Links, notes, Hints or/and solutions to 2014 Mathcounts state harder problems.
2014, 2015 Mathcounts state are harder
Sprint round:
#14 :
Solution I :
(7 + 8 + 9) + (x + y + z) is divisible by 9, so the sum of the three variables could be 3, 12, or 21.
789120 (sum of 3 for the last three digits) works for 8 but not for 7.
21 is too big to distribute among x, y and z (all numbers are district),
thus only x + y + z = 12 works and z is an even number
__ __ 0 does't work (can't have 6 6 0 and the other pairs all have 7, 8 or 9)
264 works (789264 is the number)
Solution II :
789000 divided by the LCM of 7, 8 and 9, which is 504 = 1565.47...
Try 504 * 1566 = 789264 (it works)
The answer is
264.
#18:
Watch
this video from Mathcounts mini and use the same method for the first question,
you'll be able to get the answer. It's still tricky, though.
#23 : Drop the heights of the two isosceles triangles and use similar triangles to get the length of FC.
Then solve.
#24:
The key is to see 2
10 is 1024 or about 10
3
2
30 = ( 2
10 )
3 or about (10
3 )
3about 10
9 so the
answer is
10 digit.
#25:
As you can see, there are two Pythagorean Triples : 9-12-15 and 9-40-41.
Base (40-12) = 28 gives you the smallest area.
The answer is 28 * 18 =
504
#26 :
Let there be A, B, C three winners.
There are 4 cases to distribute the prizes.
A B C
1 1 5 There are 7C1 * 6C1 * \( \dfrac {3!} {2!}\) = 126 ways -- [you can skip the last part for C
because it's 5C5 = 1]
1 2 4 There are 7C1* 6C2 * 3! = 630
1 3 3 There are 7C1 * 6C3 * \( \dfrac {3!} {2!}\) = 420
2 2 3 There are 7C2 * 5C2 * 3 (same as above)
Add them up and the answer is
1806.
If you can't see why it's \( \dfrac {3!} {2!}\) when there is one repeat, try using easier case to help you understand.
What about A, B two winners and 4 prizes ?
There are 2 cases, 1 3 or 2 2, and you'll see how it's done.
#27 : Read this and you'll be able to solve this question at ease, just be careful with the sign change.
Vieta's Formula and the Identity Theory
#28: There are various methods to solve this question.
I use binomial expansion :
\(11^{12}=\left( 13-2\right)^{12}=12C0*13^{12}\)+ \(12C1*13^{11}*2^{1}\)+...
\(12C11*13^{1}*2^{11}\)+ \(12C12*2^{12}\)
Most of the terms will be evenly divided by 13 except the last term, which is \(2^{12}\) or
4096, which, when divided by 13, leaves a remainder of 1.
Solution II :
\(11\equiv -2\left ( mod13\right)\) ; \((-2)^{12}\equiv 4096\equiv 1\left ( mod13\right)\)
Solution III :
Or use Fermat's Little Theorem (Thanks, Spencer !!)
\(11^{13-1}\equiv 11^{12}\equiv1 (mod 13)\)
Solution II :
Using dimensional change and ratio, proportion.
Cut the cone and observe the shape.
The circumference of the larger circle is 20pi (10 is the radius) and the base of
the cone circle circumference is 10pi (5 is the radius), which means that the cut-off cone shape is a half circle because it's \(\dfrac {10\pi } {20\pi }\) or \(\dfrac {1 } {2 }\) of the larger circle. (180 degrees)
To find the part that is the area of the frustum not including the top and bottom circles,
you use the area of the half circle minus the area of the smaller half circle.
Since the volume ratio of the smaller cone to larger cone = 2 to 3, the side ratio of the
two radius is \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).
Using this ratio, we can get the radius of the smaller circle as 10 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\) and the radius of the top circle of the frustum as 5 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).
Now we can solve this :
\(\dfrac {1 } {2 }\)\(\left[ 10^{2}\pi -\left( 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\right) ^{2}\pi \right] \) + \(5^{2}\pi +\left( 5\times \dfrac {\sqrt [3] {2}} {\sqrt {3}}\right) ^{2}\pi \) = about 176 (after you round up)ional change and ratio, proportion.
Cut the cone and observe the shape.
The circumference of the larger circle is 20pi (10 is the radius) and the base of
the cone circle circumference is 10pi (5 is the radius), which means that the cut-off cone shape is a half circle because it's \(\dfrac {10\pi } {20\pi }\) or \(\dfrac {1 } {2 }\) of the larger circle. (180 degrees)
To find the part that is the area of the frustum not including the top and bottom circles,
you use the area of the half circle minus the area of the smaller half circle.
Since the volume ratio of the smaller cone to larger cone = 2 to 3, the side ratio of the
two radius is \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).
Using this ratio, we can get the radius of the smaller circle as 10 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\) and the radius of the top circle of the frustum as 5 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).
Now we can solve this :
\(\dfrac {1 } {2 }\)\(\left[ 10^{2}\pi -\left( 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\right) ^{2}\pi \right] \) + \(5^{2}\pi +\left( 5\times \dfrac {\sqrt [3] {2}} {\sqrt {3}}\right) ^{2}\pi \) = about 176 (after you round up)
Solution III : Another way to find the surface area of the Frustum is :
median of the two half circle [same as median of the two bases] * the height [difference of the two radius]
\(\dfrac {1} {2}\left( 2\times 10\pi + 2\times 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\pi \right)\)* \(\left( 10-10\times \dfrac {\sqrt [3] {2}} {\sqrt [3]{3}}\right)\)
.
![\[\dfrac{r}{5}=\dfrac{s}{10}=\dfrac{\sqrt{r^2+h^2}}{10}.\]](https://latex.artofproblemsolving.com/2/0/5/205a8d45594c47d92393e07cffca02d3d3a5b415.png)
![\[10r=5\sqrt{r^2+h^2}\implies 100r^2=25r^2+25h^2\implies 75r^2=25h^2\implies 3r^2=h^2\implies h=r\sqrt{3}.\]](https://latex.artofproblemsolving.com/c/3/4/c34a4033b4515ae88c64fda9f900d84924094aa6.png)
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.![\[\dfrac{125\sqrt{3}}{3}-\dfrac{\pi\times r^2h}{3}=\dfrac{125\sqrt{3}}{9}\implies 125\sqrt{3}-r^2h=\dfrac{125\sqrt{3}}{3}\implies r^2h=\dfrac{250\sqrt{3}}{3}.\]](https://latex.artofproblemsolving.com/d/2/e/d2eb3f35bb46a600fd5db854308e1f7844964947.png)
![\[r^3\sqrt{3}=\dfrac{250\sqrt{3}}{3}\implies r=\sqrt[3]{\frac{250}{3}}.\]](https://latex.artofproblemsolving.com/a/d/9/ad9d39850b5a5aa905a1e879a6b5bfebc1804496.png)
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.![\[37.207+138.477=175.684\approx\boxed{176}.\]](https://latex.artofproblemsolving.com/9/b/9/9b9d7bd69023b24516b90a9c51dbce0f9fffbe08.png)
