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*** Updated ***

2356       4 x 3 x 2  = 24 numbers     (this is 4 P 3 )

  now suppose there are two 5's    there are 3 ways to do this...the remaining spot can be 23 or 6    3 * 3

       suppose there are three 5's      one way

   now suppose there are two 6's    there are three wys to place them...remaining spot can be 2 3 or 5    3 * 3

24 + 3*3  + 1 + 3* 3 =  43 numbers    ???         *** edited ***

2103, 2104, 2105, 2106, 2107, 2108, 2109

Since we can't have any numbers in the 2020-2029, section, 2030 is next. We can't have anything below or equal to 2033 because it would eventually run into 2033. But with 2034, there is just not enough space to finish. Same for the next. The 2090s would be 2093, 2094, 2095, 2096, 2097, 2098, and 2099.

So we go into the next century. We cant have 1 or 2 so 2103 and on will work.

Nope

5 - 5+10 does not equal 15

6 - 6+12 does not equal 16

anything in between - 6+ either 10 or 11 (doesnt matter does not equal 16)

This will Answer your Question 

First, realize that \(x^2-x=x(x+1)\). Also, realize that the prime factorization of 2010 is \(2\cdot3\cdot5\cdot67\). Therefore we can rewrite f(100) and f(101) like this:

\(f(100)=100\cdot99+2\cdot3\cdot5\cdot67 = 10(10\cdot99+3\cdot67) \\f(101)=101\cdot100+2\cdot3\cdot5\cdot67=10(10\cdot101+3\cdot67)\)

Since we cannot factor further, the answer to this question is  \(\boxed{10}\)

WJ/WK

[(2, 3, 5), (2, 3, 6), (2, 5, 5), (2, 5, 6), (2, 6, 6), (3, 5, 5), (3, 5, 6), (3, 6, 6), (5, 5, 5), (5, 5, 6), (5, 6, 6)],  >Total distinct combinations = 11

Convert the above combinations into permutations as follows:

[6 + 6 + 3 + 6 + 3 + 3 + 6 + 3 + 1 + 3 + 3] = 43 permutations. They can be listed as follows:

[(2, 3, 5), (2, 3, 6), (2, 5, 3), (2, 5, 5), (2, 5, 6), (2, 6, 3), (2, 6, 5), (2, 6, 6), (3, 2, 5), (3, 2, 6), (3, 5, 2), (3, 5, 5), (3, 5, 6), (3, 6, 2), (3, 6, 5), (3, 6, 6), (5, 2, 3), (5, 2, 5), (5, 2, 6), (5, 3, 2), (5, 3, 5), (5, 3, 6), (5, 5, 2), (5, 5, 3), (5, 5, 5), (5, 5, 6), (5, 6, 2), (5, 6, 3), (5, 6, 5), (5, 6, 6), (6, 2, 3), (6, 2, 5), (6, 2, 6), (6, 3, 2), (6, 3, 5), (6, 3, 6), (6, 5, 2), (6, 5, 3), (6, 5, 5), (6, 5, 6), (6, 6, 2), (6, 6, 3), (6, 6, 5)]   >Total distinct permutations = 43

235

236

253

263

255

266

265

256

Since there is also only 1 3 there are also 8 numbers that start at 3.

So we have 16 so far

For 6 we have

623

632

625

652

635

653

655

626

662

636

663

665

656

Then for 5 we have

523 

532

525

552

535

553

565

556

566

526

536

563

562

555

So I think that's it, 16+ the 6 column+ the 5 column

2356    choose 3      12 ways

   now suppose there are two 5's    there are 3 ways to do this...the remaining spot can be 23 or 6    3 * 3

       suppose there are three 5's      one way

   now suppose there are two 6's    there are three wys to place them...remaining spot can be 2 3 or 5    3 * 3

12 + 3*3  + 1 + 3* 3 =   31 numbers    ???      Just my guess....  Maybe MMM will do better/correctly !  

One point will be between J and K...the other will be to the right of K   ( or outside of the segment on the K side)

    what ratio are we to calculate?.....it got truncated....

(2^-2) * 2³⁰  = ( 2²)^x

2²⁸ = 2^2x      equate exponents

28 = 2x

x = 14

$C(2,1)= \dfrac{2!}{(2-1)!}= \dfrac{2}{1} = 2

The function  we are working with is

f(t)  =-16t^2 + 48t  + 64  =   answer C

(A )  the  max  height  occurs    at       -48 / (2 -16)  =  -48/ -32  =  1.5  seconds

This height   is     -16(1.5)^2  + 48(1.5) + 64   =   100ft

(B)  To find when it  hits the  ground

-16t^2   + 48t  + 64  =   0        divide through by -16

t^2  - 3t  - 4   =  0    factor

(t - 4) ( t +1)   = 0

The first factor set to  0  gives us a positive time

t - 4    =  0  

t = 4 seconds

(D)       -16t^2   +  48t  +  64  >   96        subtract  96 from both sides

-16t^2  + 48t  - 32  >  0      divide  trough  by  -16, chage the  dirction of the  inequality  sign

t^2  - 3t  + 2  <   0       factor

(t - 2)( t - 1)  <  0

This is  true when   t is between   1  and  2  seconds

x= x₀ + vot + 1/2 at²

ht = 64 + 48t - 16.1t²           Max occurs at   t = - b/2a =  -48 / 2(-16.1)

   use the value of t determined above  in the equation to calculate max height

also use the equation to calculate when above 96ft like this

  -16.1t^2 +48t + 64 > 96

  -16.1t^2 + 48t -32 >0     use quadratic formula to find t values...there will be two...the time between is when above 96ft

I can see that the latex link is from AoPS, so I'm assuming that your question is from AoPS. You should not be asking those questions here. Please try to solve it on your own. 

Via pythag theorem ...and knowing each leg is the same length...

l^2 + l^2 = 196(2)

l^2 = 196

l = 14  units

APPEARS to be 1/2 of the area of the circle with radius 12....      1/2 pi (12²) =..........   units²

What's the question? 

One problem.....three  different solution methods  !!!

The total  of the 10 values   =   10(85)   = 850

So

[ 850 - 76 - 70  ]  / [ 10 - 2 ] = 

704/ 8  =

88      when 70 , 76   are removed

"Of"  means "Multiply"

21%  =  .21

So...just calculate

.21 * 387

See the following image :

Draw perpendicular CE  from  the top right vertex of the trpezoid to the base

We have  right  triangle  CED with  CE = 2r  , ED =  (35-22) = 13     and  the  hypotenuse CD  =  (22 - r) + ( 35 - r)  = (57 - 2r)

So

ED^2 +CE^2  = CD^2

13^2 +  (2r)^2   =  ( 57  -2r)^2

Solving this for positive r  gives us    r  = 770/57

The   area of the  trapezoid = (1/2) (2r) (sum of bases)  =  (1/2) (2 * 770/57) ( 57)  =  770

The perimeter =   2(770/57) + (35 + 22)  + 57 - 2(77/57)   =  114

3 of these scores are for Cyprian.

Therefore; 

His scores must add up to 180 for the mean to be 90.

His scores must be 85, 92 and 94.

The remaining values add to 278/3 = 92.6 recurring, round as needed.