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to get the number of divisors an integer has we multiply the powers of the primes +1. for example 20 is 2^2*5^1 so the total number of divisors is (2+1)(1+1) or 6. so we need to reverse this process for 5. 5 is prime so it has to be 1*5 or p^4. there can only be \(\boxed 1\) prime that divides a terrific positive integer. so much for being terrific ;P

Let the center of the small circle = M

Draw MN  perpendicular to OA

We have right triangle ONM

MN^2 + ON^2 = OM^2

r^2 + 3^2  = (4 - r)^2

r^2 + 9 = r^2 - 8r + 16

8r =  7

r = 7/8

A terrific number must be a perfect square since it has an odd number of divisors. 

It also has 1 set of divisors other than 1 and itself. 

\(1,4,9\) all don't work since they only have 3 divisors. 

However, \(16\) has exactly 5 factors. We have \(1,2,4,8,16\)

Thus, our answer is 16. 

Thanks! :)

If a number is divisble by 6, then it is also divisble by \(1,2,3,6\)

Now, let's take a look at factors of 6. 

We know that \(12\) has 6 factors, as we have \(1,2,3,4,6,12\)

18 also 6 factors, as we have \(1,2,3,6,9,18\)

However, 24 has exactly 8 factors. We have

\(1,2,3,4,6,8,12,24\)

The prime factors are \(2,3\), so our answer is just 2. 

So there are 2 prime factors. 

Thanks! :)

Well, every number technically has a perfect square divisor 1. 

This means that the smallest positive integer that has exactly 2 perfect square divisors is the next smallest pefect square. 

Thus, 4 is the smallest. 

We have 

\(1 = 1^2\\ 4=2^2\)

Thus, our answer is just 4. 

Thanks! :)

24 =  2^3 * 3^1

No. of divisors  = (3 + 1) (1 + 1)  =  4 * 2  =  8

So, in order to calculate the list, we can use a handy trick. We simply take the largest number, multiply it by that number +1, and divide by 2. 

We have

\(\frac{200 \cdot 201}{2}=20100\)

The prime factorization of \(20100\) is 

\(20100 = 2^2 \cdot 3 \cdot 5^2 \cdot 67\)

Thus, the greatest prime divisor is 67. 

Thanks! :)

1+2+3...198+199+200 is 50*201 since you can pair up the first and last, 2nd and 2nd to last, etc. 

201 is 3*67 and 50 is 2*5^2. the greatest prime factor is \(\boxed{ 67}\) :P

See the new image

The line containing BF will   intersect the quarter-circle at H

So  BH =  the radius of the quarter-circle = 4

And FH  is  just the radius of our small  circle = r

So

BH  -  FH  =  4 - r  =  BF

ahhhhh... i was interpreting that mark had to be in the committee. Thanks :)

right but how are you sure that when you extend it it will form a nice straight line all the way to B? I get everything else :)

We have 15 people....choose any  2

So

C(15,2) =  105 ways

Since b is a constant, it must not contain any powers. This means b has no overall effect on the degree. 

Thus, let's take a look at f(x) and g(x). 

The degree of \( f(x) + b \cdot g(x)\) is essentially the larger degree for f(x) and b(x). 

f(x) has a degree of 5, because the leading term is \(2x^5\)

g(x) has a degree of 6, because the leading term is \(x^6\)

Thus, the degree of the polynomial must have a degree of 6. 

So the smallest possible degree is 6. 

Thanks! :)

pleeeez use latttexxxxxx

\(f(x) + b \cdot g(x),\space f(x) = 2x^5 - 6x^4 - 4x^3 + 12x^2 + 7x - 5,\space g(x) = x^6 - 17x^5 + 2x^4 + 6x^3 + 11x^2 - 8x + 1.\)

im pretty sure its just 6 since no matter what you do (multiplication wise) to g(x) it will always contain the \(x^6\) term. im probrably overlooking some major thing.

If we extended BF  through to the edge of the quarter-circle.....it's length would =  4 ....but we are subtracting off the radius of the circle, r.  So BF =  4 - r  and it  is the hypotenuse of right triangle  FGB

Hope that makes sense !!!

i tried doing that but i wasnt sure that the 4-r (like the quarter circle radius would connect to the center of the circle)would be a straight line can you explain to me how to be sure

so we set a vertex of a triangle at the center of the small circle and another vertex at the center of the square circumcribing the quarter circle and another vertex the corner of the quarter circle. then the triangle then becomes a right triangle since 45+45 is 90 (draw a picture it becomes apparent im just kinda bad at explaning things) so the legs of this right triangle are r*sqrt2 and 2*sqrt2 and the hypotenuse is the hypotenuse of another right triangle with legs of r and r+2. so after squareing all the things according to the pythagorean theoream we get 2r^2+8=2r^2+4r+4. so the radius is \(\boxed{1}\). at least according to my calculationnnssss *nerd emoji*. that took me a while. 😮‍💨YIKES AFTER SEEING CPHILLS ANSWER I AM VERY UNCONFIDENT OF MY SOLUTION PLEASE TELL ME WHERE I GOT WRONG.

                         x^2

x^2 -2x - 5   [   x^4 - 2x^3 +  kx^2   ]

                        x^4 - 2x^3 -  5x^2

                      _________________

k + 5     must  =  0

k =  -5

In right triangle FGB

r^2  +(2 + r)^2  =  (4-r)^2 

r^2 + r^2 + 4r + 4 =  r^2 - 8r + 16

r^2 + 12r - 12 =  0              complete the square on r

r^2 + 12r + 36  = 12 + 36

(r + 6)^2 = 48          take the positive root  

r + 6  =  sqrt (48)

r =  sqrt (48) - 6  =    4sqrt (3) - 6 ≈  .928

to get the total number of divisors a number has you take the powers of the prime factors and add one then multiply. to give an example 20 is 2^2*5^1 so the total number of divisors is (2+1)(1+1) or 6. m has 12 divisors so we need to reverse this process. so 12 is 4*3 or 2*2*3. so the final answer is \(\boxed{2\space or \space 3}\)

It means that a question has been answered. However, you can still add your answer, particularly if you believe that an answer (or answers) might be incorrect.   ( I don't always check every answer closely.)

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