answer:Let the two numbers be ( x=AB ) and ( y=CD ) (where ( A, B, C, ) and ( D ) are digits). The average of ( x ) and ( y ) is ( frac{x+y}{2} ) and the number formed by writing ( x ) before the decimal point and eight times ( y ) after the decimal point is: [ AB.CD times 8 = AB + 0.CD times 8 = AB + frac{8 cdot CD}{100} = x + frac{8y}{100} = x + frac{2y}{25}. ] Setting these equal gives: [ frac{x+y}{2} = x + frac{2y}{25} ] Multiplying through by 50 to eliminate denominators: [ 25x + 25y = 50x + 4y ] [ 21y = 25x ] From this, it follows that ( y ) must be a multiple of 25. As ( y ) is a 2-digit positive integer, this means that ( y=25 ) or ( y=50 ) or ( y=75 ). Substituting each into the equation ( 21y = 25x ): - ( y=25 ), ( 525 = 25x ), ( x=21 ) - ( y=50 ), ( 1050 = 25x ), ( x=42 ) - ( y=75 ), ( 1575 = 25x ), ( x=63 ) Thus, the valid pairs are ( (21, 25) ), ( (42, 50) ), ( (63, 75) ). Among them, the smallest integer is ( boxed{21} ) when considering the pair ( (21, 25) ).

answer:Georgia originally had a total of 4 yellow + 2 black + 3 green = 9 buttons. After giving some to Mary, Georgia has 5 buttons left. Therefore, Georgia gave Mary 9 - 5 = boxed{4} buttons.

answer:Solution: (1) Let the common difference of the arithmetic sequence {a_n} be d, since the sum of the first n terms S_n satisfies S_{3}=0 and S_{5}=-5, we have begin{cases}3a_{1}+3d=0 5a_{1}+10d=-5end{cases}, solving this, we get a_{1}=1 and d=-1. Therefore, a_{n}=1-(n-1)=2-n. (2) frac {1}{a_{2n-1}a_{2n+1}}= frac {1}{(3-2n)(1-2n)}= frac {1}{2}( frac {1}{2n-3}- frac {1}{2n-1}), thus, the sum of the first n terms of the sequence left{ frac {1}{a_{2n-1}a_{2n+1}}right} is = frac {1}{2}left[(-1-1)+(1- frac {1}{3})+( frac {1}{3}- frac {1}{5})+ldots+( frac {1}{2n-3}- frac {1}{2n-1})right] = frac {1}{2}left(-1- frac {1}{2n-1}right) = frac {n}{1-2n}. Therefore, the answers are: (1) The general formula for {a_n} is boxed{a_{n}=2-n}. (2) The sum of the first n terms of the sequence left{ frac {1}{{a}_{2n-1}{a}_{2n+1}}right} is boxed{frac {n}{1-2n}}.

answer:Step 1: Consider numbers of the form 1ijk where i, j = k or i = j = k. Since each digit ranges from 0 to 9 but all should be unique except for one pair, let's handle it by groups: 1. When two digits are identical and the remaining digits are distinct and greater than the last digit. Step 2: Cases to consider: - 1xxy: Since y < x, and x ≠ 1 and y ≠ 1, x has 8 choices (2-9) and y has x - 1 choices. - 1xyx: y < x as per the new rule, x again has 8 choices, and y follows x - 1 rules. - 1yxx: y must be less than x, x has 8 choices (2-9), and y can be any number from 2 up to x - 1. For each x, y can take on values from 2 to x - 1. Add the numbers for each valid x. Step 3: Calculate for each form separately: - 1xxy: sum_{x=2}^{9}(x-1) = 1 + 2 + ldots + 8 = frac{8times9}{2} = 36 - 1xyx: same as above, sum_{x=2}^{9}(x-1) = 36 - 1yxx: ditto, sum_{x=2}^{9}(x-1) = 36 Hence, the total number is 36 + 36 + 36 = boxed{108}.