The Average of the Rest

Total of $999$ numbers is $999*999$ and total of chosen $729$ numbers is $729*729$ .

So, average of remaining numbers is $\frac{999*999-729*729}{999-729} =\frac{(999+729)(999-729)}{999-729}=999+729=1728$

The average of 999 numbers is 999: $\frac { { a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+.....{ a }_{ 999 } }{ 999 } =999$ Multiply both sides by $999$ gives us the sum of the 999 numbers.

The average of the 729 of these 999 numbers is 729: $\frac { { a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+.....{ a }_{ 729 } }{ 729 } =729$ Multiply both sides by $729$ gives us the sum of these 729 numbers.

To find the sum of the remaining 270 numbers, we subtract the sums of the 999 numbers and the 729 numbers: $({ a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+.....{ a }_{ 999 })-({ a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+.....{ a }_{ 729 })=\quad { 999 }^{ 2 }-{ 729 }^{ 2 }=466560$ .

Now divide $466560$ by $270$ to get the average of those $270$ numbers which is $\boxed{1728}$ .

The sum of the remaining numbers is $999^2-729^2$ and there are $999-729$ of them. Hence the average is: $\frac{999^2-729^2}{999-729} = 999+729 = \boxed{1728}$ by the difference of two squares.

(999 x 999)-(729 x 729)/ (999-729) =1728

We know that The Average Formula is $X = \frac{X_{1} \times f_{1} + X_{2} \times f_{2} + . . + X_{n} \times f_{n}}{f_{1} + f_{2} + . . + f_{n}}$ , So we can input the following information : $999 = \frac{729 \times 729 + (999-729) \times y}{999}$ while y is the answer of the problem. And we got $999^{2} = 729^{2} + 270 \times y$ Hence, y is $y = \frac{999^{2} - 729^{2}}{270}$ $y = \frac{(999 + 729)(999-729)}{270}$ $y = \frac{1728 \times 270}{270}$ $y = 1728 \times \frac{270}{270}$ $y = 1728 \times 1$ $y = 1728$ So, the average of the remaing numbers is $\boxed{1728}$

999 es la media de los 999 números. Si los dividimos en 2 grupos, uno de 729 números (cuya media es 729) y otro de 270 números (cuya media es "M"), la media ponderada de las medias de estos grupos tiene que ser 999.

$\frac{(729*729 + 270*M)}{999} = 999,$ de donde se obtiene $M = 1728.$

999=(a1+a2+a3+...+a999)/999, 729=(a1+a2+a3+...+a729)/729, a730+a731+a732+...+a999=(999^2-729^2)/999-729=999+729=1728

999 999-729 729=270 1728 Average=270 1728/270=1728

The sum total of 999 numbers is 998,001. The sum total of 729 numbers is 513,441. There are 270 numbers remaining and the difference of the first two statements is 466,560. Hence, the average of the remaining numbers is 466,560/270 = 1,728.

999+ 729 = 1728 answer

Sum of all 999 numbers = 999x999 = 998001 Sum of all 729 numbers = 729x729 = 531441 Average of the remaining numbers = ( 998001 - 531441)/( 999 - 729 ) =1728

The average of the 999 numbers = 999 = [(729*729 )+ ((999-729) * X)]/999 So, X= 1728.

a=999 b=729

Average of remaining numbers =

     a.a -b.b / a-b   =  a+b =  1728

calcu ray nag solve..eheheh

A very interesting, yet simple difference of squares problem. If you have the ability to see through the problem it is easy to determine that it is 999 + 729, which is the answer 1728.

Get the total of the 999 numbers by multipying 999 and 999. The total is 998001. Then, get the total of 729 chosen numbers by multiplying 729 and 729. The total is 531441. Subtract 531441 from 998001 to get the total of the remaining 270 numbers. The total of the remaining 270 numbers is 466560. Get the average of the remaining 270 numbers by dividing 466560 by 270. The answer is 1728.

If the average of 999 numbers is 999, then their sum is 999^2. If we remove 729 numbers with an average of 729, then we have a sum of 999^2 - 729^2. To find the average, we divide this number by the total number of remaining items. So, we have: (999^2 - 729^2)/(999 - 729) (999^2 - 729^2)/270 466560/270 1728

Total of 999 no.s = 999*999 = (1000-1) (1000-1) =9,98,001 using identity Similarly, total of 729 no.s is 5,31,441. Sum of remaining 270 nos. =4,66,560 Avge. or remaining no.s =4,66,560/ 270 = 1728

999 + 270 (729/270) = 1728

999*999=998001

729*729=531441

998001-531441=466560

999-729=270

466560/270=1728

For this kind of problems i found one pattern to solve the problem i.e add the given two numbers therefore 999+729 = 1728

(729/999) x 729 + (270/999) x Average=999 --> Average = ( 999^2 - 729^2 )/270

We know that the sum of all the numbers has to be 999 * 999 since 999 is the average of 999 numbers. Similarly we also know that the sum of the 729 numbers with 729 as their average has to be 729 * 729. Finally we know that there are 270 numbers left, so the total sum of those will be the average, or n, * 270. We can then set up the equation as 729^2 + 270n = 999^2. Bring the 729^2 over for a difference of squares: 270n = 999^2 - 729^2--> 270n = (999-729)(999+729)--> 270n = (270)(1728)--> n = 1728

i also used the same technique.