to:( 0 = 0 )So, it's a valid solution. But we also have the boundary condition ( p_0 = 1 ). So, ( p_0 = B = 1 ). Therefore, ( p_n = 1 ) for all ( n )?Wait, that can't be right. If ( p_n = 1 ) for all ( n ), that would mean that no matter how many heads the dragon has, Ilya will always defeat it with probability 1. But intuitively, that doesn't seem correct because there's a chance the number of heads can increase, leading to an infinite loop.Hmm, so perhaps my approach is missing something. Maybe the recurrence is not valid for all ( n geq 1 ), or perhaps I need to consider another boundary condition.Wait, actually, the recurrence ( 3 p_{n + 1} - 8 p_n + 5 p_{n - 1} = 0 ) is valid for ( n geq 1 ). But when ( n = 0 ), the dragon is already defeated, so ( p_0 = 1 ). For ( n = 1 ), we can write the equation:( 3 p_2 - 8 p_1 + 5 p_0 = 0 )But without knowing ( p_1 ) or ( p_2 ), it's hard to proceed.Alternatively, maybe I need to consider another approach. Let's think about the expected change in the number of heads. The expected number of heads after chopping one is:( E = frac{5}{12} (n - 1) + frac{1}{3} n + frac{1}{4} (n + 1) )Simplify:( E = frac{5}{12} n - frac{5}{12} + frac{1}{3} n + frac{1}{4} n + frac{1}{4} )Convert all coefficients to twelfths:( frac{5}{12} n - frac{5}{12} + frac{4}{12} n + frac{3}{12} n + frac{3}{12} )Combine like terms:( (5 + 4 + 3)/12 n + (-5 + 3)/12 )Which is:( 12/12 n - 2/12 = n - 1/6 )So, the expected number of heads decreases by 1/6 each time. Since the expectation is decreasing, it suggests that the process is transient and might not necessarily go to infinity. Hmm, but expectation alone doesn't guarantee absorption.Alternatively, maybe I can model this as a Gambler's Ruin problem. In the Gambler's Ruin, we have absorbing states at 0 and some N, and we calculate the probability of ruin. In our case, the absorbing state is 0, and the other states are transient. However, in our case, the number of heads can potentially go to infinity, so it's an infinite state space.In the standard Gambler's Ruin with absorbing barriers, the probability of ruin can be found using similar recurrence relations. Maybe I can use a similar approach here.Let me denote ( p_n ) as the probability of reaching 0 starting from ( n ). Then, for ( n geq 1 ):( p_n = frac{5}{12} p_{n - 1} + frac{1}{3} p_n + frac{1}{4} p_{n + 1} )Wait, this is the same equation I had before. So, I need to solve this recurrence with boundary conditions ( p_0 = 1 ) and ( lim_{n to infty} p_n = 0 ).Earlier, I found the general solution ( p_n = A (5/3)^n + B ). But since ( p_n ) must approach 0 as ( n to infty ), we set ( A = 0 ), leading to ( p_n = B ). But then ( p_0 = B = 1 ), so ( p_n = 1 ) for all ( n ), which contradicts the intuition.Wait, perhaps my characteristic equation is incorrect. Let me double-check.The recurrence is:( 3 p_{n + 1} - 8 p_n + 5 p_{n - 1} = 0 )Assuming ( p_n = r^n ), plugging in:( 3 r^{n + 1} - 8 r^n + 5 r^{n - 1} = 0 )Divide by ( r^{n - 1} ):( 3 r^2 - 8 r + 5 = 0 )Yes, that's correct. So, the characteristic equation is correct, with roots at ( r = 1 ) and ( r = 5/3 ).So, the general solution is ( p_n = A (5/3)^n + B ). Applying the boundary condition ( p_0 = 1 ):( 1 = A (5/3)^0 + B = A + B )So, ( A + B = 1 ).Now, applying the other boundary condition ( lim_{n to infty} p_n = 0 ). As ( n to infty ), ( (5/3)^n ) tends to infinity, so unless ( A = 0 ), ( p_n ) would go to infinity or negative infinity. But since probabilities can't be negative, ( A ) must be zero. Therefore, ( A = 0 ), which implies ( B = 1 ). So, ( p_n = 1 ) for all ( n ).But this seems counterintuitive because, as I thought earlier, there's a chance the number of heads can increase, so the dragon might never be defeated. However, the recurrence suggests that regardless of how many heads the dragon has, the probability of defeating it is always 1.Wait, maybe my intuition was wrong. Let's think about the expected number of heads. Each time, the expected number decreases by 1/6, so over time, the number of heads tends to decrease. Therefore, the probability of eventually reaching 0 might indeed be 1.But wait, in probability theory, even if the expectation decreases, it doesn't necessarily mean that the process will be absorbed. For example, in a symmetric random walk on the integers, the walk is recurrent, meaning it will return to the origin with probability 1, but if it's asymmetric, it might not.But in our case, the walk is not symmetric. Let me think about the transition probabilities.From state ( n ), the probability to go to ( n - 1 ) is 5/12, to ( n ) is 1/3, and to ( n + 1 ) is 1/4.So, the probability to move left is 5/12, right is 1/4, and stay is 1/3.The ratio of left to right probabilities is (5/12) / (1/4) = (5/12) * (4/1) = 5/3 > 1. So, the walk is biased to the left. Therefore, it's a positive recurrent Markov chain, meaning it will be absorbed at 0 with probability 1.Wait, so maybe my initial intuition was wrong, and the correct answer is indeed 1. But let me verify this with another approach.Alternatively, let's consider the probability generating function. Let ( G(s) = sum_{n=0}^{infty} p_n s^n ).But maybe that's more complicated. Alternatively, let's think about the recurrence relation again.We have:( 3 p_{n + 1} - 8 p_n + 5 p_{n - 1} = 0 )With general solution ( p_n = A (5/3)^n + B ). As ( n to infty ), ( p_n ) tends to 0 only if ( A = 0 ). So, ( p_n = B ). But ( p_0 = 1 ), so ( B = 1 ). Therefore, ( p_n = 1 ) for all ( n ).Wait, so does that mean that no matter how many heads the dragon has, Ilya will always defeat it with probability 1? That seems surprising, but given the recurrence, it's consistent.Alternatively, maybe I made a mistake in setting up the recurrence. Let me double-check.From state ( n ), after chopping, the next state is:- ( n - 1 ) with probability 5/12,- ( n ) with probability 1/3,- ( n + 1 ) with probability 1/4.Therefore, the equation is:( p_n = frac{5}{12} p_{n - 1} + frac{1}{3} p_n + frac{1}{4} p_{n + 1} )Yes, that's correct. Then, rearranged:( 3 p_{n + 1} - 8 p_n + 5 p_{n - 1} = 0 )Yes, that's correct.So, the solution is ( p_n = A (5/3)^n + B ). Applying boundary conditions:1. ( p_0 = 1 ): ( A + B = 1 )2. ( lim_{n to infty} p_n = 0 ): ( A = 0 )Therefore, ( B = 1 ), so ( p_n = 1 ) for all ( n ).Hmm, so despite the possibility of the number of heads increasing, the probability of eventual defeat is 1. That seems to be the case here.But let me test this with a small ( n ). Let's compute ( p_1 ).From the recurrence:For ( n = 1 ):( 3 p_2 - 8 p_1 + 5 p_0 = 0 )But we know ( p_0 = 1 ), and if ( p_n = 1 ) for all ( n ), then:( 3(1) - 8(1) + 5(1) = 3 - 8 + 5 = 0 )Which holds true.Similarly, for ( n = 2 ):( 3 p_3 - 8 p_2 + 5 p_1 = 0 )Again, with ( p_1 = p_2 = p_3 = 1 ):( 3(1) - 8(1) + 5(1) = 0 )Holds true.So, it seems consistent. Therefore, the probability ( p_n = 1 ) for all ( n geq 0 ). Therefore, starting from 3 heads, the probability of defeating the dragon is 1.But wait, that seems counterintuitive because there's a chance the number of heads can increase. However, the recurrence suggests that no matter how many heads there are, the probability of eventual defeat is 1. Maybe because the expected number of heads decreases each time, leading to a sure absorption at 0.Alternatively, let's think about the expected number of steps to reach 0. Even if the probability is 1, the expected time might be infinite. But the question is only about the probability, not the expected time.Therefore, based on the recurrence and the boundary conditions, the probability is indeed 1.Wait, but I recall that in some cases, even if the expected change is negative, the process might still have a positive probability of never being absorbed. For example, in a branching process, if the expected number of offspring is less than 1, the process dies out with probability 1. But if it's greater than 1, there's a positive probability of explosion.Wait, in this case, the dragon's heads can be thought of as a branching process where each head can produce 0, 1, or 2 new heads with probabilities 5/12, 1/3, and 1/4 respectively.Wait, actually, in a branching process, each individual (head) produces offspring independently. In our case, when a head is chopped off, it's replaced by 0, 1, or 2 new heads. So, the number of heads can be modeled as a branching process with offspring distribution:- 0 with probability 5/12,- 1 with probability 1/3,- 2 with probability 1/4.The expected number of offspring per head is:( mu = 0 times frac{5}{12} + 1 times frac{1}{3} + 2 times frac{1}{4} = frac{1}{3} + frac{2}{4} = frac{1}{3} + frac{1}{2} = frac{5}{6} )Since ( mu = 5/6 < 1 ), the branching process is subcritical, meaning the population dies out with probability 1. Therefore, the probability that the dragon is eventually defeated is 1.Ah, that makes sense! So, using the theory of branching processes, since the expected number of offspring is less than 1, the extinction probability is 1. Therefore, Ilya will eventually defeat the dragon with probability 1.So, putting it all together, the probability is 1.But wait, in the initial recurrence, I found that ( p_n = 1 ) for all ( n ), which aligns with this conclusion. So, both methods agree.Therefore, the probability that Ilya will eventually defeat the dragon is 1.**Final Answer**The probability that Ilya will eventually defeat the dragon is boxed{1}.

answer：First, I need to determine how much Rachel spent on lunch. She spent 1/4 of her 200 on lunch, which is 50.Next, I calculate how much she spent on the DVD. She spent 1/2 of her 200 on the DVD, which amounts to 100.Finally, I add the amounts she spent on lunch and the DVD to find the total expenditure, which is 150. Subtracting this from her initial 200 gives her remaining money, which is 50.

answer：First, I recognize that if ( a ) and ( b ) are reciprocals, then their product ( ab ) is equal to 1.Next, since ( c ) and ( d ) are opposites, their sum ( c + d ) is equal to 0.Substituting these values into the expression ( (ab)^4 - 3(c + d)^3 ), we get ( 1^4 - 3 times 0^3 ).Calculating each term, ( 1^4 ) is 1 and ( 3 times 0^3 ) is 0.Therefore, the expression simplifies to ( 1 - 0 = 1 ).

answer：To determine the total number of toothpicks needed, I'll break down the problem into two parts: calculating the horizontal toothpicks and the vertical toothpicks.First, for the horizontal toothpicks, each row of the rectangle requires 20 toothpicks. Since the width is 10 toothpicks, there are 11 rows in total. Multiplying 20 by 11 gives 220 horizontal toothpicks.Next, for the vertical toothpicks, each column of the rectangle requires 10 toothpicks. With a length of 20 toothpicks, there are 21 columns. Multiplying 10 by 21 results in 210 vertical toothpicks.Finally, adding the horizontal and vertical toothpicks together, 220 plus 210 equals 430 toothpicks in total.