answer：First, I recognize that both logarithms have the same base, which is 3. This allows me to use the logarithmic property that states ( log_b{a} - log_b{c} = log_b{left(frac{a}{c}right)} ).Applying this property, the expression ( log_3{6} - log_3{2} ) simplifies to ( log_3{left(frac{6}{2}right)} ), which is ( log_3{3} ).Since ( log_3{3} ) equals 1, the value of the original expression is 1.

answer：Okay, so I have this problem where a store sells shirts, water bottles, and chocolate bars. Each shirt costs 10, each water bottle is 5, and each chocolate bar is 1. On a particular day, they sold x shirts, y water bottles, and z chocolate bars, making a total revenue of 120. I need to find how many possible ordered triples (x, y, z) there are, given that x, y, and z are positive integers.Hmm, let me start by writing down the equation based on the given information. The total revenue is the sum of the revenue from each item. So, that would be:10x + 5y + z = 120And x, y, z are all greater than 0. So, x ≥ 1, y ≥ 1, z ≥ 1.I need to find all possible triples (x, y, z) that satisfy this equation. Since all variables are positive integers, I can approach this by fixing one variable at a time and then finding the possible values for the others.Let me think, maybe I can express z in terms of x and y. So, rearranging the equation:z = 120 - 10x - 5ySince z has to be a positive integer, 120 - 10x - 5y must be at least 1. So,120 - 10x - 5y ≥ 1Which simplifies to:10x + 5y ≤ 119Hmm, that might be useful. Alternatively, maybe I can factor out a 5 from the equation:10x + 5y + z = 120Divide both sides by 5:2x + y + (z/5) = 24But z has to be an integer, so z/5 must be an integer or a fraction. Wait, maybe that complicates things. Let me think again.Alternatively, since z is in dollars and it's 1 per chocolate bar, z can be any integer from 1 up to 120 - 10x -5y. So, for each x and y, z is determined as long as 10x + 5y is less than 120.But maybe it's better to fix x first, then find possible y and z.Let me try that approach.First, x has to be at least 1, and since each shirt is 10, the maximum x can be is when y and z are at their minimum, which is 1 each. So,10x + 5*1 + 1*1 ≤ 12010x + 6 ≤ 12010x ≤ 114x ≤ 11.4Since x is an integer, x can be from 1 to 11.So, x can take values 1, 2, 3, ..., 11.For each x, I can find the possible y and z.Let me fix x and then express the equation in terms of y and z.So, for a fixed x, the equation becomes:5y + z = 120 - 10xLet me denote 120 - 10x as some constant, say, C.So, 5y + z = CAgain, y and z are positive integers. So, for each x, I can find the number of solutions (y, z) to this equation.Let me express z in terms of y:z = C - 5ySince z must be at least 1, C - 5y ≥ 1So, 5y ≤ C - 1Which implies y ≤ (C - 1)/5But y must also be at least 1.So, for each x, C = 120 -10x, and then y must satisfy:1 ≤ y ≤ floor((C -1)/5)So, the number of possible y for each x is floor((C -1)/5) - 1 + 1 = floor((C -1)/5)Wait, actually, the number of integer solutions for y is floor((C -1)/5). Because y can be from 1 up to floor((C -1)/5).But let me verify that.Suppose C is 120 -10x.For each x, C decreases by 10. So, starting from x=1, C=110, x=2, C=100, and so on until x=11, C=10.So, for each x, C is 110, 100, 90, ..., 10.Then, for each C, the number of possible y is floor((C -1)/5). Let me compute that.Wait, let me take x=1, C=110.Then, number of y is floor((110 -1)/5) = floor(109/5) = 21.8, so 21.But wait, y must satisfy 5y ≤ 109, so y ≤ 21.8, so y can be 1 to 21, inclusive. So, 21 possibilities.Similarly, for x=2, C=100.Number of y is floor((100 -1)/5)=floor(99/5)=19.8, so 19.Wait, 5y ≤ 99, so y ≤19.8, so y=1 to 19, which is 19 possibilities.Wait, but 5*19=95, so z=100 -95=5, which is positive. So, that works.Similarly, for x=3, C=90.Number of y is floor((90 -1)/5)=floor(89/5)=17.8, so 17.So, y=1 to17, 17 possibilities.Wait, 5*17=85, so z=90 -85=5, which is positive.Wait, but hold on, when x=11, C=10.So, number of y is floor((10 -1)/5)=floor(9/5)=1.8, so 1.So, y=1, z=10 -5=5, which is positive.Wait, but let me check x=11, y=1, z=5. That's correct.Wait, but when x=11, C=10, so 5y + z=10.Since y and z are at least 1, 5y can be 5, so z=5, which is the only solution.So, for x=11, only 1 solution.Similarly, for x=10, C=20.Number of y is floor((20 -1)/5)=floor(19/5)=3.8, so 3.So, y=1,2,3.Then z=20 -5y, which is 15,10,5, all positive.So, 3 solutions.Wait, so in general, for each x, the number of y is floor((C -1)/5), where C=120 -10x.But let me see if I can find a pattern or a formula for the number of solutions.Alternatively, maybe I can express this as:For each x, the number of (y,z) pairs is equal to the number of positive integer solutions to 5y + z = C, where C=120 -10x.Since z must be at least 1, 5y must be at most C -1.So, y can be from 1 to floor((C -1)/5).Therefore, the number of solutions is floor((C -1)/5).But since C=120 -10x, which is a multiple of 5, because 120 is divisible by 5, and 10x is also divisible by 5.Therefore, C is divisible by 5, so C=5k, where k=(120 -10x)/5=24 -2x.So, C=5k, where k=24 -2x.Therefore, the equation becomes 5y + z=5k.Which can be rewritten as z=5(k - y).Since z must be at least 1, 5(k - y) ≥1.Therefore, k - y ≥1/5.But since k and y are integers, k - y must be at least 1.Therefore, y ≤k -1.But y must also be at least 1.Therefore, y can range from 1 to k -1.Thus, the number of solutions is (k -1) -1 +1= k -1.Wait, that is, the number of possible y is k -1.But k=24 -2x.Therefore, the number of solutions is (24 -2x) -1=23 -2x.But wait, hold on, let me verify.Wait, if k=24 -2x, then the number of y is k -1=23 -2x.But when x=1, k=24 -2=22, so number of y=21, which matches our earlier result.Similarly, for x=2, k=24 -4=20, number of y=19, which also matches.For x=11, k=24 -22=2, number of y=1, which is correct.So, in general, for each x, the number of solutions is 23 -2x.But wait, we have to make sure that 23 -2x is positive.So, 23 -2x ≥123 -1 ≥2x22 ≥2xx ≤11Which is consistent with our earlier result that x can be up to 11.So, for each x from 1 to11, the number of (y,z) pairs is 23 -2x.Therefore, the total number of ordered triples is the sum from x=1 to x=11 of (23 -2x).So, let's compute that.Sum = Σ (23 -2x) for x=1 to11.Let me compute this sum.First, let's write out the terms:For x=1: 23 -2(1)=21x=2:23 -4=19x=3:23 -6=17x=4:23 -8=15x=5:23 -10=13x=6:23 -12=11x=7:23 -14=9x=8:23 -16=7x=9:23 -18=5x=10:23 -20=3x=11:23 -22=1So, the sequence is 21,19,17,15,13,11,9,7,5,3,1.This is an arithmetic sequence where the first term a1=21, the last term an=1, and the common difference d=-2.The number of terms n=11.The sum of an arithmetic sequence is n*(a1 + an)/2.So, sum=11*(21 +1)/2=11*22/2=11*11=121.Wait, so the total number of ordered triples is 121.But let me verify this because sometimes when dealing with these sums, it's easy to make a mistake.Alternatively, I can compute the sum manually:21 +19=4040 +17=5757 +15=7272 +13=8585 +11=9696 +9=105105 +7=112112 +5=117117 +3=120120 +1=121Yes, that adds up to 121.So, the total number of ordered triples (x,y,z) is 121.Wait, but let me think again. Is this correct?Because when I fixed x, I found that for each x, the number of (y,z) pairs is 23 -2x, and then summed over x=1 to11.But I want to make sure that for each x, the number of solutions is correctly calculated.Wait, another way to think about it is that the equation is 10x +5y +z=120, with x,y,z ≥1.We can make a substitution to turn this into a problem with non-negative integers.Let me set x'=x-1, y'=y-1, z'=z-1. Then x', y', z' ≥0.Substituting into the equation:10(x' +1) +5(y' +1) + (z' +1)=12010x' +10 +5y' +5 +z' +1=12010x' +5y' +z' +16=12010x' +5y' +z'=104Now, we have 10x' +5y' +z'=104, with x', y', z' ≥0 integers.This might be a different way to approach the problem, but I'm not sure if it's easier.Alternatively, maybe I can consider the equation 10x +5y +z=120, x,y,z ≥1.Let me think of z as the remainder when 120 -10x -5y is divided by 1, but that might not be helpful.Alternatively, since 10x +5y must be less than or equal to 119, as z is at least 1.But I think my initial approach was correct, that for each x, the number of (y,z) pairs is 23 -2x, and summing that from x=1 to11 gives 121.Wait, but let me check for x=1:10(1) +5y +z=120 =>5y +z=110.Since y ≥1, z=110 -5y ≥1 =>5y ≤109 => y ≤21.8, so y=1 to21, which is 21 solutions.Which is 23 -2(1)=21, correct.Similarly, for x=11:10(11)=110, so 5y +z=10.y ≥1, so z=10 -5y ≥1 =>5y ≤9 => y ≤1.8, so y=1, which gives z=5. So, 1 solution, which is 23 -2(11)=1, correct.So, the formula seems to hold.Thus, the total number of ordered triples is 121.Wait, but let me think again. Is there another way to compute this?Alternatively, maybe I can model this as a Diophantine equation.The equation is 10x +5y +z=120, x,y,z ≥1.Let me consider variables x, y, z as positive integers.We can rewrite the equation as:z=120 -10x -5ySince z must be at least 1, 10x +5y ≤119.We can factor out 5:5(2x + y) ≤119So, 2x + y ≤23.8Since 2x + y must be an integer, 2x + y ≤23.So, 2x + y ≤23.Now, x and y are positive integers, so x ≥1, y ≥1.Let me define u=2x, so u is an even integer ≥2.Then, u + y ≤23.So, y ≤23 -u.But y must be at least1, so for each u, y can be from1 to23 -u.But u=2x, so x= u/2, which must be integer, so u must be even.So, u can be 2,4,6,..., up to 22, since 2x ≤23 -1=22, so x ≤11.So, u=2,4,...,22.For each u, the number of y is 23 -u.But since u=2x, x=u/2.So, the number of y is 23 -2x.Wait, that's the same as before.Therefore, for each x from1 to11, the number of y is23 -2x, and z is determined as120 -10x -5y.So, the total number of solutions is sum_{x=1}^{11} (23 -2x)=121.So, this confirms the earlier result.Therefore, the number of ordered triples (x,y,z) is121.Wait, but just to make sure, let me compute the sum again.Sum from x=1 to11 of (23 -2x).Which is:x=1:21x=2:19x=3:17x=4:15x=5:13x=6:11x=7:9x=8:7x=9:5x=10:3x=11:1Adding these up:21+19=40, +17=57, +15=72, +13=85, +11=96, +9=105, +7=112, +5=117, +3=120, +1=121.Yes, that's correct.So, the total number of ordered triples is121.I think that's the answer.**Final Answer**The number of possible ordered triples is boxed{121}.

answer：To find the sum of all positive divisors of 91, I first need to determine its prime factors.I start by checking if 91 is divisible by smaller prime numbers. Dividing 91 by 7 gives 13, which is also a prime number. Therefore, the prime factors of 91 are 7 and 13.Next, I use the formula for the sum of divisors, which is ((p^{a+1} - 1)/(p - 1)) for each prime factor (p) raised to the power (a). Since both 7 and 13 are raised to the first power, the sum of the divisors is ((7^2 - 1)/(7 - 1) times (13^2 - 1)/(13 - 1)).Calculating each part:[frac{7^2 - 1}{7 - 1} = frac{49 - 1}{6} = frac{48}{6} = 8][frac{13^2 - 1}{13 - 1} = frac{169 - 1}{12} = frac{168}{12} = 14]Finally, multiplying these results together gives the total sum of the divisors:[8 times 14 = 112]

answer：To find the greatest common divisor of 266 and 209 using the Euclidean algorithm, I start by dividing the larger number by the smaller one.First, I divide 266 by 209, which gives a quotient of 1 and a remainder of 57. This means that 266 is equal to 209 multiplied by 1 plus 57.Next, I take the previous divisor, 209, and divide it by the remainder, 57. This division results in a quotient of 3 and a remainder of 38. So, 209 is equal to 57 multiplied by 3 plus 38.I then take the last divisor, 57, and divide it by the new remainder, 38. This gives a quotient of 1 and a remainder of 19. Therefore, 57 is equal to 38 multiplied by 1 plus 19.Finally, I divide the last divisor, 38, by the remainder, 19. This division yields a quotient of 2 with no remainder, meaning that 38 is equal to 19 multiplied by 2.Since the remainder is now zero, the last non-zero remainder is 19, which is the greatest common divisor of 266 and 209.

answer：First, I recognize that the sum of the interior angles of a pentagon is 540 degrees.Since the angles are consecutive even numbers, I can represent them as ( x ), ( x + 2 ), ( x + 4 ), ( x + 6 ), and ( x + 8 ).Adding these together gives the equation ( 5x + 20 = 540 ).Solving for ( x ), I subtract 20 from both sides to get ( 5x = 520 ), then divide by 5 to find ( x = 104 ).Finally, the largest angle is ( x + 8 = 104 + 8 = 112 ) degrees.

answer：First, I need to determine the total number of fireworks that Koby and Cherie have together.Koby has 2 boxes of fireworks. Each box contains 3 sparklers and 5 whistlers. To find the total number of fireworks in Koby's boxes, I will calculate the total sparklers and whistlers separately and then add them together.For Koby:- Total sparklers = 2 boxes × 3 sparklers per box = 6 sparklers- Total whistlers = 2 boxes × 5 whistlers per box = 10 whistlers- Total fireworks for Koby = 6 sparklers + 10 whistlers = 16 fireworksCherie has 1 box of fireworks, which contains 8 sparklers and 9 whistlers. So, the total number of fireworks in Cherie's box is:For Cherie:- Total sparklers = 1 box × 8 sparklers per box = 8 sparklers- Total whistlers = 1 box × 9 whistlers per box = 9 whistlers- Total fireworks for Cherie = 8 sparklers + 9 whistlers = 17 fireworksFinally, to find the total number of fireworks that Koby and Cherie have together, I will add the totals from both Koby and Cherie:Total fireworks = Koby's fireworks + Cherie's fireworks = 16 + 17 = 33 fireworks