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There are 53 different scores that a student can get:  0, 1, 2, 3, ..., 51, 52.

So, it will take 53(1) + 1 students to guarantee that 2 students have the same score.

Likewise, it will take 53(2) + 1 students to guarantee that 3 students have the same score.

And, it will take 53(3) + 1 students to guarantee that 4 students have the same score.

To guarantee that at least 39 students to have the same score:  53(38) + 1  =  2015.

I'll think about complementary counting, haven't tried that method on this problem yet... 

Thanks for the idea

A quick summary of what hugomimihu wrote:

First find all of 7's multiples between 1-100.

Then find all of 11's multiples between 1-100.

Add the sums together.

Minus the numbers that repeat (basically find the least common multiple of 7 and 11 which is 77-- Since there is only one multiple of 77 between 1-100, you only need to minus one.)

If \(\)\(2\times 3 \times12\times14\times15\times20\times21 \) can be divided into 2 equal sets, then that means it is something's square. If you do the square root of \(2\times 3 \times12\times14\times15\times20\times21\), it equals 2520. So the answer is \(\boxed{2520}\)

If anything is unclear or unhelpful, just ask me! 

The largest possible value of M/N, M and N being 3-digit integers, would be: 999 / 111 = 9

The figure above shows three consecutive squares of sides 5, 4 and 3, respectively.

Find the area of the black region.

Hello Guest!

\(A=5^2+4^2+3^2-\frac{(5+4+3)\cdot5}{2}\\ A=25+16+9+30\\\)

\(A=80\)

  !

The length of the diagonal is 5*sqrt(2).

By long division,

1/228 = 0.0076b45076b4507b.... in base 12,

so it is repeating.

I think he/she means that ALL digits be EVEN. If that is my undestanding, then the smallest such multiple would be =288.

N=4; R=0.12;PV=6000; FV=PV*(1 + R)^N; print"FV =$",FV
FV =$ 9441.12 - Annual compounding

N=4*4; R=0.12/4;PV=6000; FV=PV*(1 + R)^N; print"FV =$",FV
FV =$ 9628.24 - Quarterly compounding

$9628.24  -  $9441.12 = $187.12 - Extra interest that she will pay if compounded quarterly

Using this diagram, you should be able to find the area of the equilateral triangle QRP.  

Area of a triangle QRP = 3√3  

a=-200; b=-200;c=-200;d=a*b+a*c+b*c;e=a^2+b^2+c^2; if(d==9 and e==18, goto6, goto7);printd, a, b, c; a++;if(a<100, goto3, 0);a=0;b++;if(b<100, goto3, discard=0; a=0;b=0;c++;if(c<100, goto3, 0)

x     y     z

 3    3    0
 4    1    1
 1    4    1
 3    0   3
 0    3   3
 1    1   4
 

Could you explain to me how you got that answer?

a = - 9  or  a = 14  and:

b = 14  or  b = - 9

The perimeter of a triangle is 8 + 2*sqrt(2).

By the angle bisector theorem, the area of triangle XYZ is 144 + 48*sqrt(3).

I believe that it doesn't make any difference if the pyramid is regular or not.

Total surface area = 1200.

Surface area of the base = 400.

Total surface area of the four sides is 1200 - 400  =  800.

1/2 of the area of each side is below the midline*; therefore 1/2·800  =  400 is below the midline.

Midline value of each side = 10   --->  area of the top base = 100.

Total surface area  =  400 + 400 + 100  =  900.